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0:04
>> MA1102 Math for Electronics II: Okay, so, so far, what do you know about egen values and eigenvectors?
0:13
>> 26F1100067 UDAY PRATAP SAROJ: Um, adjust, glanced at the the two formula like that a Lambda equals V. Lambda.
0:20
And and determinant of a minus Lambda. I has to be 0 for the eigen eigenvectors vectors to exist. Us, but, uh. But, but okay, you mean, like, conceptually, right? Um. >> MA1102 Math for Electronics II: Yeah, whatever you know. If you tell we can build up all that. >> 26F1100067 UDAY PRATAP SAROJ: Yes, so very basically, so there are like some vectors.
0:43
Okay, uh, for a given Matrix, there are certain vectors, which which can only only be. Scaled. Uh, right. >> MA1102 Math for Electronics II: Okay. >> 26F1100067 UDAY PRATAP SAROJ: And there will be no rotation. So? So that is how we. But how we Define? Uh, and. >> MA1102 Math for Electronics II: Okay, I guess you have seen the lecture and you are trying to? Kind of explain what you have seen there, but there are few things. >> 26F1100067 UDAY PRATAP SAROJ: Yes, I mean, I, I gave him.
1:18
>> MA1102 Math for Electronics II: the way you have understood. I'm not sure if you understood correctly, so let's revisit whatever you told let's revisit. Okay. >> 26F1100067 UDAY PRATAP SAROJ: Sure, sure. >> MA1102 Math for Electronics II: Okay.
1:30
So, I'm sharing my screen just a minute.
2:10
You were able to see the screen, right? >> 26F1100067 UDAY PRATAP SAROJ: Yes. >> MA1102 Math for Electronics II: Oh.
2:20
Okay, so what you said is that, I guess so, suppose. For example, so if you are having something like this? Okay, so. For a particular Matrix. Suppose if you are having a vector, and if you are, so let's say this is B, and if you are having a matrix n by n.
2:42
Then AV. It is like this, then that particular Vector. You call it as an eigenvector and whatever is the scaling happening here. That is the corresponding eigenvalue. Is this what you have understood? >> 26F1100067 UDAY PRATAP SAROJ: Yes, yes. >> MA1102 Math for Electronics II: Yeah, okay, so this is.
3:01
the geometrical perspective of eigen values and eigen vectors. I guess this is what is given as a motivation to start eigen values and eigen vectors, right? Yeah. >> 26F1100067 UDAY PRATAP SAROJ: Yes. >> MA1102 Math for Electronics II: The water angle values and eigen Vector. So, here is the formal definition. So, formal definition goes in this big a complex number Lambda. Is said to be an eigenvalue of T. If T minus Lambda, I is singular. What is mean by singular here?
3:20
>> 26F1100067 UDAY PRATAP SAROJ: A singular Matrix, right?
3:27
>> MA1102 Math for Electronics II: Yeah, singular in the meaning. What does mainly singular congular Matrix means? what? >> 26F1100067 UDAY PRATAP SAROJ: The.
3:36
>> MA1102 Math for Electronics II: Matrix here. So, suppose if T is the linear transformation?
3:44
So, for example, from R into RN, then the corresponding Matrix a minus Lambda I, the determinant should be zero. That is, what is mean by singular correct, right? And here, one thing to note, it is mentioned that Lambda is a complex number. So, in that case, our real numbers included in complex numbers. >> 26F1100067 UDAY PRATAP SAROJ: They have to. We actually, I. I was going from, like, real to complex. So, so it has to be. >> MA1102 Math for Electronics II: Yeah, so you know that?
4:14
Yeah, so if you are saying complex numbers, all real numbers are also included so it can be. A complex number X Plus, I, y, or it can be just a regular number also, so by a complex number. It does not mean that always you have an imaginary part Sometimes I imaginary part can be 0 also. So here, complex number means this includes real numbers that we have to understand. Okay. So, we also say that Lambda is an eigen value of the Matrix M of T. So, the current respective
4:46
Matrix. So, if you are talking the eigen value of the transformation then correspondingly for the Matrix. Also, is it is the same eigen value right, and however, we starting eigenvalues. You just compute the determinant of a minor Theta I. So, if you compute determinant of a minus Lambda, you will end up in a equation. Rating will end up in a polynomial. All. But if you solve it.
5:22
Okay, so this is what you do.
5:29
>> 26F1100067 UDAY PRATAP SAROJ: Yes, and then the plug plug back that Lambda to find the eigenvector. >> MA1102 Math for Electronics II: Yes, correct. So, we will see some examples, so we. >> 26F1100067 UDAY PRATAP SAROJ: Yes.
5:40
>> MA1102 Math for Electronics II: found it will also see water eigen vector. Now, let us say Lambda is an eigen value of an operator T, then the Subspace null space of T minus Lambda. I, what is mean by null Subspace null phase of T minus Lambda? I do remember that. >> 26F1100067 UDAY PRATAP SAROJ: Here. The set of? Um. That means squish to zero. And the the set of input. The setup inputs that get switched to zero. >> MA1102 Math for Electronics II: Yeah.
6:16
So if T minus Lambda i x is equal to 0, this means that TX minus Lambda X is equal to 0, which again means that TX is equal to Lambda X. >> 26F1100067 UDAY PRATAP SAROJ: Yes, yes. >> MA1102 Math for Electronics II: This null space is of this point t x is equal to Lambda. Okay, so here, suppose Lambda is an eigen value. Then, this Subspace null of T minus Lambda, I is called the eigen space of. Okay, so this space is called the eigen space, but please remember always the
6:45
corresponding eigenvector is always a non-zero Vector, which is very important.
6:52
Okay if this are non-zero power is missing, then it will not be an icon vector So even though t h equal to Lambda X is satisfied for the zero Vector, it is not an eigen Vector. Okay, one one zero. >> 26F1100067 UDAY PRATAP SAROJ: It's so. >> MA1102 Math for Electronics II: Furry word! >> 26F1100067 UDAY PRATAP SAROJ: And the origin. >> MA1102 Math for Electronics II: No. Origin.
7:22
eigenvector. 0 is an eigenvector. Element of the eigen space, but it is not an eigen Vector. You have to have to understand that really.
7:35
>> 26F1100067 UDAY PRATAP SAROJ: Okay, it is an element of the eigen space, but not an eigenvector. >> MA1102 Math for Electronics II: Yeah, because any Subspace you know that the zero Vector is there in any substances? Yeah. >> 26F1100067 UDAY PRATAP SAROJ: Yeah, that will necessarily yes. >> MA1102 Math for Electronics II: Yeah, so you can. >> 26F1100067 UDAY PRATAP SAROJ: This is the other way to look at this right, I mean, see in here here already
7:56
our determinant of a minus Lambda, I has to be zero that. That means that, um. Uh, that means there there will be some. Some redundancy of rows right in? >> MA1102 Math for Electronics II: Ex.
8:14
Oh, you can have a non. This guarantees the determinant of a minus Lambda equal to zero, will guarantee the existing of existence of a non-zero Vector in the null space of T minus. >> 26F1100067 UDAY PRATAP SAROJ: Will be greater than or equal to one. >> MA1102 Math for Electronics II: Exactly, very good observation. Yes. Yeah. Good, so you are connecting with Ragnal etprum here. So with determinant 0, you know that rank is strictly less than n, and hence reality is strictly greater than one.
8:34
>> 26F1100067 UDAY PRATAP SAROJ: Yes, and. >> MA1102 Math for Electronics II: Yeah. >> 26F1100067 UDAY PRATAP SAROJ: because we studied analy in such detail till week, four, right? So, it makes sense to build.
8:47
>> MA1102 Math for Electronics II: So now here. One thing we have to know from whatever you say given here is a
8:54
note that analogy here Dimension of that is the nality here, right? So the anality is strictly greater than one another T of T minus Lambda I or you can can in fact signality of. A minus Lambda, a derby.
9:10
You can say it is an or equal to one. It is at least one at least one non zero in nearly independent Vector will be there. There can be more than one as well. so?
9:20
In other words, you can hear that if there is a non-zero Vector V such the TV
9:28
photo Lambda, then Lambda is okay. So here, as I mentioned earlier, this non Okay, so here, as I mentioned earlier, this non-zero is very important.
9:37
All right, let's see now examples. >> 26F1100067 UDAY PRATAP SAROJ: Yes.
9:46
>> MA1102 Math for Electronics II: So, now you have P from R to R to e of X comma y equal to X plus y comma 2y now shall we first try to find what is the respective Matrix representation this T? >> 26F1100067 UDAY PRATAP SAROJ: Yes, we'll write the X and Y side by side, and then for.
10:01
Uh, yeah, for the for the base. For the canonical basis, we will using the canonical. >> MA1102 Math for Electronics II: Yeah, when you say, what is Q of one card? >> 26F1100067 UDAY PRATAP SAROJ: Uh, okay, in the in the X column. column, uh, comma zero. It will be one and zero in the First Column. Uh, and the and the second column will be wait. This will be 1 and 2, right. >> MA1102 Math for Electronics II: So, here it will be one. >> 26F1100067 UDAY PRATAP SAROJ: Yeah, and then.
10:42
>> MA1102 Math for Electronics II: Yeah, so now if we have to find the icon values we need to consider determinant of a minus Lambda, I equal to 0. Okay, so now this?
10:59
1 minus Lambda 1, 0, 2 minus Lambda. So, what is this Matrix? This is a minus Lambda. >> 26F1100067 UDAY PRATAP SAROJ: Yes. >> MA1102 Math for Electronics II: Yeah, now if we have to equate this to 0? Then what?
11:17
>> 26F1100067 UDAY PRATAP SAROJ: this will be multiplied 1 minus Lambda and 2 minus n and then subtract 0 with justice. Yeah, equals zero. >> MA1102 Math for Electronics II: So, now you simplify this. What will be Lambda Square? Minus. V. Lambda. Plus 2 is. I just finally. >> 26F1100067 UDAY PRATAP SAROJ: We can also do directly right by setting 1 minus Lambda equals 0 and. >> MA1102 Math for Electronics II: That that will?
11:49
So here, you can do Lambda minus 1 into Lambda minus 2 equal to 0, which implies Lambda equal to 1 or Lambda equal to 2. >> 26F1100067 UDAY PRATAP SAROJ: I think we reverse the science. Okay, that should also be. Okay, yeah, because? >> MA1102 Math for Electronics II: Yep. So here, what are the values? Eigen values are one and two. Okay, so now with these are the eagle. Now, how do we find the corresponding corresponding eigen vectors?
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We call this Lambda 1 and this is Lambda 2, okay Lambda 1 V, so we one is some is Lambda 2, okay Lambda 1, V 1. So, V1 is some Vector X1 y1. In.
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so we want to find what is this Vector? So, what are we going to do? We are going to write it as 1 by 0, 2.
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X 1 y1 is equal to where Lambda 1 is 1. It is again exponent variable. So, if we are solving this, what can we get? X1. Equal to X1. And 2 y 1 equal to y1. Yeah. >> 26F1100067 UDAY PRATAP SAROJ: On the right side.
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>> MA1102 Math for Electronics II: Well. >> 26F1100067 UDAY PRATAP SAROJ: Oh yeah, okay. >> MA1102 Math for Electronics II: Okay, so now.
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>> 26F1100067 UDAY PRATAP SAROJ: Um, it's. It's one, uh, sorry, 0.5. No, no, no? 1 has to be zero, right? I mean, because? >> MA1102 Math for Electronics II: Y, one has to be 0 okay. If y1 is 0, what can you say about X1? >> 26F1100067 UDAY PRATAP SAROJ: On X. Okay, it's not giving any information actually. X1 is equal to X1, so. Um, it means we have. We have a dependent.
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So any X1 will do, right? Which means this, we will be getting a line.
14:10
>> MA1102 Math for Electronics II: This is not very straightforward to you. I will give you another method. Okay, that might be the more common method. Maybe we'll do that. >> 26F1100067 UDAY PRATAP SAROJ: Okay, but but we'll also revisit the meaning here. >> MA1102 Math for Electronics II: Well, sorry. What? >> 26F1100067 UDAY PRATAP SAROJ: Will also revisit the meaning of this situation. >> MA1102 Math for Electronics II: Yeah, yes, here. I will write the other technique, so
14:28
usually the, what is the other technique we are using is a minus Lambda 1, I, V. 1, equal to 0, and then will solve this homogeneous equation, so I'll be homogeneous. Equation is always simpler, right? So, first of all, as you mentioned here, y 1 is equal to 0 is exactly fine, and what is x 1 x 1 is R betrayal? So, X1 can be any real number that is the implication you get from here, but if it is not straight forward From here, you can also see the exact same thing from ship. So from here, if
15:04
that this will imply that. If you are doing. Yeah, Lambda 1 is equal to one, so a minus I water the the resulting Matrix you will get? >> 26F1100067 UDAY PRATAP SAROJ: Okay.
15:22
Where is a by the? I mean, I can't see the okay. Okay, when one zero one two, do right? So we. We subtract one from the main diagonal elements. >> MA1102 Math for Electronics II: So, if this place you see this place Lambda, you put 1 until the values.
15:47
>> 26F1100067 UDAY PRATAP SAROJ: For Lambda will be 1. So, so one, uh, so zero and? >> MA1102 Math for Electronics II: I'm here. >> 26F1100067 UDAY PRATAP SAROJ: The diagon.
15:59
>> MA1102 Math for Electronics II: Yeah. Right, and then you have x 1 y1. So, this is equal to. 0, 0. Right. So now you see from here, what is that you observe? You again see that y one equal to 0? Right?
16:29
I'll just make a method don't think we have to do. Okay, usually this more common method, so better use this. >> 26F1100067 UDAY PRATAP SAROJ: It's just a modified version of the first one, right? >> MA1102 Math for Electronics II: Yeah, yeah, but the second version is more simple, and that is what is mostly used. >> 26F1100067 UDAY PRATAP SAROJ: Oh, okay. >> MA1102 Math for Electronics II: And you see y1 is equal to zero. So, what is that you are observing? So, you you see, it
16:55
can be x 1, comma 0, where X1 is any real number. >> 26F1100067 UDAY PRATAP SAROJ: Yes.
17:04
>> MA1102 Math for Electronics II: Okay, so if x 1 is any real number, then this is nothing but span of. One comma zero is that? >> 26F1100067 UDAY PRATAP SAROJ: Yes. >> MA1102 Math for Electronics II: Yeah, and this is your elegant space.
17:18
Right, and if this is your eigen space, any non-zero Vector from here is the eigen Vector corresponding to Lambda 1? For example, one zero is an eigenvector, so you can easily verify what is the given Matrix. The given Matrix is one one zero, two, and you take one zero. What will you get? You will get one zero and then again, zero. And zero. 0. They're making a mistake here. This was. >> 26F1100067 UDAY PRATAP SAROJ: Oh, okay, I was busy and loading down a few things. But yeah.
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>> MA1102 Math for Electronics II: Okay.
17:58
One, one one.
18:10
>> 26F1100067 UDAY PRATAP SAROJ: So, do we have different diagon vectors for the different diagon values? >> MA1102 Math for Electronics II: Sorry, what? >> 26F1100067 UDAY PRATAP SAROJ: Um, so, so here, in this example, we we found out two eigen values, right Lambda 1, 2, and Lambda?
18:26
So also corresponding to each one. Do we have a different set of eigen vectors I mean. >> MA1102 Math for Electronics II: You will have another. >> 26F1100067 UDAY PRATAP SAROJ: Yes, so, so there you go hand in hand. >> MA1102 Math for Electronics II: find next. So, the next step is finding that only. >> 26F1100067 UDAY PRATAP SAROJ: Okay. >> MA1102 Math for Electronics II: Okay, so now, so far, we have computed the eigen Vector corresponding to only, >> 26F1100067 UDAY PRATAP SAROJ:
18:47
Lambda 1. >> MA1102 Math for Electronics II: Lambda 1, which is one. So, the next thing is, you have to compute the eigenvector eigen Vector corresponding to Lambda 2 okay, but first of. >> 26F1100067 UDAY PRATAP SAROJ: Oh, this. >> MA1102 Math for Electronics II: all, I am trying to figure out is there any mistake? So? One, one zero two, multiplied with one zero. Yeah, one zero is the eigenvector.
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Yeah. >> 26F1100067 UDAY PRATAP SAROJ: 25.
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>> MA1102 Math for Electronics II: Just one column only. We are having right. So, what is the matrix multiplication? What was the Mystic you understood? So this is a 2 cross 2 Matrix, so this is a two cross. One Matrix. So, what should be the resulting Matrix is? >> 26F1100067 UDAY PRATAP SAROJ: Um, yeah. >> MA1102 Math for Electronics II: This is.
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>> 26F1100067 UDAY PRATAP SAROJ: Well, when zero I get, I will have to check one. >> MA1102 Math for Electronics II: Yes, it's simply one zero. Yes. >> 26F1100067 UDAY PRATAP SAROJ: I think you are writing an extra column of zeros. >> MA1102 Math for Electronics II: Yeah, yeah, yeah, this is the correct Matrix. Okay, I just. >> 26F1100067 UDAY PRATAP SAROJ: I don't know what?
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>> MA1102 Math for Electronics II: yeah, yeah. So, this is the correct answer. So, what is it you observe? You observe that a suppose we are calling this as V1, a V1 equal to V1. V1. Right? So this is what we observed here. Okay, so this B1 is not a unique Vector. That is something again. You need to understand V1 is not a unique Vector. You can choose any non-zero vector from the space. So, for example, you can choose minus 2, comma 0, 100 comma zero. Any Vector you can choose
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it will it will still satisfy the same context.
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>> 26F1100067 UDAY PRATAP SAROJ: But how is this evident from the second method, the homogeneous method that we use? But I mean, in the first method, it was clear, right, because? >> MA1102 Math for Electronics II: Method. We don't have any information on X, Y, right, we? So that y1 is equal to zero and what is X1. We don't have any information, so as >> 26F1100067 UDAY PRATAP SAROJ: Okay, okay. >> MA1102 Math for Electronics II: 1 is R. We eliminate any element from.
21:02
Yeah, okay. So, similarly, now let's try to find the eigenvalue corresponding to Lambda 2. So, it means now, this time, we are going to consider what is a minus 2i? So, what is a minus 2, I? So this is your a? >> 26F1100067 UDAY PRATAP SAROJ: Uh, first of all, where is AI once again? Okay, okay. And yeah, we subtract. Two from the main diagonal elements.
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So, this is the a minus 2i. -1 0. One zero.
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>> MA1102 Math for Electronics II: Yes, okay, now we have to see that. So, now, just like we did here. So, here, a minus 2 I V2 equal to 0 where we do is?
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X2, comma, y, 2. So, we need to solve. So, now, again, you have minus 1, comma One.
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Which implies minus X Plus Y2 equal to 0. So, this means your X2 is equal to Y2.
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So this means what? What is the corresponding space so X2 come?
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Yeah, so again, here you see Thanos. One {comma}, one. >> 26F1100067 UDAY PRATAP SAROJ: Yes.
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>> MA1102 Math for Electronics II: Yeah, so you see that now, any zero Vector again from here is an eigen Vector So, for example, if you can choose V2 as 1 comma 1, then what will happen 1, 0, 2? And then, here, one comma one. What is it you will get? Two, two.
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Which is 2 times of 1 comma 1, so we are getting. Got it.
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Okay, so now, what is that observing? So there is one important observation
23:17
here. The observation is that V1 and V2. So, what is V1? For example, so any
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Vector actually on the uh x axis, so one zero. So this is the vector. the vector actually. Is this one? Okay.
23:36
And we do is 1 1, right? So any Vector on this line you can actually take? So this is one one, right? So what is that You observe about V 1 and V2? They are linearly independent. V1 and 2 are linearly independent. This is an important application.
24:02
Okay, this note I am going to share it with you, so don't, uh? try to take notes. Whatever I'm writing, you don't have to write. I'm going to share this this video with you. >> 26F1100067 UDAY PRATAP SAROJ: Okay, okay.
24:19
>> MA1102 Math for Electronics II: All right. So, now, let's look at another example. Okay, so now. Uh, for this example first, the step is to find write the Matrix. What is the Matrix representation here?
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Yeah, so if you have to find P of one your? >> 26F1100067 UDAY PRATAP SAROJ: Yeah, one engine.
24:51
>> MA1102 Math for Electronics II: One, two.
25:05
Right. Okay, so this is yours. Now, what you have to do? You have to do determine end of a minus Lambda. I equal to zero, so you are doing 1 minus to 6 minus Lambda equal to zero.
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Right, what is the end of this Matrix now?
25:25
1 minus Lambda into 6 minus Lambda minus 6 is equal to 0 correct? Yeah.
25:34
Now, what? Now you have to solve this system now if you have to solve this solve this weakness system, what will you get? >> 26F1100067 UDAY PRATAP SAROJ: Uh. 6 minus. >> MA1102 Math for Electronics II: So, let's do step.
25:51
Okay, for this part, is this? After this, is it clear for you? >> 26F1100067 UDAY PRATAP SAROJ: Yes. >> MA1102 Math for Electronics II: Okay now! From here. And we can cancel out this this.
26:15
so you have Lambda Square minus 7 Lambda equal to 0, which means Lambda into Lambda minus 7 equal to zero. Lambda is equal to to 0 and equal to seven. So let's call this Lambda 1 Lambda 2. >> 26F1100067 UDAY PRATAP SAROJ: Yes. >> MA1102 Math for Electronics II: Okay, go again, an important observation here. What is determinant of a? >> 26F1100067 UDAY PRATAP SAROJ: E equals 0. >> MA1102 Math for Electronics II: All right.
26:51
is always true whenever a matrix has zero determinant. It will always have 0 as an eigen value. This is an important fact, so these facts are going from later so, but while looking at this examples, if you see this, it will be very easy to understand understand it later when we are going to see this as property separately. Okay. >> 26F1100067 UDAY PRATAP SAROJ: Yeah. >> MA1102 Math for Electronics II: Oh, fine. So now. How do you find I can write this again?
27:19
>> 26F1100067 UDAY PRATAP SAROJ: Um, we'll set. Okay, I wanted to ask will be. Consider this Lambda 1 because it is 0? >> MA1102 Math for Electronics II: Yes, he is. Even if it is equal, distill an eigenvalue, we still considered it.
27:32
For now. What is a minus Lambda 1 I? It is simply a because Lambda 1 is 0.
27:39
Right. So now a B1 is equal to 0 means. It is simply the null space of. Right, so we need to find what is there in the null spaces?
27:55
So, this is X1 plus b y1 equal to 0. And 2 x 1 plus 6y 1, equal to 0. Both are actually the same, just the multiple multiple of each other, right? >> 26F1100067 UDAY PRATAP SAROJ: Yes, yes. >> MA1102 Math for Electronics II: Really feel all this one and two, then two is simply 2 times of one. They are are just the same equations.
28:19
Okay, so your null space is simply set of all X1 comma y 1, which satisfies the X1 plus e y 1 equal to 0?
28:33
Right. This is your space. Now you have to choose some Vector from here. How will you choose a vector from here on 0 Vector? >> 26F1100067 UDAY PRATAP SAROJ: You will find the ratio of x1 by 1, so we'll set. >> MA1102 Math for Electronics II: Can you find a vector with? How do you find a vector which spans the set? >> 26F1100067 UDAY PRATAP SAROJ: Um.
28:56
>> MA1102 Math for Electronics II: Yeah. So, for example, you can put if you are putting y1 is equal to one, then what will you get for X1? >> 26F1100067 UDAY PRATAP SAROJ: Excel will be minus 3. >> MA1102 Math for Electronics II: Yeah. So, for example, you can choose.
29:12
We, even as minus 3 comma 1. So here, in fact, you can write. This is fan of comma 1. Yeah, and you can also very clear later on. Maybe after the glass, you can that a V1 equal to 0. >> 26F1100067 UDAY PRATAP SAROJ: And yeah, I mean, that is the condition we used to find it, so it has stopped. >> MA1102 Math for Electronics II: Yeah, yeah, you can verify that later. Okay, similarly, we can find what is a >> 26F1100067 UDAY PRATAP SAROJ: Okay. >> MA1102 Math for Electronics
29:38
II: minus Lambda 2i. What is a minus Lambda 2?
29:47
>> 26F1100067 UDAY PRATAP SAROJ: time, we'll subtract 7 from the diagonal diagonal from here, so this will be 1 a minus 6, 3, 2, and minus 1 here, correct. >> MA1102 Math for Electronics II: Yeah, so now here, if you have to find the eigenvectors, what you have to do? >> 26F1100067 UDAY PRATAP SAROJ: Again, we'll. We will take this a minus Lambda 2 I and then. >> MA1102 Math for Electronics II: I can hear you see that first infection and? >> 26F1100067 UDAY PRATAP SAROJ: Okay.
30:08
>> MA1102 Math for Electronics II: equation for multiples of each other. So only one equation will. So, minus six, three, two minus 1. Is zero z? So you can transit. one. Great!
30:31
You understand so far is anything not clear. Please let me know.
30:44
Yeah, so if okay, maybe I'll write it. >> 26F1100067 UDAY PRATAP SAROJ: Okay, talking about vector wise. I don't think it will make any difference so.
30:54
>> MA1102 Math for Electronics II: Plus 3y 2 is 0, this is equation 1 and 2 x 2, minus y, two equal to 0. This is equation 2. What you can observe is that equation 1, 2. -3. >> 26F1100067 UDAY PRATAP SAROJ: Yeah, he just multiply by minus 2 to get one by. >> MA1102 Math for Electronics II: You get it? Right, so I just enough if you consider one of that. So, for example, in this case.
31:20
Your eigen spaces in the fall X2, comma, Y2, such that 2 x 2 is equal to Y2. Okay, so here you have. X2 comma. 2 x 2 x 2 in. So? >> 26F1100067 UDAY PRATAP SAROJ: Yeah. >> MA1102 Math for Electronics II: One command.
31:41
Okay, so here you can write V2 as 1 comma 2. Okay, so shall you have one comma two? This is the vector you are having and the other Vector is -3 {comma}, 1.
31:58
Okay again, you can these two vectors V1 and V 2 are nearly independent. Yeah, so this is. >> 26F1100067 UDAY PRATAP SAROJ: Oh. >> MA1102 Math for Electronics II: something very important, okay. And also, another thing you can note.
32:14
So, in this case, particularly for this case in this example alone. For this Matrix, for example. You can see that V1 the span. This is. Null of? And we preserve. This is for range of it. Correct?
32:48
>> 26F1100067 UDAY PRATAP SAROJ: Well, actually. The screen resolution has lowered a lot, so I, I don't know if it's on your side or my side, like I can't. >> MA1102 Math for Electronics II: Don't see any? >> 26F1100067 UDAY PRATAP SAROJ: Yeah, if if you can maybe zoom a little? Yeah, this. This is. >> MA1102 Math for Electronics II: Okay. >> 26F1100067 UDAY PRATAP SAROJ: Oh, okay, okay, that. Okay, it was written racist. I was reading East, okay?
33:17
>> MA1102 Math for Electronics II: So, uh, we have seen some examples. So, how to identify the eigenvalues and how to find the eigenvalues and we have seen few properties also? So, till this is is it clear for you?
33:31
>> 26F1100067 UDAY PRATAP SAROJ: Yes, by the way. Um, what is the significance of this V1 V2 being linearly independent versus? >> MA1102 Math for Electronics II: And this significance is very important when we move to the next level, which is called the diagonaliz. >> 26F1100067 UDAY PRATAP SAROJ: Okay. >> MA1102 Math for Electronics II: When removed diag.
33:49
For Cornell, this is a very important property that egg and Vector independence property we call this that is egg and vectors of distinct eigen values or linearly independent. What does it mean? So if you are having two eigen values Lambda 1 and Lambda 2, which are distinct, it means Lambda 1 is not equal to so this is what we call as distinct.
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Right. So, in that the corresponding eigenvectors are linearly independent. So here, V. 1 and B to are linearly independent. So linearly independent means, what a V1 plus b is equal to zero should imply a equal to zero and b equal to to zero. This is the definition of linearly independent vectors.
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>> 26F1100067 UDAY PRATAP SAROJ: If. And there was one more thing I think in week five, uh, this algebraic and geometric multiplicity something. >> MA1102 Math for Electronics II: That comes after this. Okay. >> 26F1100067 UDAY PRATAP SAROJ: Okay, okay.
35:02
>> MA1102 Math for Electronics II: This is the definition. Actually, you can recall the definition of linear Independence. >> 26F1100067 UDAY PRATAP SAROJ: And, and it relates to. >> MA1102 Math for Electronics II: Not immediately. >> 26F1100067 UDAY PRATAP SAROJ: Okay, so it will be coming for that. Okay. >> MA1102 Math for Electronics II: Okay for now. This is just the definition of linear independence which you are applying it here. Okay, so here this is.
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>> 26F1100067 UDAY PRATAP SAROJ: Yeah.
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>> MA1102 Math for Electronics II: Okay, so here how this is happening. So now, okay, you have a V1 plus b 2 equal equal to 0, so this is what is given? Okay, and this is what you need to prove.
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Okay, so now if T of a V1 plus b v2 your considering, then you know that since T is a linear transformation, you can bring out the scalar C and B like this and like this and split the addition also in this way.
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Okay, and then the next step. What you have done pv1 is Lambda 1 V1. And TV 2 V2 correct? >> 26F1100067 UDAY PRATAP SAROJ: Yeah, yeah. >> MA1102 Math for Electronics II: Yeah. >> 26F1100067 UDAY PRATAP SAROJ: Yes. >> MA1102 Math for Electronics II: Yes, now that what you have applied. So here is a Lambda 1, V. 1 plus b, Lambda V 2. Right. Okay, now what you can do next? >> 26F1100067 UDAY PRATAP SAROJ: And then we. Okay, okay, we've written this command under.
36:36
>> MA1102 Math for Electronics II: So, this is your let's call this equation one. And this is equation 2, okay. And then, here we have done so. I have simply
36:49
multiplied Lambda 2 with equation 1 and got this, so I am calling equation. p.
36:57
Oh. Okay, so now I am going to go back minus 3 if I do 2 minus 3, what will I
37:03
times of Lambda 1 minus Lambda 2 into V1?
37:11
Now, what? I know. I know that Lambda 1 is not equal to Lambda 2 that is also given to me. Because of this Lambda 1 minus Lambda 2 is not 0. Yeah, and also it is given that V1 is non-zero. So, because of this, this part is non-zero. >> 26F1100067 UDAY PRATAP SAROJ: And. So that leaves? >> MA1102 Math for Electronics II: So the space a must be 0?
37:41
>> 26F1100067 UDAY PRATAP SAROJ: And yes, and then another. In another set of equations, we can multiply this step 3 using Lambda 1, right, and then do the other subtraction. >> MA1102 Math for Electronics II: Yeah, yeah, so you can exactly that is the trick, right? >> 26F1100067 UDAY PRATAP SAROJ: And that way we will come to B equals 0. >> MA1102 Math for Electronics II: Yeah, you will. You can control this. >> 26F1100067 UDAY PRATAP SAROJ: Okay, yeah, I think I've seen
38:02
some version of this over and over, but yeah. linearity will study in a lot of other contexts as well.
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>> MA1102 Math for Electronics II: Yeah. So in this way, you can see that V1 and V2 are linearly independent.
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Okay, fine. So the next part is that? These are some properties. If the Matrix is singular, one of the eigen values is zero. So, this is what I mentioned earlier. So, if determinant of a equal to 0, then 0 is an eigen value. I mentioned it in the problem. Previous problem, if remember, right? >> 26F1100067 UDAY PRATAP SAROJ: Yes. >> MA1102 Math for Electronics II: Yes, and if the Matrix is invertible, that is, if the Matrix determinant is non Then all I got non-zero.
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>> 26F1100067 UDAY PRATAP SAROJ: okay, so the first and second are like corollaries to each other, or what I mean, okay? >> MA1102 Math for Electronics II: You don't call it corollar. >> 26F1100067 UDAY PRATAP SAROJ: We have to separately, maybe arrive at. >> MA1102 Math for Electronics II: Yeah, so see both of this comes from. This statement.
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Product is the determinant of M of T, so it means what suppose you are having a n by n Matrix? Okay, and let's say, Lambda, 1, Lambda 2, Lambda, and are the eigen values. Of.
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Determinant a is equal to Lambda 1 into Lambda 2. So, it is the product of. >> 26F1100067 UDAY PRATAP SAROJ: The different. >> MA1102 Math for Electronics II: Of. Okay. Yes, this is what is true. Okay, this is always. Okay.
40:10
So, because of this, these two properties first two properties are coming, so you see that if determinant is 0, it means product of the eigen values is zero, and product of some numbers are zero means one of them should be zero. Right. I similarly, if the determinant is non-zero, so all the eigen values should be non-zero. So, is it clear? >> 26F1100067 UDAY PRATAP SAROJ: Yes, yes, very straightforward, actually. >> MA1102 Math for Electronics II: Yeah, yeah.
40:33
Yeah. I'm also you can see that trace of a matrix is the sum of its diagonal Okay, what is mean by trace? >> 26F1100067 UDAY PRATAP SAROJ: What does? >> MA1102 Math for Electronics II: Defined us. So for any Matrix. Sum of diagonal elements.
40:57
So, summation a i, i equal to 1 to n. Right? >> 26F1100067 UDAY PRATAP SAROJ: Is this gold? >> MA1102 Math for Electronics II: No. No, this is usually the trace. At least. >> 26F1100067 UDAY PRATAP SAROJ: Is there any name for their product of the diagonal elements? >> MA1102 Math for Electronics II: Ah, I don't think so. I'm not aware of any such terminology. Okay. >> 26F1100067 UDAY PRATAP SAROJ: And probably, does it exist? Okay?
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>> MA1102 Math for Electronics II: Yeah, so now some of eigen values of T is the trace of T. So again, this is nothing. Also, the sum of eigen values of a?
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>> 26F1100067 UDAY PRATAP SAROJ: Okay, so the either value is added to the same value as to the diagonal elements in the first. >> MA1102 Math for Electronics II: Yes, yes, okay, so you can observe these facts in their two examples we have
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seen earlier so you can see in this case. What is the determinant in this or determinant of this Matrix? >> 26F1100067 UDAY PRATAP SAROJ: Uh.
42:08
>> MA1102 Math for Electronics II: Yeah.
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And what is Lambda 1 into Lambda 2? So we have seen that this is Lambda 1 and Lambda 1 and Lambda 2. So, what is Lambda 1 into Lambda 2?
42:35
Okay, so you see determinant of a? Is Lambda 1 into Lambda 2? Okay, and what is rays of? >> 26F1100067 UDAY PRATAP SAROJ: 2 plus 1 3. >> MA1102 Math for Electronics II: 1 + 2 and again. Slam. >> 26F1100067 UDAY PRATAP SAROJ: Okay.
42:56
>> MA1102 Math for Electronics II: let's see in the next example of the things already the dominant of a is 0, and that is also very clear that it is 0 into. What is 7?
43:08
One.
43:15
Right, this is a11 a22. Some of the diagonal elements. Of.
43:29
Yeah.
43:45
And are you here? If you are familiar with this, maybe some. There is something more you can relate to. So, are you familiar with acquired equation?
43:51
Maybe I will write it in terms of, like, so X square plus a X Plus B equal to 0 If you are finding the roots of this equation. Are you familiar with water? How are the rules related to A and B? Are you familiar with this? >> 26F1100067 UDAY PRATAP SAROJ: Uh, okay, sum of rules and product of foods. >> MA1102 Math for Electronics II: Yeah. >> 26F1100067 UDAY PRATAP SAROJ: Uh, net medical.
44:14
equals zero. Uh, so in that form, some of the roots is. >> CH DYOMANI SRI JAGANNADHAN: World. >> MA1102 Math for Electronics II: Yeah, some of roots.
44:32
>> CH DYOMANI SRI JAGANNADHAN: Some of them. >> 26F1100067 UDAY PRATAP SAROJ: No, it should be minus. >> MA1102 Math for Electronics II: Yeah, yeah, this minus is missing. Yeah, so what you said is fine, but a minus is missing, right? So?
44:49
some and product of the roots. If you are matching, so here, if you see, so you have so, maybe instead if I have minus, then what you said is fine. Right? So a is the? And B is the product of the roots.
45:07
So in that way. Also, if you are having this characteristic polynomial. Here, you can verify that. So, this is the trace of. And this is the.
45:21
So, here it is 0, which is that vitaminant of a. So, this is 0, which is the determinant of. Right? But this, you can use. >> 26F1100067 UDAY PRATAP SAROJ: Like this?
45:38
>> MA1102 Math for Electronics II: Right? So this is just another observation, which you can compare with roots of quadratic equations. Okay.
45:47
Yeah.
45:59
What happened? Student left. Okay, fine. I guess he just joined in between and between and cannot catch what's going on.
46:08
Okay, so with whatever we have seen so far, can you try to answer this exercise If not, I'll help you. Yeah. >> 26F1100067 UDAY PRATAP SAROJ: 11 or okay level.
46:31
Just a second. >> MA1102 Math for Electronics II: So the meaning for this means?
46:43
Uh, Z. This is called set, minus. We call this a set minus, so this means 0 is is not. This is the meaning of this part. >> 26F1100067 UDAY PRATAP SAROJ: Okay, add except zero. >> MA1102 Math for Electronics II: Included excluded are is excluded from. >> 26F1100067 UDAY PRATAP SAROJ: 0 is excluded from our right. >> MA1102 Math for Electronics II: So, you always excluded them. >> 26F1100067 UDAY PRATAP SAROJ: Okay.
47:17
So, V. 1 and V2 are linearly independent. So, option A is correct. Which means option b is incorrect by default. Uh, and then. >> MA1102 Math for Electronics II: If they are linearly dependent, can make multiple of each other.
47:40
Yes, so in that, what is given in the third case? C option C, V. 1, equal to Alpha, we do. It means they are lying on the same line. Is it possible when they are linearly independent? >> 26F1100067 UDAY PRATAP SAROJ: No, because this is the same as saying option B, so also around. >> MA1102 Math for Electronics II: Exactly, exactly! So, can we always say distinct Tigerian values means? 0 is an eigen value. >> 26F1100067 UDAY PRATAP SAROJ: Wait.
48:09
>> MA1102 Math for Electronics II: Look at the example we have seen so far and. So, did we have distinct eigen values earlier also? >> 26F1100067 UDAY PRATAP SAROJ: Uh, yes.
48:21
Um, but this technique is related to. Uh, determinant being zero or not, right? Uh, so by that. >> MA1102 Math for Electronics II: Not conclude this from the given information, right? It is just that is. >> 26F1100067 UDAY PRATAP SAROJ: Just just just give me a few seconds. I will answer this. Also, uh, okay, we have said V1 and V2 are linearly independent. Let's take an
48:47
Okay, you dominant.
48:54
Okay, and I think.
49:05
>> MA1102 Math for Electronics II: Okay. So, here, the thing is that the values are designed. That is, all is given. So, Lambda 1 is not equal to Lambda 2. That is always given. So, for example, if you consider the first example, we see that Lambda 1 is not equal to Lambda 2 In this case, I can values are distinct, but that is not mean that they one of the eigenvalu is zero, right? >> 26F1100067 UDAY PRATAP SAROJ: Yes, exactly, exactly. So, okay, so this was a trick question.
49:27
Okay, I mean, I will be thought too wild glueless. >> MA1102 Math for Electronics II: So why it is incorrect? Is that clear? >> 26F1100067 UDAY PRATAP SAROJ: Yeah, because it has nothing to do with yeah. >> MA1102 Math for Electronics II: Yeah. >> 26F1100067 UDAY PRATAP SAROJ: With.
50:00
four is -1 and 5 are the eigen values of a. Find the trace of it. Okay, so the trace was just the sum of the that will be 4, right. >> MA1102 Math for Electronics II: Yeah, so here is correct. >> 26F1100067 UDAY PRATAP SAROJ: I mean, because we can, we can find directly using that. >> MA1102 Math for Electronics II: Yes, just some of the eigen values is the trace. So, just four correct? Okay, what about exercise 13?
50:27
>> 26F1100067 UDAY PRATAP SAROJ: Then find the determinant. Okay, so distinct eigen values right or the screens the determinant a. Okay, so a distinct eigen values right or the screens?
50:36
>> MA1102 Math for Electronics II: Maybe here. I guess the information is. >> 26F1100067 UDAY PRATAP SAROJ: Yeah, yeah, that is what I was about to say. >> MA1102 Math for Electronics II: The idea, maybe? >> 26F1100067 UDAY PRATAP SAROJ: Okay, okay.
50:54
So, the determinant is the product of. >> MA1102 Math for Electronics II: Yeah. >> 26F1100067 UDAY PRATAP SAROJ: Wait so, 4, 2016. Sorry. >> MA1102 Math for Electronics II: Okay. Okay, friend. >> 26F1100067 UDAY PRATAP SAROJ: be that will be quite simple. >> MA1102 Math for Electronics II: All right.
51:22
>> 26F1100067 UDAY PRATAP SAROJ: And I mean, I I, I'm one of the like, a career break or so last few years. No study, so maybe that is why? Okay.
51:39
>> MA1102 Math for Electronics II: Yeah, so now let's move to the next topic, which is algebraic and geometric multiplicity. Okay, so? >> 26F1100067 UDAY PRATAP SAROJ: By the way, it's about to be 8 o'clock. >> MA1102 Math for Electronics II: Oh yeah, yeah, you have the. >> 26F1100067 UDAY PRATAP SAROJ: Yeah.
51:56
>> MA1102 Math for Electronics II: That's my next minor. So, the session is only seven to eight only. So, whatever >> 26F1100067 UDAY PRATAP SAROJ: Yeah. >> MA1102 Math for Electronics II: we have covered so far, I hope that was at least clear for you and the other topics. If you want, uh, maybe we can? >> 26F1100067 UDAY PRATAP SAROJ: Yes, yes.
52:15
>> MA1102 Math for Electronics II: I can share you the lecture, which I have done in the previous terms. So, the same is taught in the previous Dome. Also, if you want the live session link can be shared later. >> 26F1100067 UDAY PRATAP SAROJ: if you can share the for this session, that will be. >> MA1102 Math for Electronics II: This. This session is going to be there in your YouTube, but it is always going to be there. Or if you want the link, I can share it right now.
52:34
>> 26F1100067 UDAY PRATAP SAROJ: Yes, yes, yes, maybe the channel or something, because from there? >> MA1102 Math for Electronics II: Yeah, just a minute.
52:46
So, I'm sharing the YouTube link here in the.
52:54
So you can follow this channel. So, in this channel, only whatever eigen sorry whatever?
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topics are discussed in the live session, everything will be streamed here only in this channel only. >> 26F1100067 UDAY PRATAP SAROJ: Okay, okay, good?
53:13
>> MA1102 Math for Electronics II: And I will be sharing. Uh, PDF in the supplementary contents of your course portal page. You have right? So, there, in supplementary content, this will be added as a live session notes. >> 26F1100067 UDAY PRATAP SAROJ: Oh yes, okay, so any sessions I missed. I can at least. Ketchup from there. >> MA1102 Math for Electronics II: Yeah. >> 26F1100067 UDAY PRATAP SAROJ: Between there and? >> MA1102 Math for Electronics II: Yeah.
53:47
>> 26F1100067 UDAY PRATAP SAROJ: the live streams record. Okay, but but there is no supplementary section for the math course. If I'm not and just checking meaning? >> MA1102 Math for Electronics II: Uh, actually. This was the first session where someone joined, so maybe there was no session. So, no notes was added, but? Uh, but this was the scenario in my, uh, session. But the other instructor, I guess she handled two sessions where one student joined. I don't know, uh,
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whether she shared the notes or not, I'll check and maybe tell her to upload and supplementary contents. >> 26F1100067 UDAY PRATAP SAROJ: Okay, may I ask you a doctor? >> MA1102 Math for Electronics II: This is. >> 26F1100067 UDAY PRATAP SAROJ: okay. So, so you are saying this is the first? Uh.
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>> MA1102 Math for Electronics II: because whatever live sessions I was there, I was just waiting for someone to join and no one joined so.
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>> 26F1100067 UDAY PRATAP SAROJ: Okay, and I, I actually I think I also ran into this issue with, and not just only this course, but also some of the other courses where Mary I joined too early or too late, and there was no one, and so I also had to link so, uh, yes, this. This thing has been going on, but yeah, okay.
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