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Eigenvalues and Eigenvectors · Acedra
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Eigenvalues and Eigenvectors
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0:06
What is happening everyone? Today we are going to look at IGEN vectors and IGEN values. We will look at how to calculate them, how to find the IGEN vectors from the IGEN values and then what those things mean and at the end we'll use this helpful little tool I built uh this visualization model of our particular example. um we'll go over that at the towards the end once we've calculated these values. So let's look at IGEN values and IGEN vectors. We'll do so using an example
0:39
to motivate everything we do here. Um we talk about theorems, definitions, corlaries, lemas, it'll be motivated by this example. Now I'm going to use a 2x two uh two dimensionals matrix mostly so we can avoid having to do larger determinant calculations a 3x3 4x4 5x5 uh I don't really want to do co-actor expansion and uh the use of technology um would be easy but I'm just going to do a handwritten example that's a 2x two matrix so it's and R2. I think that will be the easiest. Um,
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this will all scale to larger dimensions. So, it won't be that big of a deal. Again, I don't want to do a ton of co-actor expansion. I don't know if you want to see that. Let me know in the comments if you want to talk about how to calculate determinance of larger matrices. Um, we can do that. Uh, okay. So, here's our matrix A. uh picked somewhat arbitrarily and let's recall what happens when we take a matrix A and multiply it by some vector. So if I have A times some vector V,
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what we usually get out is a brand new vector. Um this vector can be different in direction and different in length.
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So what I mean by that is let's say we have a vector V and at the end I'll we'll look at this with the visual the JavaScript visual and I want to know what happens to V when I multiply it by A. So we get 4 1 2 3 times 1 0 and hopefully you remember how to multiply it's row by column. So the row of A time the column of V. So in other words, it's 4 * 1 and then we add 1 * 0 and then 2 * 1 + 3 * 0.
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And this will equal 4 + 0 2 + 0. So this is four and two. So what does that mean? Well, this is you. And so what happened here? Well, let's say one, two, three, four. One, two, three. So our vector V was this guy. I knew that was going to happen. So our vector V is on this line. 42 would be here. So somewhat like this. Now again, I'll show you this at the end using the visual. Once we calculate the igen
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values and igen vectors um we will go through that visual and I think things will make more sense. But here's what's happening. This is u which is a v and this is v. So that's what um a does to v. Now as an aside we know that we can write this this is sort of a parallelogram which will come into play in a little bit. So that's 42. It changes the direction and the magnitude obviously of V. Uh we get something that is some degrees off of the um so in other words we
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had this and it's some angle off of the x-axis. So it changed direction, it changed length. So what igen vectors and igen values do is we want to know um what are some of the special vectors that if I multiply a by that don't change directions. It only shrinks or stretches the vector itself. So what and how do we find such vectors? So let's define now we're going to let A be an N byN matrix. Obviously this is a square matrix and igen values and vectors are
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only defined for square matrices. So 3x4 or 4x3 um in context vectors and values don't make sense. So we must have an n byn matrix. And so a scalar, this is linear algebra terminology. A scaler, a number, and we call it lambda is an en value of a matrix A. If there exists a non zero nonzero is important vector v such that a * v will be equal to lambda vector or igen value times B and B is the igen vector that corresponds
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to lambda. So non zero is important because so why non zero? Well, it's because a * 0 is equal to alpha or alpha lambda time 0 is zero. And that's true for any lambda, any real number. This is true. So this is trivial. We're not interested in this, right? So we're looking for nonzero vectors that conform to this equation here. Uh, quickly here's a fun fact. If you wanted a fun fact again is a German word that means
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own or characteristic. Now in context, this should make some sense. All right, so let's move on. Uh, let's look at our equation. A * V is equal to lambda * V. And we require V to be a nonzero vector. So let's do a little bit of algebra. If we start off here, boy, I'll try to be a little bit calmer with my writing. AV is equal to lambda V. Again, I'm not happy with this V here. Now, we can subtract lambda in the vector V from both sides. Algebra works. Then we have it equal
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to zero. identity matrix is an n byn matrix where all of the col or all of the diagonals are one and everything else is zero. Um if you don't remember so I can essentially multiply that by anything and it leaves it unchanged. That's why we call it identity. So I'm going to put that guy here and it has the same dimension as a. So if n if a is a 2x two the identity matrix is a 2x2 in this case a is n by n and the identity has to be n byn and now we have this lovely equation.
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So a minus lambda * the identity matrix times the vector v is equal to zero. This is a homogeneic or homogeneous system. It has a nonzero solution if and only if the coefficient matrix is singular. Well, what's singular mean? So a matrix is singular if and only if the determinant of a is zero. It's another way of saying it's non invertible. Right? We can't invert a. So a is non invertible or not invertible. Is it not or not? I'm going to say not invertible.
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All right. So if something is not invertible, we know the determinant is equal to zero. So for this part up here, remember we have this equation and we'll need this later. But we can develop what's called a characteristic equation by looking at the determinant of the result of this a minus lambda i and we set that equal to zero and we call this the characteristic equation. This is how we find the ien values. So let's piece some stuff together and then we'll
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look at the example. So what we're saying here is that we have the determinant of the a minus lambda i set that equal to zero take the determinant that will give us the uh characteristic equation which will give us the igen values. So remember a here is 4 1 2 3 I and a 2x two be 1 0 0 1 and let's recall that the determinant of a 2x two let's say of a b c and d and this is why I picked the 2x two it's much easier to calculate the
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determinant of this will be a * d minus b * D. So it's the multiplication of the diagonal uh time or subtracted from the multiplication of the opposite diagonal. So remember we have the determinant of a minus lambda. We don't that's we're trying to find I and we're saying that's zero. So let's plug in what we have. AS41 2 3 minus some lambda times the identity matrix. I'm going to do all steps so it's clear where it comes from. So I'm
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going to multiply each entry by lambda. So we get lambda 0 0 lambda. Now I'm going to perform the operation which is subtraction. So we have 4 - lambda 1 - 0 2 - 0 and 3 - lambda. So clean this up. We get 4 - lambda 1 2 3 - lambda
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Now we take the determinant of this matrix. Remember the ter determinant is four in this case 4 - lambda * 3 - lambda subtracted from 2 * 1. Remember we want that to equal zero. We want to find lambdas where that's zero.
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So let's let's uh simplify. So we're going to foil. We get 12 minus 4 lambda. Lambda looks bad. - 3 lambda plus lambda^ 2 - 2 is equal to 0. Now let's write this in a more familiar quadratic form.
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So this is our characteristic equation. We solve this quadratic for lambda and we get our igen values. So solve we're going to factor. You can use a quadratic equation obviously. Um but we're going to factor treating lambda as a variable. We get two numbers that multiply give me 10 that add give me -7. We get this using zero product principle. We could set each of those equal to zero because a product equals zero one or both are zero. So that means that lambda equals 5 and 2.
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So this is our en values. Awesome.
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Let me say something here that I think is important. What is a determinant anyway? We just took a determinant, set it equal to zero and solve. That means a determinant is zero for these values, right? So what is a determinant? The absolute value of a determinant, not the magnitude, but the the positive determinant is essentially an area or in larger um let's say three dimensions a volume that scales. So what we saw here is we had this guy and we had this parallelogram.
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the area of that is in indeed the determinant of the matrix that results. So that's a little bit um necessary to talk about because when the determinant is zero, right? So if I say the determinant is area or volume. So when the determinant of a equals zero, the matrix
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A squashes everything to either a point or let's say onto a line. The visual will help here. When when we get to the end, we'll show you the visual and that will make more sense. But keep that in mind. Right? So with determinate zero that it sort of squashes everything to either a point or it puts it on to a line. In this case in our example we're going to find those lines. So we have we have two values. So we want to find the igen vectors.
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associated with each. And what does that mean? That means we're solving this equation. Essentially, we're solving a minus lambda i times some v= z. Here is the igen vector. Here is our lambda. So what we're really going to do is just plug that in. Let's recall a or one two three and uh lambda again here we'll do five first and lambda i will be zero 50 05 right okay so we have a minus lambda i is 41 2 3 minus 50 05. This is for the lambda 5G value. So we're
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going to find an igen vector associated with that value. Now do subtraction. We get -1 2 1 -2. Okay. So now what do we do? Well, we take this. This is the inside of that parenthesis which is -1 2 1 -2 times v and we set it equal to zero. Now we solve for v and we have our vector. So we we have to abstract v. We don't know what v is. So we're just going to label it. Let's say it's xy that
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equals zero. I think more uh thorough would be to write it as zero. So we're going to do the same process. I'm going to do row by column. So we get -x + y = 0 for the top and then we get 2x + - 2 y is equal to zero. So we see we get negative x becomes positive. We move it over. We could do the same thing with y and 2x. And this is just y = x. So if we say x is free, since they both relate, I can pick whatever I want for x. I'm going to
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make it simple and say x is one, which means y is one. So v the solution here is 1 one.
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We could write this parametrically as 1 1 * t where t is non zero. So this is the igen vector associated with lambda 5. Let's do the same thing for two. Let's recall we have 4 1 2 3 minus lambda time the identity matrix. In this case, we have lambda equ= 2 or 2 0 2 and we do this. 4 - 2 is 2. 1 - 0 is 1. 2 - 0 is 2. 2 - 2 is 0. That should be 3 - 1, not zero. Right? So again, we're solving this
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This is the inside here and v well I would say u so we don't confuse ourselves. This is a different vector u. So we have 21 * some u. Let's call it a. No, let's not do that. Let's just say xy since we're on the same coordinate system. That's equal to 0 0. So if I do the same row by column multiplication, we get 2x + y = 0. Same thing here. So we get y = -2x. y = -2x. Same thing I can consider one of these a free variable. I'm going to let x equal 1. That means y is equal to
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-2. So we have 1 -2 as our igen vector. in this case u and parametric wise we can say t t is non zero. So those are our igen vectors and our igen values. Now I'm going to write it as a space and say that all well we'll look at it we'll look at the visual. All right so I'm going to associate e sub lambda with the space. So e sub5 that's our igen value of five will equal the span of the vector 1 1 which if we want to write it parametrically is 1 1 * t is a real number.
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So 2 is a span of 1 -2 1 -2 * t for t and r. So say this is such that t is an r. So geometrically this is a line y= x. Geometrically this is a y and y = -2x. So this is our igen space. Our igen uh values are 5 and 2. Our igen vectors here are 1 one and one. So that process is how you find igen values and igen vectors. Now let's look at this in the visualization tool and we can get a sense of of what these do. U remember the vector is going to be unchanged directionally.
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It's just going to stretch or shrink the vector itself. So let's look at that now. Okay. So our tool here is you can see the matrix A that we've been using in our examples. Here our value five is in blue. Our value 2 is in green. This will remind us the igen space associated with five and two. Remember the line y= x and the line y= -2x. All right. So if I put my cursor in this box, it's going to give us a vector v. So I control V and then it shows me what happens
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when I apply A. So the transformation of A onto V gives us this new vector in this direction. Uh let's see if I can put it to where we uh started. I think I had a the vector one. It's going to be tough to put. I'm just going to leave it here. Uh one zero was the vector and then we applied a and we had this guy over here. So that is uh roughly four and two, right? So it's not on the blue line
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or the green line. Uh let me illustrate some more. And you also see there's an angle associated with it. So no matter where I put V, it's telling me the calculation of AV and how it changes. And you can see how it's always different in direction and different in magnitude or length. So these special vectors we talked about at the beginning are these igen values and igen vectors where the direction is not changed. So if I put the V on the blue line you'll see I get another vector
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on the blue line and the scaling of the resulting vector AV will always be this is where five comes in a magnitude five larger. So if I scroll if I make it smaller but AV is always five larger and same in the other direction it'll always be five larger. So the same thing happens on the igen vector associated with the igen value two. Now this only gets larger by a factor of two. So you can see it grows much small or not as quickly as five and it goes
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in both directions. I coded this pretty fine so like any slight movement of the cursor getting off the line but so you can kind of see here it's interesting uh it'll tell us when we're in the ien space somewhat generally we're sort of approximating. So this vector V is an igen vector which results in this A scaled by two. Again here's five scaled by five. Here's two scaled by two and everything else changes direction and length. So it's a nice visualization. Um
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it's useful. You can change the matrix. It has to stay in two dimensions of course. uh you can change the igen values which will change the igen vectors and the igen space. If this is something you're interested in and want this I can send you the code. Um so let me know if that's something you want let me know in the comments. But that is it. That's all I really wanted to talk about for the igen vectors values. Hopefully that all made sense. Let me know in the
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comments if you need anything clarified. I think I covered everything I wanted to cover. Uh again, thanks for watching and have a great night and I will see you next time.
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