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Finding eigenvectors and eigenspaces example

Finding the eigenvectors and eigenspaces of a 2x2 matrix. Created by Sal Khan.
Video transcript
In the last video, we started with the 2 by 2 matrix A is equal to 1, 2, 4, 3. And we used the fact that lambda is an eigenvalue of A, if and only if, the determinate of lambda times the identity matrix-- in this case it's a 2 by 2 identity matrix-- minus A is equal to 0. This gave us a characteristic polynomial and we solved for that and we said, well, the eigenvalues for A are lambda is equal to 5 and lambda is equal to negative 1. That's what we saw in the last video. We said that if you were trying to solve A times some eigenvector is equal to lambda times that eigenvector, the two lambdas, which this equation can be solved for, are the lambdas 5 and minus 1. Assuming nonzero eigenvectors. So we have our eigenvalues, but I don't even call that half the battle. What we really want is our eigenvectors and our eigenvalues. So let's see if we can do that. So if we manipulate this equation a little bit and we've manipulate it in the past. Actually, we've even come up with this statement over here. We can rewrite this over here as the 0 vector is equal to lambda times my eigenvector minus A times my eigenvector. I just subtracted Av from both sides. We know lambda times some eigenvector is the same thing as lambda times the identity matrix times that eigenvector. So all I'm doing is rewriting this like that. You multiply the identity matrix times an eigenvector or times any vector, you're just going to get that vector. So these two things are equivalent. Minus Av. That's still going to be able to the 0 vector. So far all I've done is manipulated this thing. This is really how we got to that thing up there. You factor out the v so to speak because we know that matrix vector products exhibit the distributive property. And we get lambda times the identity matrix minus A times my eigenvector have got to be equal to 0. Or another way to say it is, for any lambda eigenvalue, and let me write it for any eigenvalue lambda, the eigenvectors that correspond to that lambda, we can call that the eigenspace for a lambda. So that's a new word, eigenspace. Eigenspace just means all of the eigenvectors that correspond to some eigenvalue. The eigenspace for some particular eigenvalue is going to be equal to the set of vectors that satisfy this equation. Well, the set of vectors that satisfy this equation is just the null space of that right there. So it's equal to the null space of this matrix right there. The null space of lambda times the identity matrix. And by an identity matrix minus A. And so everything I've done here, this is true-- this is the general case. But now we can apply this notion to this matrix A right here. So we know that 5 is an eigenvalue. Let's say for lambda is equal to 5, the eigenspace that corresponds to 5 is equal to the null space of? Well, what is 5 times the identity matrix? It's going to be the 2 by 2 identity matrix. 5 times the identity matrix is just 5, 0, 0, 5 minus A. That's just 1, 2, 4, 3. So that is equal to the null space of the matrix. 5 minus 1 is 4. 0 minus 2 is minus 2. 0 minus 4 is minus 4. And then, 5 minus 3 is 2. So the null space of this matrix right here-- and this matrix is just an actual numerical representation of this matrix right here. The null space of this matrix is the set of all of the vectors that satisfy this or all of the eigenvectors that correspond to this eigenvalue. Or, the eigenspace that corresponds to the eigenvalue 5. These are all equivalent statements. So we just need to figure out the null space of this guy is all of the vectors that satisfy the equation 4 minus 2, minus 4, 2 times some eigenvector is equal to the 0 vector. And the null space of a matrix is equal to the null space of the reduced row echelon form of a matrix. So what's the reduced row echelon form of this guy? Well, I guess a good starting point-- let me keep my first row the same, 4 minus 2. And let me replace my second row with my second row plus my first row. So minus 4 plus 4 is 0. 2 plus minus 2 is 0. Now, let me divide my first row by 4 and I get 1, minus 1/2. And then I get 0, 0. So what's the null space of this? This corresponds to v. This times v1, v2-- that's just another way of writing my eigenvector v-- has got to be equal to the 0 vector. Or another way to say it is that my first entry v1, which corresponds to this pivot column, plus or minus 1/2 times my second entry has got to be equal to that 0 right there. Or, v1 is equal to 1/2 v2. And so if I wanted to write all of the eigenvectors that satisfy this, I could write it this way. My eigenspace that corresponds to lambda equals 5. That corresponds to the eigenvalue 5 is equal to the set of all of the vectors, v1, v2, that are equal to some scaling factor. Let's say it's equal to t times what? If we say that v2 is equal to t, so v2 is going to be equal to t times 1. And then, v1 is going to be equal to 1/2 times v2 or 1/2 times t. Just like that. For any t is a member of the real numbers. If we wanted to, we could scale this up. We could say any real number times 1, 2. That would also be the span. Let me do that actually. It'll make it a little bit cleaner. Actually, I don't have to do that. So we could write that the eigenspace for the eigenvalue 5 is equal to the span of the vector 1/2 and 1. So it's a line in R2. Those are all of the eigenvectors that satisfy-- that work for the equation where the eigenvalue is equal to 5. Now what about when the eigenvalue is equal to minus 1? So let's do that case. When lambda is equal to minus 1, then we have-- it's going to be the null space. So the eigenspace for lambda is equal to minus 1 is going to be the null space of lambda times our identity matrix, which is going to be minus 1 and 0, 0, minus 1. It's going to be minus 1 times 1, 0, 0, 1, which is just minus 1 there. Minus A. So minus 1, 2, 4, 3. And this is equal to the null space of-- minus 1, minus 1 is minus 2. 0 minus 2 is minus 2. 0 minus 4 is minus 4 and minus 1 minus 3 is minus 4. And that's going to be equal to the null space of the reduced row echelon form of that guy. So we can perform some row operations right here. Let me just put it in reduced row echelon form. So if I replace my second row plus 2 times my first row. So I'll keep the first row the same. Minus 2, minus 2. And then my second row, I'll replace it with two times-- I'll replace it with it plus 2 times the first. Or even better, I'm going to replace it with it plus minus 2 times the first. So minus 4 plus 4 is 0. And then if I divide the top row by minus 2, the reduced row echelon form of this matrix right here or this matrix right here is going to be 1, 1, 0. So the eigenspace that corresponds to the eigenvalue minus 1 is equal to the null space of this guy right here It's the set of vectors that satisfy this equation: 1, 1, 0, 0. And then you have v1, v2 is equal to 0. Or you get v1 plus-- these aren't vectors, these are just values. v1 plus v2 is equal to 0. Because 0 is just equal to that thing right there. So 1 times v1 plus 1 times v2 is going to be equal to that 0 right there. Or I could write v1 is equal to minus v2. Or if we say that v2 is equal to t, we could say v1 is equal to minus t. Or we could say that the eigenspace for the eigenvalue minus 1 is equal to all of the vectors, v1, v2 that are equal to some scalar t times v1 is minus t and v2 is plus t. Or you could say this is equal to the span of the vector minus 1 and 1. So let's just graph this a little bit just to understand what we just did. We were able to find two eigenvalues for this, 5 and minus 1. And we were able to find all of the vectors that are essentially-- or, we were able to find the set of vectors that are the eigenvectors that correspond to each of these eigenvalues. So let's graph them. So if we go to R2, let me draw my axes, this is my vertical axis. That's my horizontal axis. So all of the vectors that correspond to lambda equal 5 are along the line 1/2, 1. Or the span of 1/2, 1. So that is 1. That is 1. So you go 1/2 and 1 just like that. So that's that vector, spanning vector. But anything along the span of this, all the multiples of this, are going to be valid eigenvectors. So anything along that line, all of the vectors when you draw them in standard position, point to a point on that line. All of these vectors, any vector on there is going to be a valid eigenvector and the corresponding eigenvalue is going to be equal to 5. So you give me this guy right here. When you apply the transformation, it's going to be five times this guy. If this guy is x, t of x is going to be five times this guy. Whatever vector you give along this line, the transformation of that guy, the transformation is literally, multiplying it by the matrix A. Where did I have the matrix A? The matrix A right up there. You're essentially just scaling this guy by 5 in either direction. This is for lambda equal 5. And for lambda equals 1, it's the span of this vector, which is minus 1, 1. Which looks like this. So this vector looks like that. We care about the span of it. Any vector that when you draw in standard position lies, or points to, points on this line, will be an eigenvector for the eigenvalue minus 1. So lambda equals minus 1. Let's say you take the spanning vector here. You apply the transformation, you're going to get minus 1 times it. So if this is x, the transformation of x is going to be that right there. Same length, just in the opposite direction. If you have this guy right here, you apply the transformation, it's going to be in the same spanning line just like that. So the two eigenspaces for the matrix-- where did I write it? I think it was the matrix 1, 2, 3-- 1, 2, 4, 3. The two eigenvalues were 5 and minus 1. And then it has an infinite number of eigenvectors, so they actually create two eigenspaces. Each of them correspond to one of the eigenvalues. And these lines represent those two eigenspaces. You give me any vector in either of these sets and they're going to be an eigenvector. I'm using the word vector too much. You give me any vector in either of these sets, and they will be an eigenvector for our matrix A. And then, depending on which line it is, we know what their transformation is going to be. If it's going to be on this guy, we take the transformation, the resulting vector's going to be five times the vector. If you take one of these eigenvectors and you transform it, the resulting transformation of the vector's going to be minus 1 times that vector. Anyway, we now know what eigenvalues, eigenvectors, eigenspaces are. And even better, we know how to actually find them.