Video: Calculating Eigenvalues of a 2 Γ— 2 Matrix

Calculate the eigenvalues of 𝐴 = [βˆ’1, 0 and 0, βˆ’1].

04:15

Video Transcript

Calculate the eigenvalues of the matrix 𝐴 equals negative one, zero, zero, negative one.

Well, first when we take a look at our matrix, we can see that it’s a two-by-two matrix. Well, we know that an 𝑛 by 𝑛 matrix will always have 𝑛 eigenvalues. So therefore, we know that with our matrix being a two-by-two matrix, we’re gonna have two eigenvalues. Okay, great, so we know that’s what we’re looking for. But what can we do to work out the eigenvalues? So to help us solve this problem and find our eigenvalues, we have a relationship. And that one is that the determinant of the matrix 𝐴 minus πœ†πΌ, where 𝐼 is the identity matrix ⁠— well, this is equal to zero. And It’s the πœ† in our relationship that’s gonna help us calculate our eigenvalue.

So therefore, using our relationship, we can say that the determinant of the matrix 𝐴, which is negative one, zero, zero, negative one, minus πœ† multiplied by the identity matrix, which is the matrix one, zero, zero, one, is just gonna be equal to zero. And as we said, the identity matrix is one that’s in this form. But we have these terms as one. So the diagonal running from the top left to bottom right is one. So therefore, if we multiplied by the scaler, which was our πœ† , we’re gonna get negative one, zero, zero, negative one minus πœ†, zero, zero, πœ†. And we want to find the determinant of this. So once we carried out the subtraction of our matrices, we’re gonna get the determinant of the matrix. Negative one minus πœ†, zero, zero, negative one minus πœ† is equal to zero.

So when we look at a two-by-two matrix and we want to find the determinant of that, if you remind ourselves what we do, well, we multiply diagonally and then subtract. So we have π‘Žπ‘‘ minus 𝑏𝑐. So if we have the matrix π‘Ž, 𝑏, 𝑐, 𝑑 the determinant of that is π‘Žπ‘‘ minus 𝑏𝑐. So therefore, if we do that, we’re gonna get negative one minus πœ† multiplied by negative one minus πœ† minus zero equals zero. So then, if we distribute the first parentheses over the second parentheses, we’re gonna get β€” well, first of all, we’re gonna get one plus πœ†. That’s cause negative one multiplied by negative one is one and negative one multiplied by negative πœ† is just πœ†. So it’s one plus πœ†. Then we’ve got plus another πœ† and then plus πœ† squared. Then this is equal to zero. So then if we tidy this up, we get πœ† squared plus two πœ† plus one equals zero.

So now what we need to do is solve to find πœ† cause it’s gonna give us our eigenvalues. Well now to solve this, what we can do is factor because we can factor our expression because we’ve got πœ† squared plus two πœ† plus one. Well, we need two values. They’re gonna multiply together to give us one and add together to give us two. So therefore, we’re gonna get πœ† plus one multiplied by πœ† plus one is equal to zero. And that’s because if we have one multiplied by one is one; one add one is two. So they satisfy this. So now what we need to do is solve this to find our πœ† value. Well, what we’re gonna do is set one of our parentheses equal to zero. And that’s because to get a result of zero, we need to have zero multiplied by something to get zero. So I’m gonna have πœ† plus one equals zero.

So therefore, πœ† is gonna be equal to negative one. And actually, as you can see in this one, we’ve got πœ† plus one multiplied by πœ† plus one. This is same as πœ† plus one squared. So we’ve got repeated roots. However, we were expecting two eigenvalues cause we showed that earlier. There should be two eigenvalues for two-by-two matrix. But we’ve got a repeated eigenvalue, so repeated root. So therefore, what we say is that our eigenvalue, or eigenvalues, is negative one. But negative one is degenerate. And it’s degenerate because both of our eigenvalues are identical because we’ve got, as we said, repeated roots.

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