## Transcript

Okay, so our eigenvalue lambda = a + bi for A leads to an eigenvector z = x + iy, and we now know that the span of {x, y} is an invariant subspace under A. What next? Well, we will end up using the real vectors x and y to form a matrix (instead of the actual eigenvectors, as before), but to see why, we want to start by looking at the case when A is a 2-by-2 matrix. Why? Well, not only is the span of {x, y} a subspace of Rn, but it is specifically a 2-dimensional subspace of Rn, with B = {x, y} as a basis. In the case when n = 2, the only 2-dimensional subspace of R2 is R2 itself, so we get that the set B is a basis for R2. Well, this means that the matrix P whose columns are x and y can be thought of as a change of coordinates matrix, from standard coordinates to B-coordinates. And thus, P is an invertible matrix, but moreover, we know that (P-inverse)AP will be the matrix for the linear mapping Ar, but with respect to B-coordinates instead of standard coordinates.

To look at this further, let’s step back a bit. We’ll define the linear mapping L from R2 to R2 by L(r) = Ar. So the standard matrix for L will be the matrix A, and the standard matrix for L with respect to B-coordinates will be the matrix whose columns are the B-coordinates of L(x) and the B-coordinates of L(y). Well, this will be the B-coordinates of Ax and the B-coordinates of Ay.

So we need to find the B-coordinates of Ax and Ay. But we can recall from our work in showing that the span of B is an invariant subspace that Ax = ax – by, and Ay = bx + ay. Well, then we see that the B-coordinates of Ax is the vector [a; -b], and the B-coordinates of Ay is the vector [b; a]. And so we have that the matrix for L with respect to B is the matrix [a, b; -b, a]. But remember that bracket-[L]B is the matrix for the linear mapping Ax with respect to B-coordinates, and so we have that bracket-[L]B = (P-inverse)AP.

And so we’ve ended up with a situation similar to diagonalization. We use the eigenvectors to find an invertible matrix P, and (P-inverse)AP is a matrix built using the eigenvalues. Our matrix [a, b; -b, a] is known as a real canonical form for A.

Let A be a 2-by-2 real matrix with eigenvalue lambda = a + ib, where b is not equal to 0. The matrix [a, b; -b, a] is called a real canonical form for A.

Let’s look at an example. In lecture 3q, we found that lambda = 1 + 2i is an eigenvalue for the matrix A = [3, 4; -2, -1], with corresponding eigenvector [-1 – i; 1], which we can now write as the vector [-1; 1] + i[-1; 0]. So in this case, we have that a = 1, and b = 2, that the vector x is [-1; 1], and the vector y is [-1; 0], so we must have that the matrix C = [1, 2; -2, 1] is a real canonical form for A, and that the matrix P = [-1, -1; 1, 0] is such that (P-inverse)AP = C. Note that you can easily calculate that P-inverse = [0, 1; -1, -1] and then compute the product (P-inverse)AP to verify that it is, in fact, [1, 2; -2, 1].

So, things seem to work quite nicely in the 2-by-2 case, but we definitely don’t have that {x, y} is a basis for R3. However, we do know that the set {v, x, y} is a basis for R3, where v is the eigenvector for the real eigenvalue. For if A is a 3-by-3 matrix, then its characteristic polynomial is a degree-3 polynomial, which must have at least one real root. In fact, since we know that complex roots of this polynomial will come in conjugate pairs, either A will have three real eigenvalues (counting multiplicity) or one real eigenvalue and two complex eigenvalues that are conjugates.

Having already looked at the case where A has only real eigenvalues, let’s now see what happens when A has one real eigenvalue mu with eigenvector v, and complex eigenvalues (a plus-or-minus ib) with eigenvectors (x plus-or-minus iy). Our theorem still applies, so we know that the span of {x, y} is a 2-dimensional subspace of R3. But this is where we make use of the fact that Span{x, y} does not contain any real eigenvectors, and thus, specifically does not contain v.

So, if we recall the technique for expanding a linearly independent set to a basis, we can start with the linearly independent set {x, y} and add the vector v not in the span of {x, y}, and we know that the resulting set {v, x, y} is linearly independent. And since we have a linearly independent set with three vectors, by my two-out-of-three rule, we know that this set is a basis for R3.

Since the set B = {v, x, y} is a basis for R3, the matrix P whose columns are v, x, and y is the change of coordinates matrix from standard coordinates to B-coordinates. And this means that we still have that (P-inverse)AP is the matrix for the linear mapping Ar with respect to B-coordinates. As we did in the 2-by-2 case, let’s use our knowledge of A to figure out what (P-inverse)AP should be.

So we’ll let L be the linear mapping from R3 to R3 defined by L(r) = Ar, so that bracket-of-[L] = A. Well, then bracket-[L]B is the matrix whose columns are the B-coordinates of L(v), the B-coordinates of L(x), and the B-coordinates of L(y), which are simply the B-coordinates of (Av), the B-coordinates of (Ax), and the B-coordinates of (Ay).

So we need to find these B-coordinates. Now, we still know that Ax = ax – by, so the B-coordinates of (Ax) will be the vector [0; a; -b], and we know that Ay = bx + ay, so the B-coordinates of (Ay) will be [0; b; a]. And since Av = (mu)v, we see that the B-coordinates of (Av) are [mu; 0; 0]. So we have that (P-inverse)AP equals the matrix [mu, 0, 0; 0, a, b; 0, -b, a]. And this is a real canonical form for a 3-by-3 matrix A with one real eigenvalue and two complex eigenvalues.

Again, let’s look at this in an example. In lecture 3q, we found that the matrix A = [3, 0, 1; 2, 1, 2; -1, 0, 3] had an eigenvalue 1 with corresponding eigenvector [0; 1; 0], and complex eigenvalues (3 plus-or-minus i) with corresponding eigenvectors [negative-plus 5i; 2 negative-plus 6i; 5], which we can now write as [0; 2; 5] plus-or-minus i[-5; -6; 0]. Again, you want to make sure you’ve got your negative signs all matched up. Don’t just think that because you say “plus or minus” it doesn’t matter which order they come in. But then, we’ll know that the matrix [1, 0, 0; 0, 3, 1; 0, -1, 3] is a real canonical form for A, and that the matrix [0, 0, -5; 1, 2, -6; 0, 5, 0] is a change of coordinates matrix that can bring A into this real canonical form.

Once we get into larger matrices, we end up with more potential combinations of real and complex eigenvalues, including the possibility of repeated complex roots. The mathematics involved with these new complications are beyond the scope of this course, so we will only look at 2-by-2 and 3-by-3 matrices.