Lesson: Diagonalizing a Real Matrix Over \( \mathbb{C} \)

Question 2

1 point

Transcript

Now let’s look at what happens when we diagonalize a matrix with real entries over C. So we can look at diagonalizing the matrix A = [3, 4; -2, -1]. First, we’ll need to find the eigenvalues by finding the roots of the characteristic polynomial, determinant of (A – (lambda)I). This will be equal to the determinant of the matrix [3 – lambda, 4; -2, -1 – lambda], which equals (3 – lambda)(-1 – lambda) + 8, which we find equals lambda-squared – 2(lambda) + 5. We can use the quadratic formula to find the roots of this equation, and see that lambda must equal 1 plus-or-minus 2i.

Now let’s find the eigenspaces for these eigenvalues. For lambda = 1 + 2i, the eigenspace is the nullspace of (A – (lambda)I), which is this matrix. To find a basis for the nullspace, we need to row reduce the matrix. First, I’ll interchange rows 1 and 2, and then I’ll multiply row 1 by -1/2. Lastly, I will replace row 2 with (row 2 + (-2 + 2i)(row 1)), and I end up with this matrix in reduced row echelon form. So the eigenvectors for lambda = 1 + 2i satisfy the equation z1 + (1 + i)z2 = 0. This means that the eigenspace is the span of the single vector {[-1 – i; 1]}.

For lambda = 1 – 2i, the eigenspace is the nullspace, again, of (A – (lambda)I), which is now this matrix. To find a basis for the nullspace, we need to row reduce the matrix. Again, I’ll start by interchanging rows 1 and row 2, and then multiplying row 1 by -1/2. Now I will end by replacing row 2 with (row 2 + (-2 – 2i)(row 1)), and I end up with this matrix. So this means the eigenvectors for lambda = 1 – 2i satisfy the equation z1 + (1 – i)z2 = 0, and this means that the eigenspace is the span of the single vector {[-1 + i; 1]}.

So we have that the vector [-1 – i; 1] is an eigenvector for lambda = 1 + 2i, while the vector [-1 + i; 1] is an eigenvector for the eigenvalue 1 – 2i, so this means that the matrix P = [-1 – i, -1 + i; 1, 1] is such that (P-inverse)AP = D, which is the matrix [1 + 2i, 0; 0, 1 – 2i].

To a certain extent, this situation is unsatisfying. While we couldn’t diagonalize this matrix over R, diagonalizing to a matrix with complex entries isn’t what we really want, either. But there are some useful things we can notice from this example. The first is that the eigenvalues are complex conjugates of each other. This is not surprising since the eigenvalues of a matrix with real entries are the solutions to a polynomial with real coefficients, and our theorem already pointed out that complex solutions to polynomials over R will come in conjugate pairs. But there is something new to realize at this point, and that is that the eigenvectors for lambda = 1 – 2i are vector conjugates of the eigenvectors for lambda = 1 + 2i. So, not only do our eigenvalues come in conjugate pairs, but the eigenvectors come in conjugate pairs as well. We’ll prove this for all matrices with real entries right after we introduce the following definition.

Let A be an m-by-n matrix. We define the complex conjugate of A, A-conjugate, by the ijth entry of A-conjugate is the conjugate of the ijth entry of A.

And now here’s our theorem. Suppose that A is an n-by-n matrix with real entries, and that lambda = a + bi, where b does not equal 0, is an eigenvalue of A with corresponding eigenvector z. Then lambda-conjugate is also an eigenvalue of A, with corresponding eigenvector z-conjugate.

Let’s look at the proof. First, we’ll let A be an n-by-n matrix with real entries, and let lambda be an eigenvalue of A with corresponding eigenvector z. So, Az = (lambda)z, and taking the complex conjugate of both sides gives us that the conjugate of (Az) equals the conjugate of ((lambda)z).

First, let’s look at the vector w, equal to the conjugate of (Az). Well, then w is a vector in Cn, and we know that the kth entry of w will be the conjugate of the kth entry in (Az), which we can write as the sum from j = 1 to n of ((ajk)zk). Now, since one of the properties of the complex conjugate is that the conjugate of (z1 + z2) equals (the conjugate of z1) + (the conjugate of z2), and another is that the conjugate of (z1z2) is (the conjugate of z1) times (the conjugate of z2), we see that we can go ahead and distribute our conjugate through our sum and product to see that wk equals the sum from j = 1 to n of ((ajk-conjugate)(zk-conjugate)). But this means that w = (A-conjugate)(z-conjugate). So that is to say, we have shown that the conjugate of (Az) equals (A-conjugate)(z-conjugate).

Now let’s look at the right side of our equality. Let’s let the vector u be the conjugate of ((lambda)z). Well, then u is a complex vector, and the kth component of u is the conjugate of ((lambda)zk), which, again, is just (lambda-conjugate)(zk-conjugate). So this means that, as a vector, the vector u equals (the conjugate of lambda)(z-conjugate). So that is to say, we’ve shown that the conjugate of ((lambda)z) equals (lambda-conjugate)(z-conjugate).

Now combining our results of about w and u, we now see that (the conjugate of A) times (the conjugate of z) equals the conjugate of (Az), which equals the conjugate of ((lambda)z), which equals (the conjugate of lambda) times (the conjugate of z). This means that the conjugate of lambda is an eigenvalue for the conjugate of A, with corresponding eigenvector conjugate-of-z.

Now, it’s worth noting that, so far in this proof, we’ve not made use of the fact that A is a matrix with real entries. So that means that, so far, all of these results hold true for any matrix A with complex entries. But to get the result we want now, we will make use of the fact that A is a matrix with real entries to see that (the conjugate of A) equals A, which means that we have shown that the conjugate of lambda is an eigenvalue for A, with corresponding eigenvector conjugate-of-z as desired.

Let’s diagonalize the matrix A = [3, 0, 1; 2, 1, 2; -1, 0, 3] over the complex numbers. First, we need to find the eigenvalues by finding the roots of the characteristic polynomial, determinant of (A – (lambda)I). Well, this will be the determinant of the matrix seen here. Expanding along the first row, we’ll get that this equals ((3 – lambda) times the determinant of the submatrix [1 – lambda, 2; 0, 3 – lambda]), our second term in the first row is 0, so we don’t need to worry about the determinant of the submatrix, but then we’ll have to add (1 times the determinant of the submatrix [2, 1 – lambda; -1, 0]). For our first product, we’ll get (3 – lambda)((1 – lambda)(3 – lambda))—thankfully, we have a 0 term there to make that first determinant easy. And again, in our second submatrix, we also have a 0, so the determinant of that matrix is simply a (1 – lambda). I can factor out a (1 – lambda) from both of those terms, leaving me with (3 – lambda)(3 – lambda) + 1. At that point, I will go ahead and multiply the rest out, and I’ll see that my characteristic polynomial is (1 – lambda)(10 – 6(lambda) + lambda-squared).

So because (1 – lambda) is a factor of the characteristic polynomial, we know that lambda = 1 is an eigenvalue of A. To find the other eigenvalues of A, we can use the quadratic formula to find the roots of 10 – 6(lambda) + lambda-squared. Now, be careful when you plug into the quadratic formula to keep track of your a and your c because, frequently, the characteristic polynomial is written somewhat backwards. So in this case, we still see that our b is a -6, but our a is the 1 in front of our lambda-squared, and our c is the 10. Well, this will become (6 plus-or-minus the square root of -4)/2, which is the same as 3 plus-or-minus i. So the eigenvalues of A are lambda = 1, 3 + i, 3 – i.

Now let’s find their eigenspaces. The eigenspace for lambda = 1 is the nullspace of (A – (lambda)I), which would be this matrix. To find the nullspace, we need to row reduce. As we’re simply row reducing a real matrix, I’ll skip to the end, and we see we end up with this matrix. So the eigenvectors of lambda = 1 are the vectors that satisfy z1 = 0 and z3 = 0, which we can write as the span of the single vector {[0; 1; 0]}.

The eigenspace for lambda = 3 + i will again be the nullspace of (A – (lambda)I), which now is this matrix. To find the nullspace, we will need to row reduce this matrix. And now since our matrix has complex terms, I will go through all the row reduction steps. First, I multiply the first row by i, then I replace row 2 with (row 2 – 2(row 1)), and I replace row 3 with (row 3 + row 1), and finally, I multiplied row 2 by (1/5)(-2 + i), and I ended up with this matrix in reduced row echelon form. So the eigenvectors of lambda = 3 + i are the vectors that satisfy z1 + (i)z3 = 0 and z2 – (2/5 – (6/5)i)z3 = 0. If we replace the variable z3 with the parameter alpha, we see that the general solution to this system is that the vector [z1; z2; z3] = [(-i)(alpha); (2/5 – (6/5)i)(alpha); alpha]. I could pull the alpha out in front, but I can sort of replace my alpha with a (1/5)(alpha), or in other words, replace, multiply everything inside my vector by 5 to get rid of those pesky fractions, and see that this is equivalent to (alpha)[-5i; 2 – 6i; 5]. And so this means that the eigenspace for lambda = 3 + i is the span of the single vector {[-5i; 2 – 6i; 5]}.

To find the eigenspace for lambda = 3 – i, we note that 3 – i is the conjugate of (3 + i), and we’ll use our theorem to see that the eigenspace for lambda = 3 – i will simply be the conjugate of the eigenspace for lambda = 3 + i. That is to say, it’s the set of all z-conjugate where z is in the eigenspace of 3 + i. Well, we can write this as the span of the vector {[conjugate-(-5i); conjugate-(2 – 6i); conjugate-5]}, which is the span of the vector {[5i; 2 + 6i; 5]}.

And so we have that [0; 1; 0] is an eigenvector for the eigenvalue lambda = 1, [-5i; 2 – 6i; 5] is an eigenvector for the eigenvalue lambda = 3 + i, and [5i; 2 + 6i; 5] is an eigenvector for the eigenvalue lambda = 3 – i. And this means that the matrix P, seen here, is such that (P-inverse)AP equals our diagonal matrix D, which will be [1, 0, 0; 0, 3 + i, 0; 0, 0, 3 – i].

So the value in our theorem was that we no longer had to go through all the row reduction steps to find the eigenspace for lambda = 3 – i. Instead, we could simply get our answer using the work we did for lambda = 3 + i.

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