Lesson: Eigenvectors

Question 2

1 point


Our next goal in this course is to use complex numbers to help us diagonalize real matrices. That is to say, we want to look at diagonalizing matrices with real entries, but consider them as complex matrices so that we can find all the roots to their characteristic polynomial. So, we should first look at how to diagonalize any complex matrix, and in order to do that, we need to look at eigenvalues and eigenvectors over C.

Let L from Cn to Cn be a linear mapping. If for some lambda in C, there exists a non-zero vector z in Cn such that L(z) = (lambda)z, then lambda is an eigenvalue of L, and z is called an eigenvector of L that corresponds to lambda. Similarly, a complex number lambda is an eigenvalue of an n-by-n matrix A with complex entries with corresponding eigenvector z in Cn, where z does, cannot equal 0 vector, if Az = (lambda)z.

As you can see, the definition remains unchanged from the one we used in R, except that our vectors, matrices, and scalars are now in C instead of R. Moreover, this is yet another place where the theory we developed in R translates directly to our situation in C. And so we will use the exact same techniques to find eigenvalues, eigenvectors, and to diagonalize matrices as we did in R. Of course, simply doing these calculations with the complex numbers can make things different. Let’s look at an example.

Let A be the matrix [2, 2 – i; -1 + i, -1]. Find an invertible matrix P and a diagonal matrix D such that (P-inverse)AP = D. As before, the columns of P will be made from eigenvectors of A, and so we need to find these eigenvectors. But to find the eigenvectors, we first need to find the eigenvalues. These will be the solutions to the characteristic polynomial determinant of (A – (lambda)I).

So to compute the determinant of (A – (lambda)I), we will compute the determinant of the matrix [2 – lambda, 2 – i; -1 + i, -1 – lambda]. Well, this is (2 – lambda)(-1 – lambda) – (-1 + i)(2 – i), and if we multiply all this out, we get the polynomial lambda-squared – lambda + (-1 – 3i). We can use the quadratic formula to find the roots of this polynomial. Again, we simply have lambda = (-b plus-or-minus the square root of (b-squared – 4ac))/(2a). In this case, our b is a -1, our a is a 1, and our c is a -1 – 3i. But we’ll see that we end up with (1 plus-or-minus the square root of (5 + 12i))/2.

But what is the square root of (5 + 12i)? Well, we need to use de Moivre’s Formula to find this. Note the use of plus/minus symbol is still valid here, since de Moivre’s Formula tells us that the two square roots of any complex number z will be 2(pi)/2, or pi apart from each other, which is the same as multiplying by the scalar -1. So we only need to use de Moivre’s Formula to find one root of (5 + 12i). But we still need to put it into polar form first. To that end, we see that our modulus r will equal the square root of (5-squared + 12-squared), which is the square root of 169, which is 13. And this means that we need theta to be such that cosine(theta) = 5/13 and sine(theta) = 12/13. Since our cosine and sine values are both positive, theta is in the first quadrant, so we can set theta equal to either (cosine-inverse)(5/13) or (sine-inverse)(12/13). And then we’ll have that (5 + 12i)-to-the-(1/2) = (13-to-the-(1/2))(cosine(((cos-inverse)(5/13))/2) + i(sine(((cos-inverse)(5/13))/2))). Again, if you keep all of that within your calculator without doing any rounding off, you’ll see that our square root is 3 + 2i. And so we have that lambda = (1 plus-or-minus (3 + 2i))/2, which means it equals 2 + i or -1 – i.

Now that we have the eigenvalues for A, we need to find the eigenspaces for these eigenvalues. For lambda = 2 + i, we need to find the nullspace of this matrix, which we will do by row reducing. First, we’ll multiply the first row by an i, then we can replace the second row by (row 2 + (1 – i)(row 1)), and we have our reduced row echelon form matrix. And so we see that our eigenspace consists of all solutions to the equation z1 + (1 + 2i)z2 = 0, which we can write as the span of the single vector {[-1 – 2i; 1]}.

For lambda = -1 – i, we need to find the nullspace of this matrix, which equals this. And again, this means row reducing our matrix. First, I’ll go ahead and swap rows 1 and rows 2, and then we can multiply row 1 by (1/2)(-1 – i). Last, we replace row 2 with (row 2 + (-3 – i)(row 1)), and we’re left with this matrix in reduced row echelon form. And so we see that our eigenspace consists of all solutions to the equation z1 + (1/2 – (1/2)i)z2 = 0, which we can write as the span of the single vector {[-1 + i; 2]}.

Now that we know the eigenvalues and their eigenspaces, we know that the matrix P whose columns are the basis vectors for our eigenspace, is an invertible matrix such that (P-inverse)AP = D, where D is a diagonal matrix whose diagonal entries are the eigenvalues corresponding to the eigenvectors used in P. In this particular example, this means that P will be the matrix [-1 – 2i, -1 + i; 1, 2], and D is the matrix [2 + i, 0; 0, -1 – i].

And that was just a 2-by-2 matrix. Since diagonalizing general complex matrices is not our goal, we will satisfy ourselves with this example and move on to looking at matrices with real entries.

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