Mathematics Prelims

June 26, 2009

The Characteristic Polynomial

Filed under: Algebra,Linear Algebra — cjohnson @ 7:18 pm

Last time we defined the eigenvalues and eigenvectors of a matrix, but didn’t really discuss how to actually calculate the eigenvalues or eigenvectors; we said that if it so happened that your matrix was similar to a diagonal matrix, the non-zero entries of the diagonal matrix were the eigenvalues, and the columns of the change-of-basis matrix were the eigenvectors.  Now we’re going to discuss how to find the eigenvalues using the matrix’s characteristic polynomial.

Notice that if \lambda is an eigenvalue of A with associated eigenvector v we have the following.

Av = \lambda v

\implies Av - \lambda v = 0

\implies (A - \lambda I) v = 0

Of course, v = 0 satisfies this equation, but that’s a trivial solution.  For any other, non-trivial, solution we’d require that A is non-singular, and so \text{det}(A - \lambda I) = 0.  Thus if \lambda is an eigenvalue of A, we must have \text{det}(A - \lambda I) = 0.

Now suppose that \omega is such that \text{det}(A - \omega I) = 0.  Then there is a non-trivial solution to (A - \omega I) u = 0, so Au = \omega u, and \omega is an eigenvalue.  We’ve shown that \lambda is an eigenvalue of A if and only if \text{det}(A - \lambda I) = 0.  Furthermore, \text{det}(A - \lambda I) is a polynomial in \lambda (this is obvious if A is 1 \times 1, and inductively we can show that this is true for n \times n matrices).  This means that with the characteristic polynomial, the problem of finding eigenvalues is reduced to finding the roots of a polynomial.

As an example, suppose

\displaystyle A = \left[ \begin{array}{ccc} 1 & -1 & 2 \\ 2 & 2 & -3 \\ 3 & 5 & 7 \end{array} \right]

Then the characterstic polynomial is

\displaystyle \text{det} \left( \left[ \begin{array}{ccc} 1 - \lambda & -1 & 2 \\ 2 & 2 - \lambda & -3 \\ 3 & 5 & 7 - \lambda \end{array} \right] \right) = -\lambda^3 + 10\lambda^2 - 34 \lambda + 60

\displaystyle = -(\lambda-6) \, (\lambda^2 - 4 \lambda + 10)

\displaystyle = (\lambda - 6) \, (\lambda - (2 - i \sqrt{6})) \, (\lambda - (2 + i \sqrt{6}))

So we see that the eigenvalues are 6, 2 - i \sqrt{6}, and 2 + i \sqrt{6} (notice the last two are complex conjugates of one another).

Now, once we’ve found the eigenvalues, the next step is to find the eigenvectors.  Since

Av = \lambda v

\implies (A - \lambda I) v = 0

what we want is to find the nullspace of A - \lambda I, since these are all the vectors that A - \lambda I will take to zero.  In our particular example, for \lambda = 6,

(A - \lambda I) v = 0

\implies \left( \left[ \begin{array}{ccc} 1 & -1 & 2 \\ 2 & 2 & -3 \\ 3 & 5 & 7 \end{array} \right] - \left[ \begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array} \right] \right) \left[ \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array} \right] = 0

\implies \left[ \begin{array}{ccc} -5 & -1 & 2 \\ 2 & -4 & -3 \\ 3 & 5 & 1 \end{array} \right] \left[ \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array} \right] = 0

Now we take the row-reduced echelon form of this matrix, since it shares the same null space:

\implies \left[ \begin{array}{ccc} 1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{1}{2} \\ 0 & 0 & 0 \end{array} \right] \left[ \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right]

This tells us that

\text{NS}(A - 6I) = \left\{ \left[ \begin{array}{c} \frac{1}{2} v_3 \\ -\frac{1}{2} v_3 \\ v_3 \end{array} \right] : v_3 \in \mathbb{C} \right\}

So the eigenvectors associated with the eigenvalue \lambda = 6 are the multiples of \left[ 0.5, \, -0.5, \, 1 \right]^T.  We’d repeat the above process with \lambda = 2 \pm i \sqrt{6} to find the other eigenvectors.

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2 Comments »

  1. Nice post :).

    Comment by watchmath — June 26, 2009 @ 11:27 pm | Reply

  2. very useful…10Q

    Comment by mani — July 18, 2009 @ 6:55 am | Reply


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