# 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.