# Find the eigenvalues and associated eigenvectors for this matrix

$$A = \begin{pmatrix} 1 & -1 & 4 \\ 3 & 2 & -1 \\ 2 & 1 & -1\end{pmatrix}$$

I solved for the determinant of matrix $A$ and got the following:

$\det\begin{bmatrix} 1-\lambda & -1 & 4\\ 3 & 2-\lambda & -1 \\ 2 & 1 & -1-\lambda \end{bmatrix} = -\lambda^{3}+2\lambda^{2}+5\lambda-6$

How do I find the eigenvalues from this? Once I have the eigenvalues, do I get the eigenvectors from the bases?

• Solve for $\lambda$ by letting $det(A-\lambda I)= 0$. You may want to refer to the following link-tutorial.math.lamar.edu/Classes/DE/LA_Eigen.aspx – John_dydx Nov 30 '15 at 0:21
• Hint: A reasonable first guess at the roots of the polynomial is factors of its constant term. – amd Nov 30 '15 at 1:43

To find eigenvalues, note that you want to find the solutions to the characteristic equation given by $\det (A - \lambda I) = 0$

As you noted, we have the equation $-\lambda^3 + 2\lambda^2 + 5 \lambda - 6 = 0$ Note that this is just $-(\lambda - 3)(\lambda + 2)(\lambda -1) = 0$ so your eigenvalues are $\lambda_1 = 3, \lambda_2 = -2,$ and $\lambda_3 = 1$.

Now, to get the eigenvectors, you want to look for $v$ such that $Av = \lambda v$ or $(A- \lambda I) v = 0$. For $\lambda_1$, $A- \lambda I = \begin{pmatrix} -2 & -1 & 4 \\ 3 & -1 & -1 \\ 2 & 1 & -4 \end{pmatrix}$ so row reducing gives $\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{pmatrix}$ In other words, for your eigenvector $v_1 = (a_1,a_2,a_3)$, \begin{align*} a_1 - a_3 &= 0 \\ a_2 - 2a_3 &= 0 \end{align*} Let $a_3 = t$, then $a_1 = t$ and $a_2 = 2t$ so a basis for the eigenspace corresponding to $\lambda_1$ is given by $\{\begin{pmatrix} 1 \\ 2 \\ 1 \\ \end{pmatrix} \}$.

So, for $\lambda_1 =3$, we get a corresponding eigenvector $(1,2,1)^T$. Try finding the eigenvectors for $\lambda_2$ and $\lambda_3$!

• I think I'm having issues when it comes to factoring a cubed polynomial. Are there any tips you can give me? – juliodesa Nov 30 '15 at 0:55
• It's mostly just practice! You can start off with the rational root theorem (given that most of the time you'll be given nice enough eigenvalues) en.wikipedia.org/wiki/Rational_root_theorem which tells you that the only rational roots of the above characteristic equation are $\pm 1,2,3,6$ so you can try to reduce your cubic to a quadratic after finding a root and use the quadratic formula afterward. There's a messy formula for a root that you might learn as you get into Galois theory but it's not really applicable. – Alvin Jin Nov 30 '15 at 1:01