1
$\begingroup$

Struggling with this eigenvector problems. I've been using this SE article (Finding Eigenvectors of a 3x3 Matrix (7.12-15)) as a guide and it has been a very useful, but I'm stuck on my last case where $\lambda=4$.

Q: Find the eigenvalues $\lambda_1 < \lambda_2 < \lambda_3$ and corresponding eigenvectors of the matrix

$$A=\begin{bmatrix}-2 & 3 & 0 \\ 0 & -1 & -10 \\ 0 & 0 & 4\end{bmatrix}=\begin{bmatrix}-2-\lambda & 3 & 0 \\ 0 & -1-\lambda & -10 \\ 0 & 0 & 4-\lambda\end{bmatrix}$$

The eigenvalues for the $A$ matrix are $\lambda_1=-2$, $\lambda_2=-1$, $\lambda_3=4$ respectively.

Case $\lambda=-2$

$$A-2I_3=\begin{bmatrix}-2-(-2) & 3 & 0 \\ 0 & -1-(-2) & -10 \\ 0 & 0 & 4-(-2)\end{bmatrix}=\begin{bmatrix}0 & 3 & 0 \\ 0 & 1 & -10 \\ 0 & 0 & 6\end{bmatrix}$$

$$rref(A-2I_3)=\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}$$

$x_1$ is our free variable

$$\vec{v_1}=\begin{bmatrix}1\\0\\0\end{bmatrix}$$

Case where $\lambda=-1$.

$$A-I_3=\begin{bmatrix}-2-(-1) & 3 & 0 \\ 0 & -1-(-1) & -10 \\ 0 & 0 & 4-(-1)\end{bmatrix}=\begin{bmatrix}-1 & 3 & 0 \\ 0 & 0 & -10 \\ 0 & 0 & 5\end{bmatrix}$$

$$A^\prime=rref(A-I_3)=\begin{bmatrix}1 & 3 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$

Finding the $ker(A-I_3)$ we find $x_2$ and $x_3$ as our free variables. Solving for each variable we get $x_1=3x_2$ and $x_2=1$.

$$\vec{v_2}=\begin{bmatrix}3 \\ 1 \\ 0\end{bmatrix}$$

Case $\lambda=4$

$$A+4I_3=\begin{bmatrix}-2-4 & 3 & 0 \\ 0 & -1-4 & -10 \\ 0 & 0 & 4-4\end{bmatrix}=\begin{bmatrix}-6 & 3 & 0 \\ 0 & -5 & -10 \\ 0 & 0 & 0\end{bmatrix}$$

$$rref(A-I_3)=\begin{bmatrix}1 & -\frac{1}{2} & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 0\end{bmatrix}$$

This is where I'm stuck. I checked wolframalpha and it says my last eigenvector should be $\vec{v_3}=\begin{bmatrix}-1 \\ -2 \\ 1\end{bmatrix}$ and I'm not sure how it can be that. Thanks in advance!

$\endgroup$
2
$\begingroup$

The equations corresponding to that row-reduced form at the end are

$$ x - y/2 = 0 \\ y + 2z = 0 $$

Since $z$ is a free variable, you can pick $z = 1$ and back-substitute to get $y = -2$, and then use this in the first equation to get $x = 1$. Of course, any nonzero multiple of this vector is also an eigenvector for $4$.

$\endgroup$
  • $\begingroup$ Awesome, I see the process that was used here. Many thanks for the assistance! $\endgroup$ – djthoms May 12 '16 at 3:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.