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Struggling with this eigenvector problems. I’ve been using this S.E. article as a guide and it has been 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 $\lambda=-2$

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

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

$x_1$ is our free variable

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

Case where $\lambda=-1$.

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

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

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 $\lambda=4$

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

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

This is where I’m stuck. I checked Wolfram Alpha 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!

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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 $4$.

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  • $\begingroup$ Awesome, I see the process that was used here. Many thanks for the assistance! $\endgroup$
    – djthoms
    May 12, 2016 at 3:03

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