2
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For this matrix

$$\begin{bmatrix} 3 & 0 & 0 \\ 1 & 2 & 0 \\ -4 & 5 & -1 \end{bmatrix}$$

The Eigenvalues = $-1, 2, 3$

The Eigenvectors I got were $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 0 \\ \frac35 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 4 \\ 4 \\ 1 \end{bmatrix}$.

However, only the last pair, $3$ and $\begin{bmatrix} 4 \\ 4 \\ 1 \end{bmatrix}$ were correct. I've looked over and over my algebra, but I don't see where I went wrong.

EDIT: For anyone in the future, the Eigenvalues/vectors I got were right, but I used the wrong matrix. I should have multiplied the Eigenvector with the matrix $$A=\begin{bmatrix} 3 & 0 & 0 \\ 1 & 2 & 0 \\ -4 & 5 & -1 \end{bmatrix}$$but instead I used the wrong matrix, (A-lambda*I) and substituted the corresponding Eigenvalue into lambda.

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  • $\begingroup$ math.meta.stackexchange.com/q/5020/306553 mathjax references. $\endgroup$ – Siong Thye Goh Aug 9 '17 at 2:31
  • $\begingroup$ For your future questions, if you want someone to show you where you went wrong, show your work so that we don’t have to guess how you came up with your results. $\endgroup$ – amd Aug 9 '17 at 6:47
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Checking if the condition of $Av=\lambda v$ holds:

$$\begin{bmatrix} 3 & 0 & 0 \\ 1 & 2 & 0 \\ -4 & 5 &- 1 \end{bmatrix}\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ -1\end{bmatrix}=-\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$$

$$\begin{bmatrix} 3 & 0 & 0 \\ 1 & 2 & 0 \\ -4 & 5 &- 1 \end{bmatrix}\begin{bmatrix} 0 \\ 0.6 \\ 1\end{bmatrix}=\begin{bmatrix} 0 \\ 1.2 \\ 2\end{bmatrix}=2\begin{bmatrix} 0 \\ 0.6 \\ 1\end{bmatrix}$$

They are indeed the right eigen pair.

Remark: In case the confusion is due to obtaining different answer from a book. Eigenvectors are not unique.

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