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Find all the eigenvalues and associated eigenvectors for the given matrix:

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

So I went about finding the determinant by writing (5-$\lambda$) $\begin{bmatrix}2 &0 &-3\\0 & 2 &1 \\ 0 & 0 &3\\\end{bmatrix}$ Then I calculated the 3x3 determinant to be (5-$\lambda$)+2(6+0) and got the 17-$\lambda$ is this correct? Also where do I plug in this eigenvalue of 17 to find the eigenvector?

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Where does the eigenvalue $17$ come from? Keep going by expanding the determinant and in that way, keep factoring.

The matrix is in upper triangular form and that's good news, because the determinant is then just the product of the diagonal elements! So: $$\det\begin{bmatrix}5-\lambda &1 &-1& 0\\0 & 2-\lambda &0 &3\\ 0 & 0 &2-\lambda &1 \\0 & 0 &0 &3-\lambda\end{bmatrix} = (5-\lambda)(2-\lambda)^2(3-\lambda)$$ This means the eigenvalues are $5$, $2$ and $3$.

For each of the eigenvalues, plug back in and solve the associated homogeneous system of equations. For $\lambda = 5$, solve: $$\begin{bmatrix}0 &1 &-1& 0\\0 & -3 &0 &3\\ 0 & 0 &-3 &1 \\0 & 0 &0 &-2\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix}0\\0\\0\\0 \end{bmatrix} $$ From the last equation ($-2x_4 = 0$), it is clear that $x_4 = 0$. Substitute into the third equation to get $x_3=0$ and in the second to get $x_2=0$. From the first equation, $x_1$ remains as a free variable so vectors of the form $(x_1,0,0,0)$ are eigenvectors associated with the eigenvalue $5$; pick e.g. $(1,0,0,0)$.

Do the same for the other eigenvalues. Can you take it from here?

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    $\begingroup$ ah makes sense! Idk how I really got 17 to be honest but I just knew it was really wrong. Thank you! $\endgroup$ – Lil Apr 26 '16 at 14:13
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    $\begingroup$ You're welcome! $\endgroup$ – StackTD Apr 26 '16 at 14:14
  • $\begingroup$ @Lil If you stare at that first matrix equation long enough, you should be able to convince yourself that the eigenvalues of a triangular matrix are its diagonal entries. $\endgroup$ – amd Apr 26 '16 at 17:11

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