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I have the following problem:enter image description here

I know how to compute the eigenvectors given the matrix and then finding eigenvalues. I could turn A into a triangular matrix and then compute for lambdas, but I wanted to know if there was another procedure by using the eigenvectors and A to find the eigenvalues.

Any help is appreciated.

Thanks!

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    $\begingroup$ Refer back to the definition of eigenvalues and eigenvectors: $M$ a matrix; we say $v$ is an eigenvector of $M$ with eigenvalue $\mu$ if $Mv=\mu v$. $\endgroup$ – FireGarden May 12 '16 at 14:28
  • $\begingroup$ and you need just to multiply the first line of $A$ by $v$ and compare with first component of $v$ $\endgroup$ – G Cab May 12 '16 at 14:31
  • $\begingroup$ first line as in column or row? $\endgroup$ – idelara May 12 '16 at 14:33
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$$ \begin{bmatrix} 13 & 2 & -18\\ 14 & 1 & -18\\ 10 & 2 & -15 \end{bmatrix}\cdot \begin{bmatrix} -1\\ -1\\ -1 \end{bmatrix}= \begin{bmatrix} 3\\ 3\\ 3 \end{bmatrix}=-3\cdot \begin{bmatrix} -1\\ -1\\ -1 \end{bmatrix}. $$

Thus, $-3$ is an eigenvalue of $A$. Try the same strategy for the other two eigenvectors.

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If you multiply the matrix and the eigenvector $(A*v)$, you obtain $\lambda*v$, so from that you will see what is the actual lambda. Just a consequence of the definition.

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  • $\begingroup$ Do I do this by columns? say v1 (eigenvector) times a (1st column of A) $\endgroup$ – idelara May 12 '16 at 14:32
  • $\begingroup$ yes, just a standard operation :) $\endgroup$ – Jan Sila May 12 '16 at 14:35

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