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How can I find $a$ and $b$ if I have half of the matrix ($2\times 2$), one eigenvector, and one eigenvalue? The matrix is $2\times 2$:

$$ A=\left( \begin{array}{cc} 6 & a \\ 5 & b \end{array} \right)\,,$$

and the eigenvector is $(-4,-6)$ associated to this eigenvalue $\lambda=6$.

Sorry for my English. I'm Chilean.

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  • $\begingroup$ By definition, what does it mean for $6$ to be an eigenvalue? $\endgroup$ – Shubham Johri Dec 14 '18 at 22:11
  • $\begingroup$ the 6 is the eigenvalue of the eigenvector (-4,6) $\endgroup$ – Nelson Aguilera Dec 14 '18 at 22:12
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    $\begingroup$ Use the definition of $\lambda$ i.e. $Av=\lambda v$. You are given $\lambda$ and $v$. $\endgroup$ – Yadati Kiran Dec 14 '18 at 22:13
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Since $\lambda = 6$ is an eigenvalue and $v = [-4 \;| -6]^\mathbf{T}$ is an eigenvector, then , by definition, it must be :

$$Av = \lambda v \implies \begin{pmatrix}6 & a \\ 5 & b \end{pmatrix}\begin{pmatrix} -4\\-6\end{pmatrix}= 6\begin{pmatrix} -4\\-6\end{pmatrix} \implies \begin{cases} -24 -6a = -24 \\ -20 - 6b = -36 \end{cases} $$

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  • $\begingroup$ @NelsonAguilera Your calculations are wrong. Solve the equations yielded more carefuly. It's just algebra. $\endgroup$ – Rebellos Dec 14 '18 at 22:24
  • $\begingroup$ Now I can understand the question. Thank you for help me. $\endgroup$ – Nelson Aguilera Dec 14 '18 at 22:25
  • $\begingroup$ @egreg Thanks a lot for the heads up! $\endgroup$ – Rebellos Dec 14 '18 at 22:46
  • $\begingroup$ @egreg thanks :) $\endgroup$ – Nelson Aguilera Dec 14 '18 at 22:48
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Consider $$ u=\begin{pmatrix} 6 \\ 5 \end{pmatrix} \qquad x=\begin{pmatrix} a \\ b \end{pmatrix} \qquad v=\begin{pmatrix} -4 \\ -6 \end{pmatrix} $$ Then your data can be rewritten as $$ -4u-6x=6v $$ and therefore $$ x=-\frac{2}{3}u-v $$

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