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

Question

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.

Answer 1

Since $\lambda = 6$ is an eigenvalue and $v = [-4 \;| -6]^\mathbf{T}$ is an eigenvector, then , by definition, it must be :

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

Answer 2

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 $$ x=-\frac{2}{3}u-v $$