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

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.

• By definition, what does it mean for $6$ to be an eigenvalue? – Shubham Johri Dec 14 '18 at 22:11
• the 6 is the eigenvalue of the eigenvector (-4,6) – Nelson Aguilera Dec 14 '18 at 22:12
• Use the definition of $\lambda$ i.e. $Av=\lambda v$. You are given $\lambda$ and $v$. – Yadati Kiran Dec 14 '18 at 22:13

Since $$\lambda = 6$$ is an eigenvalue and $$v = [-4 \;| -6]^\mathbf{T}$$ is an eigenvector, then , by definition, it must be :
$$Av = \lambda v \implies \begin{pmatrix}6 & a \\ 5 & b \end{pmatrix}\begin{pmatrix} -4\\-6\end{pmatrix}= 6\begin{pmatrix} -4\\-6\end{pmatrix} \implies \begin{cases} -24 -6a = -24 \\ -20 - 6b = -36 \end{cases}$$
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 $$x=-\frac{2}{3}u-v$$