Linear Algebra - Finding Matrix from Eigenspaces

Question

I tried equating -1*{1,-5} = {1,-5}A = {-1,5} and -4{-1,6} = {-1,6}*A = {4,-24} and think I'm on the right track but don't know what to do next...

Answer 1

If all else fails, and you can't think of anything cleverer, you can fall back on brute force. Let $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\;.$$ You know that $$\begin{bmatrix}-1\\5\end{bmatrix}=-\begin{bmatrix}1\\-5\end{bmatrix}=A\begin{bmatrix}1\\-5\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1\\-5\end{bmatrix}=\begin{bmatrix}a-5b\\c-5d\end{bmatrix}$$ and $$\begin{bmatrix}4\\-24\end{bmatrix}=-4\begin{bmatrix}-1\\6\end{bmatrix}=A\begin{bmatrix}-1\\6\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}-1\\6\end{bmatrix}=\begin{bmatrix}-a+6b\\-c+6d\end{bmatrix}\;.$$

This gives you two systems of linear equations in two unknowns each:

$$\left\{\begin{align*}a-5b&=-1\\-a+6b&=4\end{align*}\right.$$ and

$$\left\{\begin{align*}c-5d&=5\\-c+6d&=-24\;.\end{align*}\right.$$

These are readily solved for $a,b,c$, and $d$.

Added: For example, if you add the two equations of the first system to each other, you get $(a-5b)+(-a+6b)=-1+4$, or $b=3$; substituting that into the first equation gives you $a-15=-1$, so $a=14$. For safety's sake you can check that the second equation is satisfied: $-14+6\cdot3=-14+18=4$.