<s>[INST] Matrices $\begin{pmatrix} a &b \\ c &d \end{pmatrix}\begin{pmatrix} x\\y \end{pmatrix}=k\begin{pmatrix}x\\y \end{pmatrix}$ If $\begin{pmatrix}
a &b \\ 
c &d 
\end{pmatrix}\begin{pmatrix}
x\\y \end{pmatrix}=k\begin{pmatrix}x\\y \end{pmatrix}$, prove that $k$ satisfies the equation $k^2-(a+d)k+(ad-bc)=0$.
If the roots of this quadractic equation are $\alpha$ and $\beta$, find the value of $\alpha+\beta$ and $\alpha \beta$ in terms of $a,b,c$ and $d$. Hence, or otherwise, prove that $\begin{pmatrix}
a &b \\ 
c &d 
\end{pmatrix}\begin{pmatrix}
b &b \\ 
\alpha+a &\beta-a 
\end{pmatrix}=\begin{pmatrix}
b &b \\ 
\alpha+a &\beta-a 
\end{pmatrix}\begin{pmatrix}
\alpha &0 \\ 
 0&\beta 
\end{pmatrix}$
Can anyone give me some hints for solving this question? Thanks
 [/INST] we can rewrite the matrix equation as s system of two linear equations. they are $$\begin{align} (a-k)x + by &= 0\\cx + (d-k)y &=0\end{align}$$
multiply the first equation by $(d-k)$, the second one by $b$, and subtracting gives $$\left((a-k)(d-k)-bc\right)x = 0.$$
we have two choices: (a) $x = 0$ or (b) $$(a-k)(d-k)-bc = 0.\tag 1$$
the first choice will lead to $y \neq  0, b = 0, d = k$ which also satisfies $(1).$ 
therefore it is necessary for $k$ to satisfy $(1)$ called the characteristic equation of the matrix $\pmatrix{a&b\\c&d}.$
rewriting $(1)$ as a quadratic in $k,$ we have $$k^2 - (a+d)k + ad - bc = (k - \alpha)(k-\beta) = 0 $$ equating coefficients you get $$\alpha + \beta = a + d, \quad \alpha \beta = ad - bc.  $$
you can verify the last matrix equation by just multiplying it out.
</s>