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

• you left out one more, and important, condition $x^2 + y^2 \neq 0.$ – abel May 30 '15 at 11:41
• $AX=kX\\(A-kI)X=0$ so $k$ is an eigenvalue – Khosrotash May 30 '15 at 11:42
• @darya: Obviously this exercise would never be asked in these words of someone who knows what an eigenvalue is. – Henning Makholm May 30 '15 at 11:55

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.

$Ax=kx$

Roots of the given equations are the eigen-values of $A$.

So sum of roots = trace (A) = a+b

Product of roots = det (A)= ad-bc