# Find the Eigenvectors of $\begin{pmatrix}1&-P\\P &-Q\end{pmatrix}$

I want to find the eigenvectors of this matrix:

$\begin{pmatrix}1&-P\\P &-Q\end{pmatrix}$

First I found the eigenvalues:

$det(A-\lambda_1)$

$det\begin{pmatrix}1-\lambda&-P\\p&-Q-\lambda\end{pmatrix}=(1-\lambda)(-Q-\lambda)+P^2=-Q-\lambda+Q\lambda+\lambda^2+P^2-Q$

$=\lambda^2+(Q-1)\lambda+P^2-Q$

$\implies \lambda_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(Q-1)\pm \sqrt{(Q-1)^2-4P^2+4Q}}{2}$

I know that I can find the eigenvectors by solving:

$(A-\lambda_k I)x=0$

but I am just getting wrong results. Wolfram Alpha says that the matrix has the following eigenvectors: LINK

Could someone show me how I can calculate the eigenvectors?

• You find eigenvectors by solving $$(A - \lambda_k I)x = 0$$ which may be what you meant, but isn't what you typed. – Ben Grossmann Nov 18 '15 at 19:25
• Yeah I meant that. I was being sloppy. Sorry. – bluemoon Nov 18 '15 at 19:28
• As far as I can see, your eigenvalue solutions are now in accordance with the ones calculated by WolframAlpha. So we are one step closer. :-) You then end up with the two systems $(A-\lambda_1 I)x_1 = 0$ and $(A-\lambda_2 I) x_2 = 0$ for the eigenvectors $x_1$ and $x_2$ which can be solved by Gauss elimination. Because of the expression for the $\lambda$ being unwieldy this is more a challenge about making no mistake along the road than determining the solution $x_i$ for a homogenous linear system $(A - \lambda_i I) x_i = 0$. – mvw Nov 18 '15 at 22:01
• @mvw Sorry for the late reply. I tried to solve it today and I always end up with the following equation $$\begin{pmatrix}1-\lambda_1 &-P \\P&-Q-\lambda_1\end{pmatrix}\longrightarrow \begin{pmatrix}1-\lambda_1&-P \\0&0\end{pmatrix} \\ \implies (1-\lambda_1)x_1=Px_2 \iff x_1=\frac{P}{1-\lambda_1}x_2 \implies \text{eigenvector:}\begin{pmatrix}\frac{P}{1-\lambda_1}\\1\end{pmatrix}$$ But if I plug in my eigenvalue $\lambda_1$ I get exactly the reciprocal of what wolfram alpha says $x_1$ should be. I don't know where I am making a mistake. – bluemoon Nov 19 '15 at 16:57
• @ bluemoon, are $P$ and $Q$ matrices here? if yes, then can we calculate det$(A-\lambda I)$ in the way you did? – rationalize Nov 20 '15 at 5:51

Suppose that $\lambda \neq 1$ is an eigenvalue. Then $$\pmatrix{1 - \lambda & -P\\ P & -Q - \lambda}$$ is singular, so that the top and bottom rows are multiples. Verify that the vector $$v = \pmatrix{P \\ 1 - \lambda}$$ must be an eigenvector associated with $\lambda$.

• I don't see how that is an eigenvector because: $$\begin{pmatrix}1-\lambda&-P\\P&-Q-\lambda\end{pmatrix}\begin{pmatrix} P \\1-(\lambda) \end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix} \\ \iff \begin{pmatrix}0 \\ P^2+\lambda^2+(Q-1)\lambda-Q \end{pmatrix}\not=\begin{pmatrix}0\\0\end{pmatrix}$$ – bluemoon Nov 18 '15 at 19:31
• Why are you so sure that $P^2+\lambda^2+(Q-1)\lambda-Q \neq 0$? Remember, $\lambda$ is an eigenvalue. The top and bottom row are multiples. – Ben Grossmann Nov 18 '15 at 19:34
• I am sorry if this is elementary linear algebra but why is it important that the top and bottom row are multiples? Is $$P^2+\lambda^2+(Q-1)\lambda-Q$$ the characteristic polynomial? Is the polynomial always zero? – bluemoon Nov 18 '15 at 19:39
• Write that the entries of the top row are $a$ and $b$. Write that the entries of the eigenvector are $x_1$ and $x_2$. Then we know that $ax_1 + bx_2 = 0$. Now, there is a $k$ such that the entries of the bottom row are $ka$ and $kb$. We may then state that $$(ka)x_1 + (kb)x_2 = k(ax_1 + bx_2) = k0 = 0$$ So, $(x_1,x_2)$ must be mapped to $(0,0)$. – Ben Grossmann Nov 18 '15 at 19:43
• And yes: the definition of an eigenvalue tells us that if you put $\lambda$ in to the characteristic polynomial, you should get $0$. – Ben Grossmann Nov 18 '15 at 19:44

I believe this part went wrong: $$-Q-\lambda+Q\lambda+\lambda^2+P^2 = \lambda^2+(Q-1) \lambda + P^2 - Q \ne \lambda^2+(1-Q)\lambda+P^2 - Q$$

• Thanks. I edited my opening post. It was just a mistake while copying the question from my notes. I am still struggling to find the eigenvectors. Maybe you can give me a hint. – bluemoon Nov 18 '15 at 19:19
• I edited my answer. It is still different. – mvw Nov 18 '15 at 19:20
• I fixed the $Q-1$ part too. – bluemoon Nov 18 '15 at 19:24
• So you got your problem solved? – mvw Nov 18 '15 at 20:28
• No, not yet. I am still talking to Omnomnomnom in the comment section but there are several things I don't quite understand yet. I kind of have more questions than before. :D – bluemoon Nov 18 '15 at 20:31