Well what are the eigenvalues?
We are asked to prove the following:
Given a 2x2 matrix:
|a b|
|c d|
Prove or disprove that the two eigenvectors are of the form:
[a-lambda,c]^T and [b,d-lambda]^T
I have been messing around for quite some time now, and I cant see how this works out. I have found the better known one, that if b=/=0 then [b,lamba-a] and if c=/=0 then [lambda-d,c], but this one eludes me. I thought that maybe it couldnt be proved, and that it was false, but all the 2x2 matrices that I have tried it on worked, so I am stumped...
That's not what I meant, I meant, what are the eigenvalues in terms of a, b, c, d.
But another way: A vector $\displaystyle \begin{align*} \mathbf{x} \end{align*}$ is an eigenvector of matrix $\displaystyle \begin{align*} \mathbf{A} \end{align*}$ if the transformation from multiplying by $\displaystyle \begin{align*} \mathbf{A} \end{align*}$ is the same as that if $\displaystyle \begin{align*} x \end{align*}$ was multiplied by a scalar value, $\displaystyle \begin{align*} \lambda \end{align*}$.
So $\displaystyle \begin{align*} \mathbf{A}\mathbf{x} = \lambda \mathbf{x} \end{align*}$.
Therefore, evaluate $\displaystyle \begin{align*} \mathbf{A}\mathbf{x}\end{align*}$, where $\displaystyle \begin{align*} \mathbf{A} = \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] \end{align*}$, and $\displaystyle \begin{align*} x \end{align*}$ is one of the given possible eigenvectors. See if you get $\displaystyle \begin{align*} \lambda \mathbf{x} \end{align*}$.
If x= [x,y]^T, then A*X=[ax+by,cx+dy]^T
Lambda*x would be [lambda*x,lambda*y]^T, which suggests hat ax+by=lambda*x, and cx+dy=lambda*y
Is this the line of reasoning you mean? Because if I solve for x on one of them, and y on the other, and then plug in one of them into the other, I get that x=(lambda^2+ad)/(cb-d*lambda+a)