Proofing an eigenvector method

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...

Re: Proofing an eigenvector method

Well what are the eigenvalues?

Re: Proofing an eigenvector method

It is supposed to be a general proof, so lambda is the eigenvalues.

Re: Proofing an eigenvector method

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 $\displaystyle \begin{align*} \lambda \mathbf{x} \end{align*}$.

Re: Proofing an eigenvector method

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)

Re: Proofing an eigenvector method

No, choose x to be one of the vectors you have been asked to check, not a general x.

Re: Proofing an eigenvector method

Ok, I have tried that now, but it doesnt seem to add up. I end up with that Ax=[a^2-lambda*a+bc , ac-lambda*c+dc]^T, but lambda*x=[a*lambda-lambda^2 , c*lambda]^T

Re: Proofing an eigenvector method

Well that should be enough to disprove the claim then...