# Proofing an eigenvector method

• May 15th 2014, 03:40 AM
Latsabb
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...
• May 15th 2014, 03:57 AM
Prove It
Re: Proofing an eigenvector method
Well what are the eigenvalues?
• May 15th 2014, 03:58 AM
Latsabb
Re: Proofing an eigenvector method
It is supposed to be a general proof, so lambda is the eigenvalues.
• May 15th 2014, 04:08 AM
Prove It
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*}.
• May 15th 2014, 04:26 AM
Latsabb
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)
• May 15th 2014, 04:54 AM
Prove It
Re: Proofing an eigenvector method
No, choose x to be one of the vectors you have been asked to check, not a general x.
• May 15th 2014, 09:16 AM
Latsabb
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
• May 16th 2014, 08:26 PM
Prove It
Re: Proofing an eigenvector method
Well that should be enough to disprove the claim then...