# Hard eigenvector/eigenvalue proof

• Nov 25th 2010, 12:44 PM
chr91
Hard eigenvector/eigenvalue proof
Attachment 19859

Struggling with this:

I started by saying A is invertible if and only if Ax = 0 has a trivial solution.

So there exists x = 0 where Ax = 0

0 = lamda. 0 = 0

I'm not quite sure this proves that lamda is not equal to zero though?!
• Nov 25th 2010, 01:54 PM
dwsmith
$\displaystyle Ax=\lambda x$

Multiple both sides by A inverse

$\displaystyle A^{-1}Ax=A^{-1}\lambda x\rightarrow x=A^{-1}\lambda x\rightarrow \frac{1}{\lambda}x=A^{-1}x$
• Nov 26th 2010, 04:52 AM
HallsofIvy
Quote:

Originally Posted by chr91
Attachment 19859

Struggling with this:

I started by saying A is invertible if and only if Ax = 0 has a trivial solution.

You've pretty much got everything backwards here. Ax= 0 for x= 0 for any A. What is true is that A is invertible if and only if there is NO non-trivial solution. That is if x= 0 is the only x such that Ax= 0.

Saying that $\lambda$ is an eigenvalue of A means there exist non-zero x such that $Ax= \lambda x$. Use the fact that A is invertible to multiply both sides by $A^{-1}$.

Quote:

So there exists x = 0 where Ax = 0

0 = lamda. 0 = 0

I'm not quite sure this proves that lamda is not equal to zero though?!
• Nov 26th 2010, 07:08 AM
Ackbeet
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