0
$\begingroup$

Let A be a square matrix, show that any non-zero linear combination of two eigenvectors $v$ and $w$, corresponding to the same eigenvalue, is also an eigenvector.

First I'll show what I did..

1) Let $V$ and $W$ be eigenvectors of A with corresponding eigenvalue $\lambda$

2) If we use the characteristic equation for both eigenvectors, we get...

$Av = \lambda v$
$Aw = \lambda w$

3) Now if we use $v$ and $w$ as vectors for a linear combination with the eigenvalue $\lambda$ we get

$\lambda v + \lambda w = x$

4) now we just have to prove that $x$ is an eigenvector with corresponding eigen value $\lambda$

this is where im stuck.. i need someone to give me a useful hint on what to do next..

did i even take the proper steps here?

any help will be appreciated

$\endgroup$
  • $\begingroup$ the matrix multiplication is linear, thus for any vectors $v,w$ and any scalar $c$ we have that $A(cv+w)=cAv+Aw$. Apply the linearity of $A$ to your case and, voilá! $\endgroup$ – Masacroso Jan 2 '18 at 20:31
3
$\begingroup$

First remember that $A$ is linear, so we have:

$$A(av+bw) = A(av)+A(bw) = aAv+bAw =a(\lambda v) +b(\lambda w)= \lambda (av+bw)$$

so if $v$ and $w$ are eigenvectors with eigenvalue $ \lambda $ then $av+bw$ is also eigenvector with eigenvalue $\lambda $.

$\endgroup$
  • $\begingroup$ Thank you very much, i understand now! $\endgroup$ – Soon_to_be_code_master Jan 2 '18 at 20:50
1
$\begingroup$

Note that

$$A( v + w)= \lambda v + \lambda w=\lambda(v+w)$$

thus

$$A( av + bw)= \lambda av + \lambda bw=\lambda(av+bw)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.