1. ## eigen vector problem....

Q:Let u and v be the eigen vectors of A corresponding to eigen values 1 and 3 respectively.Prove that u+v is not an eigen vector of A?

2. $Au=u$
$Av=3v$
$A(u+v)=u+3v$
Think

3. Originally Posted by Gamma
$Au=u$
$Av=3v$
$A(u+v)=u+3v$
Think
really very silly problem...

4. Originally Posted by Mathventure
Q:Let u and v be the eigen vectors of A corresponding to eigen values 1 and 3 respectively.Prove that u+v is not an eigen vector of A?
Since (1,u) and (3,v) are eigenpairs, then $Au=u$ and $Av=3v$, where $u,v\neq0$

Suppose that $u+v$ is an eigenvector. Thus $A(u+v)=\lambda(u+v)$ where $u+v\neq0$

But $A(u+v)=Au+Av=u+3v$ and this is equal to $\lambda u+\lambda v$. Comparing both sides, we get $1=\lambda=3\implies 3=1$, which is impossible!

It is now evident that $u+v$ has no eigenvalue, and thus it is not an eigenvector of A.

5. Originally Posted by Mathventure

Q:Let u and v be the eigen vectors of A corresponding to eigen values 1 and 3 respectively.Prove that u+v is not an eigen vector of A?
let's do the general case: suppose $Au=ru, \ Av=sv,$ where $u,v$ are non-zero vectors and $r \neq s$ are scalars. we'll show that there's no scalar $t$ satisfying $A(u+v)=t(u+v).$

so suppose there's such $t.$ then $ru+sv=t(u+v)$ and thus $(r-t)u=(t-s)v.$ call this 1). we also have $r(r-t)u=A(r-t)u=A(t-s)v=s(t-s)v,$ which with 1) gives us:

$r(r-t)u=s(r-t)u,$ and hence $r=t$ because $u \neq 0$ and $r \neq s.$ this with 1) gives us $r=s,$ which is a contradiction!

Note: the above is a trivial thing if you know that distinct eigenvalues correspond to linearly independent eigenvectors.