# Math Help - Eigenvalue Problem

1. ## Eigenvalue Problem

A is a simetric metrices nxn. so $v\in R^n$ and $v\neq 0$
so $(\lambda I -A)^2 v=0$ for some $\lambda$

prove that for the same $v$
$(\lambda I -A)v =0$

how i tried to solve it:
i just collected data from the given.
simetric matrices is diagonizable.
$B=(\lambda I -A)$
we were given that $B^2v=0$
so $B^2v \bullet v=0$ (dot product is also v)
so v is orthogonal to $B^2v$

what to do now?

2. ## Re: Eigenvalue Problem

The condition specified in the problem does not say anything about v.
Can you post the question exactly as it was given to you?

3. ## Re: Eigenvalue Problem

Originally Posted by Isomorphism
The condition specified in the problem does not say anything about v.
Can you post the question [COLOR="rgb(0, 0, 0)"]exactly[/COLOR] as it was given to you?
v is some vector in R^n which solves both homogeneus systems
i forgot to put v
in the square system

4. ## Re: Eigenvalue Problem

The solution goes along the lines of:

$(\lambda I - A)^2 v = 0 \implies v^T (\lambda I - A)^2 v = 0 \implies ||(\lambda I - A)v||^2 = 0 \implies (\lambda I - A)v = 0$

5. ## Re: Eigenvalue Problem

cant understand why you add tranposed vector
and going to norm

6. ## Re: Eigenvalue Problem

for any 0-vector (nx1 matrix) u (= 0), and any other row-vector $w^T$ (1xn matrix),

$w^Tu = w^T0 = 0$.

we are given that the vector $(\lambda I - A)^2v = 0$ for some $\lambda$.

so multiplying it by $v^T$ will also give a 0 1x1 matrix, that is, the number 0.

remember, A is symmetric so λI - A is symmetric too: $(\lambda I - A)^T = \lambda I - A$.

thus $0 = v^T(\lambda I - A)^2v = v^T(\lambda I - A)^T(\lambda I - A)v$

$= ((\lambda I - A)v)^T(\lambda I - A)v) =\ <(\lambda I - A)v,(\lambda I - A)v>\ = ||(\lambda I - A)v||^2$.

you don't need to pass to the norm, you can just use the positive-definiteness of the inner product.

7. ## Re: Eigenvalue Problem

There is no answer to 'why' generally. If you did not understand 'how' any of the implications are true, then I can help you.

You can do the same proof in another way. Start with the norm step and then expand it. Is that meaningful?

8. ## Re: Eigenvalue Problem

Originally Posted by transgalactic
A is a simetric metrices nxn. so $v\in R^n$ and $v\neq 0$
so $(\lambda I -A)^2 v=0$ for some $\lambda$

prove that for the same $v$
$(\lambda I -A)v =0$

how i tried to solve it:
i just collected data from the given.
simetric matrices is diagonizable.
$B=(\lambda I -A)$
we were given that $B^2v=0$
so $B^2v \bullet v=0$ (dot product is also v)
so v is orthogonal to $B^2v$

what to do now?
this method is very hard for me.

i asked my prof for a tip,and he said that i should find the diagonolized form of $\lambda I$-A
p is simetric so its diagonisable

D- $\lambda I$= $P^-1( A-\lambda I) P$

and then i should put it in the equation i need to prove
$(\lambda I -A)v =0$

9. ## Re: Eigenvalue Problem

Originally Posted by Deveno
for any 0-vector (nx1 matrix) u (= 0), and any other row-vector $w^T$ (1xn matrix),

$w^Tu = w^T0 = 0$.

we are given that the vector $(\lambda I - A)^2v = 0$ for some $\lambda$.

so multiplying it by $v^T$ will also give a 0 1x1 matrix, that is, the number 0.

remember, A is symmetric so λI - A is symmetric too: $(\lambda I - A)^T = \lambda I - A$.

thus $0 = v^T(\lambda I - A)^2v = v^T(\lambda I - A)^T(\lambda I - A)v$

$= ((\lambda I - A)v)^T(\lambda I - A)v) =\ <(\lambda I - A)v,(\lambda I - A)v>\ = ||(\lambda I - A)v||^2$.

you don't need to pass to the norm, you can just use the positive-definiteness of the inner product.
ok but there is a transformation from zero vector into 1x1 zero vector
its not the same.
??

why
$w^Tu = w^T0 = 0$
????
we are not given that u is zero

then we got to the step of
$||(\lambda I - A)v||^2$
how to prove that for the same v
$(\lambda I - A)v=0$
from the norm
?

10. ## Re: Eigenvalue Problem

in your original post, you said it was given that:

$(\lambda I - A)^2v = 0$.

the whole expression " $(\lambda I - A)^2v$" can be our $u$, and we are going to multiply it by the matrix $v^T$.

then we re-arrange the terms for our matrix product, to see that it is just an inner product of $(\lambda I - A)v$ with itself.

the inner product of any vector with itself is positive...unless, our vector $(\lambda I - A)v$ is the 0-vector,

which is what we hope to prove.

11. ## Re: Eigenvalue Problem

There is another way to proof.

$\vec{v}\neq 0,\;(\lambda I_{n} - A)^2\vec{v}=0\rightarrow \det[(\lambda I_{n} - A)^2]=\det(\lambda I_{n} - A)^2=0\rightarrow \det(\lambda I_{n} - A)=0 \rightarrow \exists\vec{w},\; (\lambda I_{n} - A)\vec{w}=0$

So this method fits for $(\lambda I_{n} - A)^{n}\vec{v}=0$