Help me prove something on eigenvectors?

[Rijad Hadzic]

Homework Statement

Prove that if the eigenvalues of a matrix A are $\lambda_1 ... \lambda_n$ with corresponding eigenvectors $x_1...x_n$ then $\lambda^m_1...\lambda^m_n$ are eigenvalues of A^m with corresponding eigenvectors x_1...x_n

Homework Equations

$Ax= \lambda x$

The Attempt at a Solution

So I start with

$Ax= \lambda x$

I think I am trying to prove

$A^mx= \lambda^m x$

correct?

If so I proceed:

$A^{m-1}Ax= \lambda^m x$

$A^{m-1}\lambda x= \lambda^m x$

and basically this will continue... but I'm not sure how to write this out to get it to

$\lambda^m x= \lambda^m x$

?

I don't get how I'm going to be able to set

$A^{m-1}$ = to $\lambda^m$

 

[andrewkirk]

Do you know how to use mathematical induction?

If so, prove the statement
$$P(m):\quad A^m x=\lambda^m x$$
by induction on the power $m$.

$P(m)$ is trivially true for $m\in\{0,1\}$. So what you need to do is prove that if it is true for $m=k$, with $k\geq 0$, then it is true for $m=k+1$.

 

[Rijad Hadzic]

Ok guys I'm stupid. I actually plugged numbers into the exponents and its starting to make sense now. As A goes down, lambda goes up, and eventually it comes to lambda = lambda

 

[Rijad Hadzic]

Do you know how to use mathematical induction?

If so, prove the statement
$$P(m):\quad A^m x=\lambda^m x$$
by induction on the power $m$.

$P(m)$ is trivially true for $m\in\{0,1\}$. So what you need to do is prove that if it is true for $m=k$, with $k\geq 0$, then it is true for $m=k+1$.

I do not know how to use mathematical induction. Is it common to learn this before your first Linear Algebra course? I will read the wiki article right now though.

 

[fresh_42]

You cannot start with $A^m x_i = \lambda^m_i x_i$ and proceed from there, if this is what you want to show. It must be your last line, not the first. So either you prove it by induction or you go with $A^m x_i = A^{m-1}(Ax_i) = A^{m-1} (\lambda_i x_i) = \ldots = \lambda_i^m x_i$ and use properties of matrix multiplication. Since $m$ is a fixed finite number, an argument "and so on" is equally good.

 

[andrewkirk]

I do not know how to use mathematical induction. Is it common to learn this before your first Linear Algebra course? I will read the wiki article right now though.

Yes. Where I come from, mathematical induction is taught in high school, whereas linear algebra is not taught until uni, or in the most advanced late high school classes. Mathematical induction is one of the most commonly used types of proofs there is.

 

[Rijad Hadzic]

You cannot start with $A^m x_i = \lambda^m_i x_i$ and proceed from there, if this is what you want to show. It must be your last line, not the first. So either you prove it by induction or you go with $A^m x_i = A^{m-1}(Ax_i) = A^{m-1} (\lambda_i x_i) = \ldots = \lambda_i^m x_i$ and use properties of matrix multiplication. Since $m$ is a fixed finite number, an argument "and so on" is equally good.

That makes sense. It must be the last line since that is what we are trying to prove.

Thanks for all the repsonses guys. I will continue my studies.

 

[Mark44]

I do not know how to use mathematical induction.

There are three parts in proving that $A^mx = \lambda^m x$ using induction.
1) Show that this statement $Ax = \lambda x$ is true, which is trivially true here.
2) Assume that $A^kx = \lambda^k x$
3) Show that $A^kx = \lambda^k x$ being true implies that $A^{k + 1}x = \lambda^{k + 1} x$ is also true. In other words, use the statement in 2) to show that the statement in 3) must also be true.