Author Topic: moe's maths questions (Read 921 times) Tweet Share

moekamo · « **on:** April 29, 2011, 11:55:11 am »

hey guys , just a thread for maths questions that im bound to have

anyway, i need help with this eigenvector/eigenvalue proof:

Suppose the matrix $A$ has eigenvectors $\vec{v}$ with corresponding eigenvalues $\lambda$ . Show that
$\vec{v}$ is an eigenvector of $A^n$ .

This is what i tried so far:

We have $A\vec{v}=\lambda \vec{v}$ , suppose the eigenvalue of $A^n$ is $\alpha$ and the eigenvector is $\vec{w}$ .

Then we have

$\begin{align}<br />\alpha \vec{w} &= A^n \vec{w} \\<br />&=A^{n-1} A \vec{w} \\<br />&=A^{n-1} \lambda \vec{v}<br />\end{align}$

but then i dont know what to do next... how to show $\vec{w}=\vec{v}$ and relate $\lambda$ to $\alpha$ ?

can someone give me a tip to complete it, i dont want the full answer...

latex isnt showing in the preview, i dont know why, sorry about that and thanks for your help

Mao · « **Reply #1 on:** April 29, 2011, 12:20:53 pm »

Try this:

A.v = lambda v

A^2 v = A (lambda v) = lambda^2 v

A^n v = lambda^n v

moekamo · « **Reply #2 on:** October 19, 2011, 11:18:28 pm »

ok, so i have this vector calc proof to do, i need to show that(using tensor notation):

$\tilde{\nabla} (\tilde{u} \cdot \tilde{v}) = (\tilde{u} \cdot \tilde{\nabla} )\tilde{v} + (\tilde{v} \cdot \tilde{\nabla})\tilde{u} + \tilde{u} \times (\tilde{\nabla} \times \tilde{v}) + \tilde{v} \times (\tilde{\nabla} \times \tilde{u})$

this is my attempt so far:

$LHS=\tilde{\nabla} (\tilde{u} \cdot \tilde{v} ) = \partial _i (u_j v_j )$
$=v_j (\partial _i u_j) + u_j (\partial _i v_j )$ using the product rule.

Now heres where im confused, since i have different suffixes in the brackets, is it a cross product? so do i need to introduce the alternating tensor too? i have a feeling that i will need to introduce two of them in each term in my last line, then use the alternating tensor - kronecker delta function relation to get two terms for each, so then i can get the four terms needed in the RHS...

i also tried taking the RHS and simplifying it down to the LHS, but that was a big mess, i had suffixes everywhere and i gave up lol

so yea, a hint or two would be awesome

Mao · « **Reply #3 on:** October 20, 2011, 12:11:34 am »

Expand $\tilde{u}\times (\nabla \times \tilde{v})$ . This should give you $(\tilde{u}\times (\nabla \times \tilde{v}))_i = u_j \partial_i v_j - u_j \partial_j v_i$

Search the forums now!

We have moved!

Author Topic: moe's maths questions (Read 921 times) Tweet Share

moekamo

moe's maths questions

Mao

Re: moe's maths questions

moekamo

Re: moe's maths questions

Mao

Re: moe's maths questions

Recent Posts

User:
Password: