Hi, everyone! I'm stuck with this problem...

 

Being [latex]T\in L(\mathbb{R}^n)[/latex] a linear operador defined by [latex]T(x_1, ... ,x_n )=(x_1+...+x_n,...,x_1+...+x_n )[/latex], find all eigenvalues and eigenvectors of T.

 

By checking n=1,2,3,4 I guess the answer is:

 

λ=n, x=(1,1,1)

λ=0 (multiplicity n-1), x such as , [latex]\forall k \in \{1,...,(n-1)\}[/latex], [latex]x_k=1[/latex], [latex]x_n=-1[/latex] and [latex]x_i=0[/latex] in all other positions. For instance, for n=4, we have (1,0,0,-1), (0,1,0,-1), (0,0,1,-1).

 

My problem is... how do I prove it for the general case? I'm trying induction, but I think I'm missing something...

 

Thanks in advance!

All the entries in the matrix for T are 1. Does that help?

Hej,

 

Trying out so far it looks good. But it's hard to guess the way your working. It's always good to post some of your calculations so it's easier to help

Well I assume you also use characteristic polynomials.

 

My suggestion would be go directly for an arbitrary n and try prove your claim that way.

You already know what T looks like and so the next step would be to calculate [latex]det(T_n - \lambda E_n)[/latex]

For this Leipniz might be useful. (might be a little work, and maybe you need a small induction there but most of the stuff should annihilate)

 

I hope the resulting polynomial is somewhat 'nice' because we expect only two zeros.

 

Hope that helps you a little bit.

 

take care

yoshi

Edited July 20, 201114 yr by yoshi