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!
