0

I have tried to start with mathematical induction, but it does not seem to work. Could you please help me? Thank you in advance.

BigDikEnergy
  • 160

1 Answers1

0

One eigenvalue is $n$, with eigenvector $v_1=(1,1, ...,1)^T$. The other $n-1$ many eigenvalues are all $0$, any vector perpendicular to $v_1$ is an eigenvector.

Yuval
  • 3,439
  • 6
  • 20
  • I think one eigenvalue should be n instead of 1. And could you please give a more detailed clarification? Thank you very much! – BigDikEnergy Nov 10 '19 at 08:34
  • Yes, the first should be $n$. This matrix maps any vector $u$ to $(u\cdot v_1)v_1$, so if an eigenvalue is not $0$ then the associated eigenvector must be $v_1$. – Yuval Nov 10 '19 at 08:37