If $A = u*u^T$ where $0 \neq u \in \mathbb{R}^n$ , then find the eigenvalues of $A$ and show that $A$ is diagonalizable.

Question

If $A = u\cdot u^T$ where $0 \neq u \in \mathbb{R}^n$ , then find the eigenvalues of $A$ and show that $A$ is diagonalizable.

I mean how to find eigenvalues of $A$?

score 2 · Answer 1 · edited May 03 '19 at 14:36

2

Hints:

$u$ is an eigenvector of $A$, with which eigenvalue?
$A$ has rank $1$, so what is the nullity?

edited May 03 '19 at 14:36

J. W. Tanner

60,406

answered May 03 '19 at 14:24

Arthur

199,419

score 0 · Answer 2 · answered May 03 '19 at 14:37

Hint: Rank of $A$ is one, so zero is one of the eigenvalue. Also the dimension of the eigenspace corresponding to zero is $n-1$. Now recall that $$1\leq \text{dim}E_0 \leq m$$ where $m$ is the multiplicity of zero and $E_0$ is the eigenspsce corresponding to zero. So $m$ is either $n-1$ or $n$. Actually $m$ cannot be the later, since otherwise trace is zero. But we don't know what trace is. So in summary, there are two eigenvalues, one is zero with multiplicity $n-1$ and the other one is trace!

If $A = u*u^T$ where $0 \neq u \in \mathbb{R}^n$ , then find the eigenvalues of $A$ and show that $A$ is diagonalizable.

2 Answers2