# Rank one matrix has a nonzero eigenvalue?

• Nov 24th 2012, 12:06 AM
wesleybrown
Rank one matrix has a nonzero eigenvalue?
How can I show that a rank one matrix has a nonzero eigenvalue?
It seems obvious, but how to prove it?
• Nov 24th 2012, 01:51 AM
girdav
Re: Rank one matrix has a nonzero eigenvalue?
Not so obvious in the case $A:=\begin{pmatrix}0&1\\0&0\end{pmatrix}$.
• Nov 24th 2012, 02:07 AM
wesleybrown
Re: Rank one matrix has a nonzero eigenvalue?
Oops! Then, is it true that not all rank one matrices are diagonalizable?
• Nov 24th 2012, 02:14 AM
girdav
Re: Rank one matrix has a nonzero eigenvalue?
Yes, the $A$ in my previous post showed.
• Nov 24th 2012, 02:19 AM
wesleybrown
Re: Rank one matrix has a nonzero eigenvalue?
If it is rank one and symmetric matrix, how do we show it is diagonalizable?
• Nov 24th 2012, 02:23 AM
girdav
Re: Rank one matrix has a nonzero eigenvalue?
It's not diagonalizable.
• Nov 24th 2012, 02:31 AM
wesleybrown
Re: Rank one matrix has a nonzero eigenvalue?
Maybe my expression is too poor.
I am talking about another matrix now, that is having rank one and it is symmetric, with these two criteria, it should be diagonalizable, and how to prove it?
• Nov 24th 2012, 03:14 AM
girdav
Re: Rank one matrix has a nonzero eigenvalue?
Sorry, it's not your expression but my reading which is poor. A symmetric matrix with real entries is diagonalizable (it's a general result, known as spectral theorem, and doesn't use the fact that the rank is 1), and the involved diagonal matrix has rank 1. So there is in this case a non-zero eigenvalue.
• Sep 4th 2014, 06:40 PM
phys251
Re: Rank one matrix has a nonzero eigenvalue?
Quote:

Originally Posted by girdav
Not so obvious in the case $A:=\begin{pmatrix}0&1\\0&0\end{pmatrix}$.

Both eigenvalues of this matrix are 0.
• Sep 5th 2014, 04:46 AM
HallsofIvy
Re: Rank one matrix has a nonzero eigenvalue?
Yes, that was the point of girdav's example.
• Sep 5th 2014, 08:20 AM
phys251
Re: Rank one matrix has a nonzero eigenvalue?
Quote:

Originally Posted by HallsofIvy
Yes, that was the point of girdav's example.

Ah!