# Eigenvalues of a rank 1 matrix?

1. Mar 31, 2013

### brownman

How come a square matrix has eigenvalues of 0 and the trace of the matrix?
Is there any other proof other than just solving det(A-λI)=0?

2. Apr 1, 2013

### JustMeDK

I might argue something like the following: By row operations, a rank 1 matrix may be reduced to a matrix with only the first row being nonzero. The eigenvectors of such a matrix may be chosen to be the ordinary Euclidian basis, in which the eigenvalues become zero's and the 11-component of this reduced matrix. As row operations are invertible, the trace is unchanged, and thus this nonzero eigenvalue equals the trace of the original matrix.

Afterthought: But that is probably erroneous, because even though the row operations are indeed invertible, they do not generally preserve the trace. So the last part of my argument fails.

A better argument seems to be the following: For a rank k matrix there exists a basis in which k of its columns are nonzero, the other ones being zero. The transformation between bases may be chosen to be orthogonal, thus preserving the trace.

Last edited: Apr 1, 2013
3. Apr 1, 2013

### jbunniii

We assume $A$ is an $n \times n$ rank one matrix. If $n > 1$, any rank one matrix is singular. Therefore $\lambda = 0$ is an eigenvalue: for an eigenvector, just take any nonzero $v$ such that $Av = 0$.

So let's see if there are any nonzero eigenvalues.

If $A$ is a rank one matrix, then all of its columns are scalar multiples of each other. Thus we may write $A = xy^T$ where $x$ and $y$ are nonzero $n \times 1$ vectors.

If $\lambda$ is an eigenvalue of $A$, then there is a nonzero vector $v$ such that $Av = \lambda v$. This means that $(xy^T)v = \lambda v$. By associativity, we may rewrite the left hand side as $x(y^T v) = \lambda v$.

Note that $y^T v$ is a scalar, and of course $\lambda$ is also a scalar. If we assume $\lambda \neq 0$, then this means that $v$ is a scalar multiple of $x$: specifically, $v = x(y^T v)/\lambda$.

Therefore $x$ itself is an eigenvector associated with $\lambda$, so we have $x(y^T x) = \lambda x$, or equivalently, $x(\lambda - y^T x) = 0$. As $x$ is nonzero, this forces $\lambda = y^T x$.

All that remains is to recognize that $y^T x = \sum_{n = 1}^{N} x_n y_n$ is the trace of $A = xy^T$.

4. Apr 1, 2013

### jbunniii

By the way, note that this does not necessarily mean that $A$ has two distinct eigenvalues. The trace may well be zero, for example
$$A = \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}$$
is a rank one matrix whose only eigenvalue is 0.