Eigenvalues of a rank 1 matrix?

[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?

 

[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.

 

[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 $A = xy^T$.

 

[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.

 

[blue_raver22]

The eigenvalues of a rank 1 matrix are indeed 0 and the trace of the matrix. This can be seen by considering the definition of a rank 1 matrix, which is a matrix that can be written as the outer product of two vectors. Let's call these vectors u and v, then the rank 1 matrix can be written as A = uv^T.

Now, let's consider the eigenvalue equation Av = λv. Substituting A = uv^T, we get (uv^T)v = λv. This can be simplified to u(v^Tv) = λv. Since v is an eigenvector, it cannot be the zero vector, so v^Tv is a non-zero scalar. This means that u must be a multiple of v, and we can write u = cv for some scalar c.

Substituting this back into the original equation, we get A = cvv^T = c(vv^T). This means that A is a scalar multiple of the outer product of v with itself, which is the definition of a rank 1 matrix.

Now, the trace of A is given by tr(A) = tr(c(vv^T)) = ctr(vv^T). Since v is an eigenvector, we know that v^Tv = 1, so tr(vv^T) = 1. Therefore, tr(A) = c.

This shows that the trace of A is indeed one of the eigenvalues, with the corresponding eigenvector being v. The other eigenvalue is 0, with the corresponding eigenvector being any vector orthogonal to v.

As for other proofs, there are various approaches that can be taken. One approach is to use the fact that the eigenvectors of a rank 1 matrix are orthogonal to each other, and thus the matrix can be diagonalized. Another approach is to use the singular value decomposition of the matrix. However, ultimately, all of these approaches will lead to the same result of the eigenvalues being 0 and the trace of the matrix.