# diagonal times rank 1 matrix still rank 1?

First let me make two statements to give my question the proper context.

1. Consider $D$ a diagonal matrix and $u v^T$ a rank 1 matrix. From my current knowledge there exist computationally cheap algorithms to diagonalize $D+uv^T$ as discussed here: maximum eigenvalue of a diagonal plus rank-one matrix

2. Diagonalizing a matrix of the form $Duv^T$ is straightforward, see here: Eigenvector of a diagonal matrix times a rank-1 matrix

Now to my question(s):

1. I want to diagonalize $D^2+Duv^T$. Is $Duv^T$ still a rank 1 matrix? I considered some examples but I am not really sure if this is a general statement.

2. This question depends on the answer of the first. If the statement is true the algorithms in statement 1 should be applicable to diagonalization of $D^2+Duv^T$, no? If it is not true are there others ways to tackle this?

Let $u=v=(1,0)^T$, and

$$D=\left(\begin{matrix}0&0\\0&1\end{matrix}\right)$$

Then

$$uv^T=\left(\begin{matrix}1&0\\0&0\end{matrix}\right)$$

And $Duv^T=0$.

Of course, if $D$ has only nonzero diagonal elements, then it has full rank and the product has rank $1$, see matrix rank properties on Wikipedia. And if $Duv^T$ has not rank $1$, it's necessarily the null matrix, so it's not a very interesting case.

• Ok thanks, so its actually very easy to write a proof from these properties... let D be of full rank ($rank(D)=n$) and use sylvester rank inequality: $1 = rank(uv^T) = rank(D)+rank(uv^T)-n \le rank(Duv^T) \le min(rank(D),rank(uv^T)) =1$ – J. Goe Apr 6 '16 at 14:02