# Given $A \in \mathcal{M}_n(\mathbb{C}) , \mbox{rank}(A) <n$ and $u\in \text{Ker}(A^*)$ nonzero, show that $\mbox{rank}(A+uu^*)=\mbox{rank}(A)+1$

I got this problem from numerical matrix analysis, I've worked on it for a while but I can't figure out how to solve it.

Given $$A \in \mathcal{M}_n(\mathbb{C}) , \mbox{rank}(A) and $$u\in \text{Ker}(A^*)$$ nonzero, show that $$\mbox{rank}(A+uu^*)=\mbox{rank}(A)+1$$. I've tried a couple things. Let $$r=\text{rank}(A)$$.

1. Thinking I might be able to use the rank nullity theorem, I showed $$\text{Ker}(A+uu^*)=\text{Ker}(A) \cap \text{Ker}(uu^*)=\text{Ker}(A) \cap \text{Span}(u)^\bot$$.
Proof: Let $$(A+uu^*)x=0, \text{then } u^*(A+uu^*)x=u^*uu^*x=\parallel u \parallel^2u^*x=0$$, so we get $$u^*x=0$$ and $$Ax=0$$. From that we have $$\text{Ker}(A+uu^*) \subseteq \text{Ker}(A) \cap \text{Span}(u)^\bot$$ and the other inclusion is obvious.

From that it follows that $$\mbox{rank}(A+uu^*) \geq \mbox{rank}(A)$$, but I haven't found a way to calculate $$\text{dim}(\text{Ker}(A) \cap \text{Span}(u)^\bot)$$, so I cant use the rank nullity theorem as I wanted.

2. I also tried to use the singular value decomposition of $$A$$.
Let $$\{x_1,\ldots,x_n\},\{y_1,\ldots,y_n\}$$ be two orthonormal basis such that if $$U=(x_1,\ldots,x_n), V=(y_1,\ldots,y_n)$$ then $$A=U \Sigma V^*$$ with $$\Sigma=\text{diag}(\mu_1, \ldots , \mu_r, 0 , \ldots , 0)$$ and $$0 < \mu_1 < \ldots < \mu_r$$ the nonzero singular values of $$A$$. Then $$A= \sum_{i=1}^r \mu_i x_iy_i^*$$ and $$A+uu^*=\sum_{i=1}^r \mu_i x_iy_i^*+uu^*$$.
So, if I were able to prove that $$\{x_1,\ldots,x_r,u\},\{y_1,\ldots,y_r,u\}$$ are both linearly independent, completing both sets to form a basis I would be able to solve the problem.

I've managed to show that $$\{x_1,\ldots,x_r,u\}$$ is l.i. (as for $$1 \leq i \leq r,$$ we have $$x_i \in \text{Im}(A)$$, and $$u \in \text{Ker}(A^*)=\text{Im}(A)^\bot$$) but I had no luck with $$\{y_1,\ldots,y_r,u\}$$.

Can anybody give me an idea about how to tackle the problem?

• note that if $\mathbf 0 \neq \mathbf u\in \text{Ker}(A^*)\cap \text{Ker}(A)$ then the statement is actually true; to prove it in such a case, I'd suggest Schur's Unitary Triangularization. – user8675309 Apr 9 at 17:16

## 1 Answer

This is not true. E.g. \begin{aligned} &A=\pmatrix{0&1\\ 0&0},\ u=\pmatrix{0\\ 1}\in\ker(A^\ast)=\ker\pmatrix{0&0\\ 1&0},\\ &\operatorname{rank}(A)=1

• Guess that explains why I had such a hard time proving the result. Should have tried some easy examples. Thanks – Unai Apr 10 at 14:19