# 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. Apr 9, 2021 at 17:16

## 3 Answers

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
– UCL
Apr 10, 2021 at 14:19

As shown in the other answer, the statement is false for general matrix, but if $$A$$ is normal so that its rank is equal to the number of nonzero eigenvalues, we can use the following result about rank-one perturbation to show that the statement holds:

Proposition. Let $$x, y \in \mathbb{C}^{n}$$, where $$x \neq 0$$, and let $$A$$ be an $$n$$-by-$$n$$ complex matrix. Suppose that $$A$$ has eigenvalues $$\lambda, \lambda_2, \dots, \lambda_{n}$$, where $$\lambda$$ is an eigenvalue associated with $$x$$. Then the eigenvalues of the rank-one update $$A + x y^{*}$$ are $$\lambda + y^{*} x, \lambda_2, \dots, \lambda_{n}$$.

Back to the question, $$(0, u)$$ is an eigenpair of $$A^{*}$$. By the proposition, $$u^{*} u$$ replaces $$0$$ as an eigenvalue of $$A^{*} + u^{*} u$$, whence $$\mathrm{rank}(A^{*} + u u^{*}) = \mathrm{rank}(A^{*}) + 1$$.

This property turned out to be false, but I did manage to prove it for a case a bit more general than Yez's answer. Didn't think it would be of much interest but I guess we lose nothing by leaving it here.

Fact 1:
Given $$A\in \mathcal{M}_n(K)$$ for any field $$K$$, it holds that $$\text{Ker}(A)\oplus \text{Im}(A)=K^n$$ if and only if $$\text{Ker}(A^2)=\text{Ker}(A)$$.

Fact 2:
Given $$A\in \mathcal{M}_n(\mathbb{C})$$, we have $$\text{Ker}(A^*)=\text{Im}(A)^\bot$$

Fact 3:
Given $$A \in \mathcal{M}_n(\mathbb{C}) , \mbox{rank}(A) and $$u\in \text{Ker}(A^*)$$ nonzero we have that $$\text{Ker}(A+uu^*)=\text{Ker}(A) \cap \text{Span}(u)^\bot$$

Now finally, given $$A \in \mathcal{M}_n(\mathbb{C}) , \mbox{rank}(A) and $$u\in \text{Ker}(A^*)$$ nonzero, if we further assume that $$\text{Ker}(A^2)=\text{Ker}(A)$$ (all diagonalizable matrices verify this, and therefore normal matrices will verify it too) then it holds that $$\mbox{rank}(A+uu^*)=\mbox{rank}(A)+1$$
Proof:
We have $$u\in\text{Ker}(A^*)\Rightarrow \text{Span}(v)\subset \text{Ker}(A^*) = \text{Im}(A)^\bot \Rightarrow \text{Im}(A) \subset \text{Span}(v)^\bot$$
Secondly, $$\text{Ker}(A)+\text{Span}(v)^\bot\supset \text{Ker}(A)+\text{Im}(A)\overset{\text{Fact 1}}{=}\mathbb{C}^n \Rightarrow \text{Ker}(A)+\text{Span}(v)^\bot=\mathbb{C}^n$$

Now considering the dimension of the last set we get $$n=\dim (\text{Ker}(A)+\text{Span}(v)^\bot)= \dim \text{Ker}(A)+\dim \text{Span}(v)^\bot - \dim (\text{Ker}(A) \cap \text{Span}(v)^\bot) \Rightarrow$$

$$n= \dim \text{Ker}(A)+n-1 - \dim \text{Ker}(A+uu^*) \Rightarrow$$

$$\dim \text{Ker}(A+uu^*)=\dim \text{Ker}(A)-1\Rightarrow \mbox{rank}(A+uu^*)=\mbox{rank}(A)+1$$ where the last implication is due to the rank-nullity theorem.