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 $A \in \mathcal{M}_n(\mathbb{C}) , \mbox{rank}(A) <n$ 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?

  • 1
    $\begingroup$ 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. $\endgroup$ – user8675309 Apr 9 at 17:16

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<n=2,\\ &\operatorname{rank}(A+uu^\ast)=\operatorname{rank}\pmatrix{0&1\\ 0&1}=1=\operatorname{rank}(A). \end{aligned}

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

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