$2$-norm of a rank-$1$ matrix

Question

I want to prove that $\|A\|_2 = \|x\|_2\|y\|_2$ given that $A = xy^T$ is a rank one matrix. This is my incomplete attempt so far, I get stuck when I need to take into account the spectral radius of the symmetric matrix: \begin{align} \|A\|_2 &= \sqrt{\rho(A^TA)} \\ & = \sqrt{\rho(xy^Tyx^T)} \\ & = \sqrt{\rho((y^Ty)xx^T)} \\ & = \sqrt{\rho(\|y\|_2^2xx^T)} \\ & \qquad\quad \vdots\\ & = \sqrt{x^Tx } \cdot\sqrt{y^Ty}\\ & = \|x\|_2\|y\|_2 \end{align}

I know $xx^T$ is a symmetric matrix and its got a scalar $\|y\|_2^2$ out in front of it. Wondering if there's some property of a rank one matrix here that I could use, or perhaps my knowledge on spectral radii is limited and that is why I cannot get to the next step? I tried to look up the spectral radius/largest eigenvalue for a symmetric matrix but what I found was beyond the scope of what I am currently studying. Not sure where to go from here.

Hint: show $\rho(\alpha A) = |\alpha| \rho(A)$ and $\rho(xx^T)=x^Tx$ — Hyperplane, Feb 14 '21 at 22:29
Note that $A^\top A$ is either rank-$0$ or rank-$1$. In either case, its spectral radius is equal to its trace. Exploit the cyclic property of the trace. — Rodrigo de Azevedo, Feb 15 '21 at 06:27
I am confused, when you multiply vectors don't you get a scalar? How do you get a matrix? With A=xyT — Ford F150 Gaming, Oct 22 '24 at 08:42

score 3 · Accepted Answer · edited Feb 15 '21 at 06:16

3

Hint: The spectral radius of an operator is by definition the magnitude of its largest eigemvalue. What can you say about the eigenvalues of a rank-one matrix?

edited Feb 15 '21 at 06:16

Rodrigo de Azevedo

23,223

answered Feb 14 '21 at 22:26

Semiclassical

18,592

From what I know, the eigenvalues of a rank one matrix is either $0$, or of the form $y^Tx$ (which is the trace). – User_13 Feb 14 '21 at 22:32
@User_13 Correct. Now apply that insight to $\rho(|y|_2^2 xx^T).$ – Semiclassical Feb 14 '21 at 22:38
Thanks, coupled with Hyperplane's comment I really see how obvious this was. Appreciate you giving me that nudge. – User_13 Feb 14 '21 at 23:11

score 1 · Answer 2 · 2021-02-14T22:57:41.040

Here is another approach for your own entertainment and understanding.

If $v\in \mathbb{R}^n$ is arbitrary, then we have from Cauchy Schwartz that $$\|Av\|_2=\big|y^Tv\big|\|x\|_2\leq\Big(\|x\|_2\|y\|_2\Big)\|v\|_2$$ So for any $v \neq 0$ $$\frac{\|Av\|_2}{\|v\|_2}\leq \|x\|_2\|y\|_2$$ This shows $\|A\|_2\leq \|x\|_2\|y\|_2$. We get equality by noting that this upper bound is attained whenever we take $v\in \text{span}(y)-\{0\}$.

$2$-norm of a rank-$1$ matrix

2 Answers2