Numerical Linear Algebra Norms of Matrices

Question

I am reading Holger Wendland's Numerical Linear Algebra and am looking through the exercises of Chapter 2.

Here is my question:

Consider the matrix $A=uv^T$ for $u \in \mathbb{R^m}$ and $v \in \mathbb{R^n}$. Show that $\|A\|_2 = \|u\|_2\|v\|_2$.

I am wondering how to prove this for the case where $m = n = 2$ and for the case where $m = n$.

So far I have been trying to use the definition of the Euclidean norm, that $\|u\|_2 = \sqrt{\sum_{i=1}^n |u_i|^2}$ (same for $v$).

Is it true that $\|uv^T\|_2 = \|u\|_2\|v\|_2$?

I've taken multiple real analysis courses but my memory fails me in this regard. The last one I took was two years ago.

Thank you!

Answer 1

Yes, it's true. Consider this definition of 2-Norm: $\Vert A \Vert_2=\sqrt{\lambda_{max}(A^TA})$, where $\lambda_{max}$ denotes largest eigenvalue of $A^TA$. If both $u$ and $v$ are not zero, then the only non zero eigenvalue of $A^TA$ is $\Vert u\Vert^2_2 \Vert v \Vert_2 ^2$ since $$(A^TA)v^Tu=\Vert u\Vert^2_2 \Vert v \Vert_2 ^2v^Tu.$$ Hence $\Vert A \Vert_2=\sqrt{\Vert u\Vert^2_2 \Vert v \Vert_2 ^2}=\Vert u\Vert_2 \Vert v \Vert_2 $.

Answer 2

$$ \|A\|_2 = \sup_{\|x\|\le 1 } \|Ax\|_2= \sup_{\|x\|\le 1 } \|uv^Tx\|_2 = \sup_{\|x\|\le 1 } |v^Tx| \cdot \|u\|_2 = \|v\|_2 \|u\|_2 $$