Submultiplicavity of spectral norm

Question

Is spectral norm (i.e. the maximum singular value of a matrix) submultiplicative? I am absolutely confused. How to express the singular value of the product of matrices in terms of that of the original matrices? Please help.

Answer 1

Well, the spetral norm is then norm induced by the euclidean norm for vectors. I'm not sure if you are already supposed to know/use that fact, vs the definition using singular values. That means, if $A,B$ are square matrices and $x$ is a compatible vector, then

$$\lVert (AB)x\rVert_2 \le \lVert (AB)\rVert_2 \lVert x\rVert_2 $$

is true for all $A,B,x$ and when fixing $AB$, there is an $x_0\neq \vec0$ where equality holds.

OTOH, we also have

$$\lVert (AB)x\rVert_2 = \lVert A(Bx)\rVert_2\lVert \le \lVert A\rVert_2\lVert (Bx)\rVert_2 \le \lVert A\rVert_2\lVert B\rVert_2\lVert x\rVert_2$$

for all $A,B,x$. Setting $x=x_0$ with the above given $x_0$, we finally have

$$\lVert (AB)\rVert_2 \lVert x_0\rVert_2 = \lVert (AB)x_0\rVert_2 \le \lVert A\rVert_2\lVert B\rVert_2\lVert x_0\rVert_2.$$

If we now divide by $\lVert x_0\rVert_2 \neq 0$, we get the submultiplicativity.

Note that this does not use anything special about the euclidean vector norm $\lVert x\rVert_2$, the proof is valid for any vector norm and the induced matrix norm.

If you want a proof of the fact that the matrix norm induced by the euclidean vector norm is indeed the spectral norm, take a look here: Proving definition of norms induced by vector norms