Suppose I have vector $\vec x \in \mathbb R^n$ and matrix $\mathbf M$ of dimension $m\times n$. Is there an alternative expression for $\lVert \mathbf M \cdot \vec x \lVert$ that includes $\lVert \vec x \lVert$?


  1. There's no equality that looks like this, no. I'll give a reason below.

  2. There's an inequality: $$ \| M \cdot x \| \le \|M \| ~ \| x \|. $$ But when you see the definition of $\|M\|$, it's pretty disappointing: $$ \|M \| = \max_{\|x\| = 1} \| M\cdot x \|. $$

What about the first part? Well, look at $$M = \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}$$.

The result of multiplying this by $[1, 0]^t$ has length 1; the result of multiplying it by $[0, 1]^t$ has length zero. And each of these vectors has length one. So you're looking for something that can multiply 1 to get 0 in one case, and can multiply 1 to get 1 in the other. There ain't no such thing. Two dimensions are more complicated than one, and matrix multiplication is more complicated than scalar multiplication.