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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 $\lVert \vec x \lVert$?

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I don't understand what you are asking. – copper.hat Mar 12 '14 at 23:32
Is there an alternative way to express $\lVert\mathbf M\cdot\vec x\lVert$ that somehow includes the value $\lVert\vec x\lVert$? Analogous to how $\lVert k\cdot\vec v\lVert=k\cdot\lVert\vec v\lVert$ for any scalar $k$ and vector $\vec v$. – Disousa Mar 12 '14 at 23:36
What do you mean by "includes"? – Sanath K. Devalapurkar Mar 12 '14 at 23:36
I mean the value $\lVert\vec x\lVert$ is included on the expression. For example, if the expression were $\lVert\mathbf M\cdot\vec x\lVert=\lVert\mathbf M\lVert\cdot\lVert\vec x\lVert$ (which is not true, but gives an idea for the kind of expression I'm looking for). – Disousa Mar 12 '14 at 23:39
Well, yes, there is, but probably this isn't what you look for:) $$\|Mx\|=\|Mx\|+\|x\|-\|x\|\,.$$ – Berci Mar 12 '14 at 23:39
up vote 7 down vote accepted
  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.

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What is that inequality called? – M. Alaggan Dec 19 '15 at 17:50
It doesn't have a name that I know's a straightforward consequence of the definition of the matrix norm. – John Hughes Dec 19 '15 at 17:54

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