Matrix Multiplication Norm I'm trying to prove a proposed theorem for my thesis and was wondering if the following property is true and can be used in my proof.
$ \bf {s ^T M sgn (s)} < \parallel \bf s \parallel \parallel M \parallel $
where the function $\bf {sgn (s)}$ is the signum (sign) function, the dimensions of the vector $\bf s$ and the matrix $\bf M$ is such that the product $ \bf {s ^T M sgn (s)}$ is a scalar and the norm operator represent the euclidian norm.
 A: Suppose that we define $\mathrm{sign}(s)$ elementwise, as follows:
$$
s = \begin{pmatrix} s_1 \\ \vdots \\ s_n \end{pmatrix} \Rightarrow
\mathrm{sign}(s) = \begin{pmatrix} \mathrm{sign}(s_1) \\ \vdots \\ \mathrm{sign}(s_n)\end{pmatrix}
$$
where we denote
$$
\mathrm{sign}(z) = \begin{cases}
  1, & z \geq 0 \\
  -1, & z < 0
\end{cases}
$$
For that case, notice that
$$
s^\top M \mathrm{sign}(s) = \mathrm{trace}(s^\top M \mathrm{sign}(s))
= \mathrm{trace}(\mathrm{sign}(s)s^\top M) \quad (\text{cyclic invariance of trace}) \\
\leq
\sqrt{\mathrm{trace}\left((\mathrm{sign}(s)s^\top)^\top \mathrm{sign}(s)s^\top\right)} \cdot
\sqrt{\mathrm{trace}(M^\top M)} \quad (\text{Cauchy-Schwarz})
$$
Now, notice that the latter quantity is exactly $\left\| M \right\|_2$, and
that
$$
\mathrm{trace}\left( s \cdot \mathrm{sign}^\top(s) \mathrm{sign}(s) s^\top) \right) = n s^\top s = n \left\| s \right\|^2
$$
since the inner product of the sign vectors is
$$
\sum_{i=1}^n [\mathrm{sign}(s_i)]^2 = n \cdot 1
$$
The above imply that the following inequality holds:
$$
s^\top M \mathrm{sign}(s) \leq
\sqrt{n} \left\| s \right\| \left\| M \right\|
$$
Therefore, you can prove your claim if you augment $\mathrm{sign}(s)$ with an appropriate normalization constant.
