I do not understand why the following property for Matrix subordinate norms holds: \begin{equation} \|AB\| \leq \|A\|\|B\| \end{equation}
Please explain clearly as I know that it should be shown by the definition: \begin{equation} \|A\| = \sup\{\|A\|: u \in \mathbb{R}^n, \|u\|=1\} \end{equation}
And the fact that: \begin{equation} \|Ax\| \leq \|A\|\|x\| \end{equation}
I understand the definition and I understand the property. I just can't seem to put them together correctly.
Hints only please!
EDIT Is the following reasoning correct:
Let $u$ be any unit vector: \begin{align} \|ABu\| \leq& \dfrac{\|ABx\|}{\|x\|}\\ \leq& \frac{\|A\|\|Bx\|}{\|x\|} \quad{\text{as $Bx$ is a vector}}\\ \leq& \frac{\|A\|\|B\|\|x\|}{\|x\|} \quad{\text{from above property}}\\ \leq& \|A\|\|B\| \end{align}
Since $u$ is any unit vector then we have: \begin{equation} \|A\|\|B\| \geq \|AB\| \end{equation}
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$\begingroup$Note, the answer below was given in response to the original form of this question which did not contain the line "Hints only please!" or the edit which follows it.
Note that for $u \in \mathbb{R}^n$ with $\|u\| = 1$, we have $$\|ABu\| = \|A(Bu)\| \leq \|A\|\|Bu\| \leq \|A\|\|B\|\|u\| = \|A\|\|B\|$$ so $\|AB\| = \sup\{\|ABu\| : u \in \mathbb{R}^n, \|u\| = 1\} \leq \|A\|\|B\|$.
$\endgroup$ $\begingroup$Your reasoning is correct. Although I didn't see the need of writing the unit vector $u$ as $u=\frac{x}{\| x\|}$. The same conclusion can be reached by applying the same trick to $u$ instead of $x$.
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