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We know that an $r \times n$ matrix $A=(\alpha_{jk})$ defines a linear operator from the vector space $X$ of all ordered $n$-tuples of numbers into the vector space $Y$ of all ordered $r$-tuples of numbers.

Suppose that any norm $\|\cdot\|_1$ is given on $X$ and any norm $\|\cdot\|_2$ is given on $Y$. A norm $\|\cdot\|$ on $Z$ (all complex $r$ by $n$ matrices) is said to be compatible with $\|\cdot\|_1$ and $\|\cdot\|_2$ if $\|Ax\|_2 \leq \|A\|\cdot \|x\|_1$.

Show that the norm defined by $\|A\|= \sup\frac{\|Ax\|_2}{ \|x\|_1}$ where $x\in X$ is compatible with $\|\cdot\|_1$ and $\|\cdot\|_2$.

Please help me, if you have any good idea about my question. I tried to firstly prove that the two norms are equivalent. But I do not.

mathsstudent
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1 Answers1

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$\newcommand{\n}[1]{\left\Vert #1 \right\Vert} \newcommand{\0}{\vec 0}$ This result is fairly trivial. Let $x_0 \in X$ be arbitrary. Then $$ \n{A} := \sup_{\substack{x \in X \\ x \ne \0}} \frac{\n{Ax}_2}{\n{x}_1} \ge \frac{\n{Ax_0}_2}{\n{x_0}_1} \implies \n{A} \n{x_0}_1\ge \n{Ax_0}_2 $$ The first inequality follows from the definition of supremum.

That $x_0$ is arbitrary gives compatibility.

PrincessEev
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