Suppose that $\Bbb R^m$ and $\Bbb R^n$ are equipped with norms $\|\cdot\|_b$ and $\|\cdot\|_a$ respectively. Show that the induced matrix norm $\|\cdot\|_{a,b}$ can be computed by the formula

$$\|A\|_{a,b} = \max\limits_{x\neq 0}\dfrac{\|Ax\|_b}{\|x\|_a}$$

and i use this definition for induced matrix

can anyone answer this question?

  • $\begingroup$ Remind us what definition you are using for the induced matrix norm since the definition I am familiar with is the one you are trying to show equivalence to. $\endgroup$
    – JMoravitz
    Commented Apr 6, 2020 at 14:39
  • $\begingroup$ i add the description you asked for. (– JMoravitz) $\endgroup$ Commented Apr 6, 2020 at 14:48
  • $\begingroup$ welcome to MSE. you are encouraged to include thoughts and attempts if any. $\endgroup$ Commented Apr 6, 2020 at 14:54

1 Answer 1


Hint: Note that $$ \frac{\|Ax\|_b }{\|x\|_a} = \left\| A\left(\frac{x}{\|x\|_a}\right)\right\|_b. $$ Now, show that $$ \begin{align*} \max\left\{\frac{\|Ax\|_b}{\|x\|_a} : x \neq 0\right\}&= \max\{\|Ax\|_b : \|x\|_a = 1\} \\ & = \max \{\|Ax\|_b: \|x\|_a \leq 1\} = \|A\|_{a,b} \end{align*} $$

  • $\begingroup$ can i get more hint or the complete solution to this question? $\endgroup$ Commented Apr 6, 2020 at 19:52
  • $\begingroup$ See my latest edit $\endgroup$ Commented Apr 6, 2020 at 20:50
  • $\begingroup$ oh i see. so in the first part since A is a linear transformation we can first divide x by it's magnitude and then do the linear transformation. and in the second part, in the first equality since the magnitude of x is equal to one, we could omit it but i don't get the next equality where max{∥Ax∥b:∥x∥a=1} = max{∥Ax∥b:∥x∥a≤1} $\endgroup$ Commented Apr 6, 2020 at 21:44
  • $\begingroup$ @mohsentajdeh Note that if $\|x\|<1$, then $\|Ax\|$ will never be equal to $\max\{\|Ax\|_b : \|x\|_a \leq 1\}$ since $\|A(x/\|x\|)\| > \|Ax\|$. $\endgroup$ Commented Apr 6, 2020 at 23:15

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