# Proof for induced matrix norm

1. Sep 17, 2015

### FOIWATER

Hi, I found a statement without a proof. It seems simple enough, but I am having trouble proving it because I am not positive about induced matrix norms. The statement is that $$||A^k|| \leq||A||^{k}$$ for some matrix A and positive integer k. I have found that the norm of a matrix is the supremum of the norm of Ax over the norm of x, but I do not know to which norms these refer?

I am assuming euclidean norms. Since Ax gives us back a vector and x is itself a vector.

So I have that:
$$||A^k|| = \sup_{||x||=1}(||A^{k}x|| : ||x||=1)$$
and
$$||A||^{k} = \sup_{||x||=1}(||Ax|| : ||x||=1)^{k}$$

Not sure what to do with them, though, any hints appreciated. I was thinking triangle inequality.. but I didn't really get anything from it. And I do not know if the triangle inequality applies to this matrix induced norm (although I think it applies to any operation that qualifies as a norm, since it defines norms).

2. Sep 17, 2015

### RUber

What if the x that gives the sup for A^k is not the same as the x that is the sup for A? A acting on any vector will always be less than or equal to its sup.

3. Sep 17, 2015

### RUber

To be more clear...
Let $X$ be your maximizer for $\|Ax\|$. Then
$\| A^k \| = \| A A A ... A \| = \sup_{\|x\|=1} \left( \| A A A ... Ax \| \right)$
Let $x_0$ be any x such that $\|x_0 \| = 1$.
Then you know that $\|Ax_0\| \leq \|AX\|=\|A\|.$
Let $x_1 = Ax_0$. By similar argument, $\|A\frac{x_1}{\|x_1\|}\|=\frac{\|Ax_1\|}{\|x_1\|} \leq \|AX\|\implies \|Ax_1\|\leq \|AX\|\|x_1\| \leq \|AX\|\|A\|=\|A\|^2 .$
And by induction, you can clearly see that no matter what, you won't be able to break the inequality.

4. Sep 17, 2015

### FOIWATER

∥AX∥=∥A∥ I am not sure of this statement? A times X should give a vector, not a matrix A?

5. Sep 17, 2015

### FOIWATER

This actually makes sense now thanks a lot!

Last edited: Sep 17, 2015
6. Sep 17, 2015

### RUber

Right, but I defined ||AX|| to be ||A|| earlier.
Of course, that is assuming that there is a maximum. You might have to keep the definition of ||A|| as the supremum to be proper...but the logic is the same.

7. Sep 17, 2015

### FOIWATER

That's good insight I hope to be able to work my way through problems like that some day