Prove the following are norms.

Question

I'd like to show that the following are norms:

1. $A \in \mathbb{R}^{n\times n}$ is invertible, so $\|{\cdot}\|\colon\mathbb{R}^n\to\mathbb{R}$ is thus defined: $\forall x \in\mathbb{R}^n$, $\|x\|_{A^2}=\|Ax\|$

2. The induced norm: $\displaystyle\|A\|=\sup_{x \in\mathbb{R}^n}\frac{\|Ax\|}{\|x\|}$

NB I am aware of the three requirements for a norm, but I am finding it somewhat challenging to show they are fulfilled in the above cases.

I'd appreciate some assistance.

Answer 1

For 2nd part, check for 3 conditions. I am going to give a very detailed, very basic argument argument here for part 2.

Condition 1:

By property of vector norm, $||Ax||\geq 0$ and $||x||\geq 0$, but since $x\not=0$, once again by vector norm property, $||x||\not=0$, so $||x||>0$. This means $\frac{||Ax||}{||x||}\geq 0$ for any $x\not=0$. So $||A||\geq 0$.

If $A=0$, then $Ax=0$ for any $x$. By property of vector norm, $||Ax||=0$ whenever $Ax=0$ so $||Ax||=0$ for any $x$. Hence $||A||=\sup\limits_{x\not=0}\frac{||Ax||}{||x||}=0$.

If $||A||=0$, then by definition of $||A||=\sup\limits_{x\not=0}\frac{||Ax||}{||x||}$ we must have $\frac{||Ax||}{||x||}\leq 0$ for all $x\not=0$. Combining this with the above inequality, we have $\frac{||Ax||}{||x||}=0$ for any $x\not=0$, hence $||Ax||=0$ for any $x\not=0$. By vector norm property again, $||Ax||=0$ only happen if $Ax=0$. Which means that $Ax=0$ for all $x\not=0$. Clearly $Ax=0$ when $x=0$. Hence $Ax=0$ for all $x$, which means $A=0$.

Condition 2:

$||\alpha A||=\sup\limits_{x\not=0}\frac{||(\alpha A)x||}{||x||}$ (by definition)

$\sup\limits_{x\not=0}\frac{||(\alpha A)x||}{||x||}=\sup\limits_{x\not=0}\frac{||\alpha (Ax)||}{||x||}$ (by linearity)

$\sup\limits_{x\not=0}\frac{||\alpha (Ax)||}{||x||}=\sup\limits_{x\not=0}|\alpha|\frac{||Ax||}{||x||}$ (by property of vector norm)

$\sup\limits_{x\not=0}|\alpha|\frac{||Ax||}{||x||}=|\alpha|\sup\limits_{x\not=0}\frac{||Ax||}{||x||}$ (by property of supremum and the fact that $|\alpha|\geq 0$)

$|\alpha|\sup\limits_{x\not=0}\frac{||Ax||}{||x||}=|\alpha|||A||$ (by definition)

Condition 3.

$||A+B||=\sup\limits_{x\not=0}\frac{||(A+B)x||}{||x||}$ (by definition)

$\sup\limits_{x\not=0}\frac{||(A+B)x||}{||x||}=\sup\limits_{x\not=0}\frac{||Ax+Bx||}{||x||}$ (by linearity)

For any $x\not=0$, then $||Ax+Bx||\leq ||Ax||+||Bx||$ by a property of vector norm, so $\frac{||Ax+Bx||}{||x||}\leq\frac{||Ax||+||Bx||}{||x||}=\frac{||Ax||}{||x||}+\frac{||Bx||}{||x||}$. Hence by the property of supremum, we have $\sup\limits_{x\not=0}\frac{||Ax+Bx||}{||x||}\leq\sup\limits_{x\not=0}(\frac{||Ax||}{||x||}+\frac{||Bx||}{||x||})$.

Now $\sup\limits_{x\not=0}(\frac{||Ax||}{||x||}+\frac{||Bx||}{||x||})=\sup\limits_{x\not=0}\frac{||Ax||}{||x||}+\sup\limits_{x\not=0}\frac{||Bx||}{||x||}$ by property of supremum.

Finally $\sup\limits_{x\not=0}\frac{||Ax||}{||x||}+\sup\limits_{x\not=0}\frac{||Bx||}{||x||}=||A||+||B||$ by definition.

Combining from the beginning $||A+B||\leq||A||+||B||$.

Now as for part 1, the argument is very similar. Really, just use the fact that $A$ is invertible to ensure that $Ax=0$ if and only if $x=0$.