Is a norm on $R^n$ linear?

Question

I was reading the book Linear Algebra Done Right by Axler. In the chapter on inner product space (Ch.6), he defines the norm of x on $R^n$ space as:

$||x|| = \sqrt{x_1^2 + ... + x_n^2}$

and says:

"The norm is not linear on $R^n$. To inject linearity into the discussion, we introduce the dot product."

I don't see why the norm is not linear.

If I check a multiplicity for $R^2$, using a scalar of 3 for example

$3||x|| =? ||3x||$

$3 \sqrt{x_1^2 + x_2^2} =? \sqrt{(3 x_1)^2 + (3 x_2)^2} $

$3 \sqrt{x_1^2 + x_2^2} =? \sqrt{9(x_1^2 + x_2)^2} $

$3 \sqrt{x_1^2 + x_2^2} = 3 \sqrt{x_1^2 + x_2^2} $

Why is the norm not linear?

Axler then says:

"Also, if $y \in R^n$ is fixed, then clearly the map from $R^n$ to $R$ that sends $x \in R^n$ to $x \cdot y$ is linear."

Why is the dot product linear if the norm isn't?

Regards, Madeleine.

Answer 1

Try checking linearity of the norm with $-3$ as the scalar instead, or by checking the additive property - you should see that it fails.

The dot product is linear in the sense that the map $x\mapsto x\cdot y$ for some fixed $y$ (independent of $x$) is linear; the map $x\mapsto x\cdot x$ is not linear.

Answer 2

Does

$$\|x+y\|=\|x\|+\|y\|\quad?$$ Find a counterexample and recall the triangle inequality.