Showing that a the function $N(x) = \sqrt{x_1^2 - x_1x_2 + 4x_2^2}$ defines a norm on $\mathbb{R^2}$

Question

I want to show that the function $N: \mathbb{R} \rightarrow \mathbb{R^2}$ defined by $\sqrt{x_1^2 - x_1x_2 + 4x_2^2}$

I know that in order to do so, I have to show the 3 following properties:

• $N(x) = 0 \iff x= 0$

• $N(\lambda x) = |\lambda|N(x)$

• $N(x+y) ≤ N(x) + N(y)$

I tried to show the frst property but I got stuck.

I don't know how to prove that $x_1^2 - x_1x_2 + 4x_2^2 = 0 \iff (x_1, x_2) = (0,0)$

For the third property I guess I'd just have to use the triangle inequality. But I don't know how to show the second property either.

Answer 1

One way: $$Q(x_1,x_2)=x_{1}^2 - x_1 x_2 + 4 x_{2}^{2}=\begin{pmatrix}x_1,\;x_2\end{pmatrix}\begin{pmatrix}{1}&{-1/2 }\\{-1/2 }&{4}\end{pmatrix}\begin{pmatrix}x_1\\{x_2}\end{pmatrix}$$ is a definite positive quadratic form, so $$\langle (x_1,x_2),(y_1,y_2) \rangle=\begin{pmatrix}x_1,\;x_2\end{pmatrix}\begin{pmatrix}{1}&{-1/2 }\\{-1/2 }&{4}\end{pmatrix}\begin{pmatrix}y_1\\{y_2}\end{pmatrix}$$ is an inner product. As a consequence, $$N(x_1,x_2)=\sqrt{\langle (x_1,x_2),(x_1,x_2) \rangle}$$ is a norm.

Answer 2

Let $x=(a,b)$ and $y=(c,d)$.

Hence, $$N(x)+N(y)=\sqrt{a^2-ab+4b^2}+\sqrt{c^2-cd+4d^2}=$$ $$=\sqrt{\left(a-\frac{b}{2}\right)^2+\frac{15}{4}b^2}+\sqrt{\left(c-\frac{d}{2}\right)^2+\frac{15}{4}d^2}\geq\sqrt{\left(a+c-\frac{b+d}{2}\right)^2+\frac{15}{4}(b+d)^2}=$$ $$=\sqrt{(a+c)^2-(a+c)(b+d)+4(b+d)^2}=N(x+y)$$ The rest is obvious, I think.