# Showing a function is not an inner product

From Axler's Linear Algebra Done Right, 6.A 1,
Show that the function that takes $$((x_1,x_2),(y_1,y_2)) \in \mathbb{R}^2 \times\mathbb{R}^2$$ to $$|x_1 y_1|+|x_2 y_2|$$ is not an inner product on $$\mathbb{R}^2$$.

My attempt:
I will verify all the properties of inner products:
Positivity
$$|x_1 x_1|+|x_2 x_2|$$ is always $$\geq0$$ so this holds.

Definiteness
$$\langle (x_1,x_2),(x_1,x_2)\rangle=|x_1 x_1|+|x_2 x_2|=0$$ only if $$|x_1 x_1|=|x_2 x_2|=0$$ so $$(x_1,x_2)=0$$

$$\langle (x_1,x_2)+(z_1,z_2),(x_1,x_2)\rangle =\langle (x_1+z_1,x_2+z_2),(x_1,x_2)\rangle=|(x_1+z_1) y_1|+|(x_2+z_2)y_2|=|x_1 y_1|+|z_1 y_1|+|x_2 y_2|+|z_2 y_2|=\langle (x_1,x_2),(y_1,y_2)\rangle +\langle (z_1,z_2),(y_1,y_2)\rangle$$

Homogeneity in first slot
$$\langle \lambda (x_1,x_2),(y_1,y_2)\rangle=\langle (\lambda x_1,\lambda x_2),(y_1,y_2)\rangle =|\lambda x_1 y_1|+|\lambda x_2 y_2|=\lambda|x_1 y_1|+\lambda|x_2 y_2|=\lambda(|x_1 y_1|+|x_2 y_2|)=\lambda\langle (x_1,x_2),(y_1,y_2)\rangle$$

Conjugate symmetry
$$\overline {\langle (y_1,y_2),(x_1,x_2)\rangle}=\overline {|y_1 x_1|+|y_2 x_2|}=\overline{|y_1 x_1|}+\overline{|y_2 x_2|}=|y_1 x_1|+|y_2 x_2|=|x_1 y_1|+|x_2 y_2|$$

It seems to me that everything holds so I am not sure why this is not an inner product.

• $|\lambda x_1 y_1|+|\lambda x_2 y_2|=\lambda|x_1 y_1|+\lambda|x_2 y_2|$ is wrong. – Martin R Dec 28 '20 at 1:54
• Small typo: You probably meant $\geq 0$ in your positivity section? – Brian Tung Dec 28 '20 at 1:56
• $|(x_1+z_1) y_1|+|(x_2+z_2)y_2|=|x_1 y_1|+|z_1 y_1|+|x_2 y_2|+|z_2 y_2|$ is also wrong. – Martin R Dec 28 '20 at 1:56
• Hi, thanks for your comments. Could you please explain why those are wrong? Thank you. – pritchard Dec 28 '20 at 2:01
• Put in some numbers, Nikolai, and see for yourself why they are wrong! – Gerry Myerson Dec 28 '20 at 2:02

HINT. $$|a+b|\neq |a|+|b|$$, $$|\lambda a|\neq \lambda |a|$$. Consider negative numbers.