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:
$|x_1 x_1|+|x_2 x_2|$ is always $\geq0$ so this holds.

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

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

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.

  • 1
    $\begingroup$ $|\lambda x_1 y_1|+|\lambda x_2 y_2|=\lambda|x_1 y_1|+\lambda|x_2 y_2|$ is wrong. $\endgroup$
    – Martin R
    Dec 28, 2020 at 1:54
  • $\begingroup$ Small typo: You probably meant $\geq 0$ in your positivity section? $\endgroup$
    – Brian Tung
    Dec 28, 2020 at 1:56
  • 1
    $\begingroup$ $|(x_1+z_1) y_1|+|(x_2+z_2)y_2|=|x_1 y_1|+|z_1 y_1|+|x_2 y_2|+|z_2 y_2|$ is also wrong. $\endgroup$
    – Martin R
    Dec 28, 2020 at 1:56
  • $\begingroup$ Hi, thanks for your comments. Could you please explain why those are wrong? Thank you. $\endgroup$ Dec 28, 2020 at 2:01
  • $\begingroup$ Put in some numbers, Nikolai, and see for yourself why they are wrong! $\endgroup$ Dec 28, 2020 at 2:02

1 Answer 1


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


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