This is a question from the textbook Elementary Linear Algebra: Applications Version which I am really struggling with.

Let $u = (u_1, u_2, u_3)$ and $v = (v_1, v_2, v_3)$. Show that the following expressions does not define an inner product on $\mathbb{R}^3$, and list all inner product axioms that fail to hold.

a) $\langle u, v\rangle = u_1^2v_1^2 + u_2^2v_2^2 + u_3^2v_3^2$

a) $\langle u, v\rangle = u_1v_1 - u_2v_2 + u_3v_3$

I have been told that the solution for a) is that axioms 2) and 3) do not hold but I do not understand how to prove that. I don’t think I understand the premise of the question, how can $\langle u, v\rangle = u_1^2v_1^2 + u_2^2v_2^2 + u_3^2v_3^2$ if $u = (u_1, u_2, u_3)$ and $v = (v_1, v_2, v_3)$, wouldn’t it rather be $\langle u, v\rangle = u_1v_1 + u_2v_2 + u_3v_3$?

The Axioms listed in the text book are as follows:

  1. $\langle u, v\rangle = \langle v, u\rangle$ (Symmetry axiom)
  2. $\langle u + v, w\rangle = \langle u, w\rangle + \langle v, w\rangle$ (Additivity axiom)
  3. $\langle ku, v\rangle = k\langle u, v\rangle$ (Homogeneity axiom)
  4. $\langle v, v\rangle \geq 0$ and $\langle v, v\rangle = 0$ if and only if $v = 0$ (Positivity axiom)
  • 3
    $\begingroup$ In each of these expressions, $\langle \cdot, \cdot \rangle$ denotes a new pairing (a function of two vectors that produces a scalar) that has nothing to do with the standard Euclidean inner product. $\endgroup$
    – catherine
    Oct 31, 2022 at 6:12

1 Answer 1


The wording from the comment by @catherine helped me realise how to solve this. Below is how I was able to show that axiom 2 did not hold.

When: \begin{equation*} u = (u_1, u_2, u_3) \end{equation*} \begin{equation*} v = (v_1, v_2, v_3) \end{equation*} \begin{equation*} \langle u, v\rangle = u_1^2v_1^2 + u_2^2v_2^2 + u_3^2v_3^2 \end{equation*}

LHS: \begin{equation*} \langle u + v, w\rangle = (u_1+v_1)^2w_1^2 + (u_2+v_2)^2w_2^2 + (u_3+v_3)^2w_3^2 \end{equation*} \begin{equation*} \langle u + v, w\rangle = (u_1^2+2u_1v_1+v_1^2)w_1^2 + (u_2^2+2u_2v_2+v_2^2)w_2^2 + (u_3^2+2u_3v_3+v_3^2)w_3^2 \end{equation*} \begin{equation*} \langle u + v, w\rangle = u_1^2w_1^2+2u_1v_1w_1^2+v_1^2w_1^2 + u_2^2w_2^2+2u_2v_2w_2^2+v_2^2w_2^2 + u_3^2w_3^2+2u_3v_3w_3^2+v_3^2w_3^2 \end{equation*}

RHS: \begin{equation*} \langle u, w\rangle + \langle v, w\rangle = u_1^2w_1^2+v_1^2w_1^2 + u_2^2w_2^2 + v_2^2w_2^2 + u_3^2w_3^2 + v_3^2w_3^2 \end{equation*}

\begin{equation*} \langle u, w\rangle + \langle v, w\rangle \neq \langle u + v, w\rangle \end{equation*}

Therefore Axiom 2 does not hold. A similar method was utilised to demonstrate Axiom 3 does not hold.


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