This is a question from textbook.

Determine if the vector space of all $2 \times 2$ matrices is a inner product. Let $A$ and $B$ be $2\times 2$ matrices then $\langle A, B \rangle = a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4$.

My understanding is that this is not an inner product because it does not satisfy P4 $\langle v, w \rangle \geq 0$ for all $v$ and $w$. Suppose we let $A =\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 \\ 0 & 0 \end{bmatrix} $ then $\langle A, B \rangle = -1$ therefore it is not an inner product. However, the text solution states that it is an inner product. What am I misunderstanding here? I apologize for the bad syntax.

  • 3
    $\begingroup$ The inequality you are trying to remember is $(u,u)\geq 0$, the same element multiplied by itself. $\endgroup$
    – Marja
    Aug 5, 2017 at 12:39
  • $\begingroup$ Thank you, I totally missed that. $\endgroup$
    – samuel
    Aug 5, 2017 at 12:40
  • 4
    $\begingroup$ That operation is going to be an inner product. The space of $2\times2$ matrices is the same as $\mathbb{R}^4$ by sending $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ to $(a,b,c,d)$. That formula is just the standard inner product on $\mathbb{R}^4$. $\endgroup$
    – Marja
    Aug 5, 2017 at 12:44

1 Answer 1


Let $A$ be a non-zero matrix with entries $[A]_{ij} = a_{ij}$ Then $\langle A, A \rangle = \langle \begin{pmatrix} a_{11} & a_{12} \\a_{21} & a_{22}\end{pmatrix}, \begin{pmatrix} a_{11} & a_{12} \\a_{21} & a_{22}\end{pmatrix} \rangle = a_{11}^2 + a_{12}^2 + a_{21}^2 + a_{22}^2 > 0$

this proves the positive definiteness.

To understand why we require positieve definiteness, in any kind of inner vector space, either real or complex, you must remember that the norm must always be a positive number (as this represents a distance, and distances can't be negative, can they?). Since $\Vert v \Vert = \sqrt{\langle v,v \rangle}$, it makes sense to ask positive definiteness, or the number under the square root is possibly negative, something we don't want.


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