inner product trouble

Question

Problem statement:

let $A$, $B$ be matrices in $\mathbb{M}_{n \times n}(\mathbb{C})$. We define $ \langle A|B \rangle = \operatorname{tr}\left(A \bar{B}\right)$. Then is the map an inner product?

For a map to be an inner product we need to have the following conditions to hold:

1. $\langle A|A \rangle \ge 0 , \forall A \in \mathbb{M}_{n \times n}(\mathbb{C})$
2. $\langle A|A \rangle = 0 \iff A = 0 , \forall A \in \mathbb{M}_{n \times n}(\mathbb{C})$
3. $\langle A|B \rangle = \langle B|A \rangle^{*} , \forall A , B \in \mathbb{M}_{n \times n}(\mathbb{C})$
4. $\langle(\alpha A + \beta B ) | C \rangle = \alpha \langle A|C \rangle + \beta \langle B|C \rangle , \forall A, B, C \in \mathbb{M}_{n \times n}(\mathbb{C}) , \forall \alpha , \beta \in \mathbb{C} $

We can see easily that the $3^{\text{rd}}$ & $4^{\text{th}}$ are true, but what about the first two? If there is any counterexample for the two conditions, kindly mention them and also the thought process of searching the example.

Answer 1

Let $\langle A|B\rangle:=\mathrm{tr}(AB^*)$.

Note that the following product gives $\mathbf{a}_i\cdot\mathbf{b}_i$ on the main diagonal: $${AB}^*=\begin{pmatrix}--\mathbf{a}_1--\cr \vdots\cr--\mathbf{a}_n--\end{pmatrix}\begin{pmatrix}\vdots&&\vdots\cr\overline{\mathbf{b}_1}&\cdots&\overline{\mathbf{b}_n}\cr\vdots&&\vdots\end{pmatrix}=\begin{pmatrix}\mathbf{a}_1\cdot\overline{\mathbf{b}_1}&&\cr&\ddots&\cr&&\mathbf{a}_n\cdot\overline{\mathbf{b}_n}\end{pmatrix}$$ (I'm trying to convey that the $\mathbf{a}_i$ vectors are rows of $A$, while $\mathbf{b}_i$ are columns. The last matrix need not have zeros in the off-diagonal.)

So $\mathrm{tr}({AA}^*)=\sum_i\mathbf{a}_i\cdot\overline{\mathbf{a}_i}=\sum_{ij}|a_{ij}|^2$.

It follows that $\mathrm{tr}({AA}^*)\ge0$ and $\mathrm{tr}(AA^*)=0\iff(\forall i,j,\ |a_{ij}|=0)\iff A=0$.

The other two axioms are true by properties of the trace, but for completeness, $$\mathrm{tr}(BA^*)=\mathrm{tr}(AB^*)^*=\overline{\mathrm{tr}(AB^*)}$$

$$\mathrm{tr}((\alpha_1A_1+\alpha_2A_2)B^*)=\mathrm{tr}(\alpha_1A_1B^*+\alpha_2A_2B^*)=\alpha_1\mathrm{tr}(A_1B^*)+\alpha_2\mathrm{tr}(A_2B^*)$$

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Note: $\mathrm{tr}(A\bar{B})$ is not an inner product on square matrices. Pick any real $A\ne0$ such that $A^2=0$, for example $A=\begin{pmatrix}O&I\cr O&O\end{pmatrix}$. Then $\mathrm{tr}(A\bar{A})=0$ but $A\ne0$.