Solution: We recall that $\left\langle \cdot, \cdot \right\rangle $ on a vector space $V$ is a real inner product if
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$\left\langle v,v \right\rangle > 0$ for any $v \neq 0$.
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$\left\langle v,v \right\rangle = 0$ if and only if $v = 0$.
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For any $v,w \in V$, $\left\langle v,w \right\rangle = \left\langle w,v \right\rangle $.
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For any $\lambda \in \mathbb{R} $ and $v_1,v_2,v_3 \in V$ $\left\langle \lambda v_1 + v_2, v_3 \right\rangle = \lambda \left\langle v_1, v_3 \right\rangle + \left\langle v_2, v_3 \right\rangle $.
We will verify each property for each options.
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$\left\langle A,B \right\rangle = \operatorname{trace}\left( A \right) \operatorname{trace}\left( B \right) $.
This is NOT an inner product. Take
\[
A = \begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix}.
\]
Then,
\[
\left\langle A,A \right\rangle = \operatorname{trace}\left( A \right) \operatorname{trace}\left( A \right) = 0,
\]
but $A \neq 0$. Thus, this is not an inner product.
-
$\left\langle A,B \right\rangle = \operatorname{trace}\left( AB \right) $.
This is
TRUE. We recall that for any real symmetric matrix all the eigenvalues are real and the matrix is diagonalizable.
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Let $A$ be any real symmetric matrix. Then,
\begin{align*}
\left\langle A, A \right\rangle & = \operatorname{trace}\left( A^2 \right) \\
& = \sum_{\lambda }\lambda ^2 > 0,
\end{align*}
where $\lambda$ is an eigenvalue of $A$.
-
If $\left\langle A,A \right\rangle = 0$, then $\operatorname{trace}\left( A^2 \right) = 0$, which means all the eigenvalues are zero. Since $A$ is diagonalizable and all the eigenvalues are zero, it is a zero matrix. Thus, $A = 0$.
-
For any real symmetric matrices $A,B$,
\[
\left\langle A,B \right\rangle = \operatorname{trace}\left( AB \right) = \operatorname{trace}\left( BA \right) = \left\langle B,A \right\rangle .
\]
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Also, the linearity follows from the linearity property of $\operatorname{trace}$. That is, for any $\lambda \in \mathbb{R} $ and for any symmetric matrices $A,B,C$
\begin{align*}
\left\langle \lambda A + B, C \right\rangle & = \operatorname{trace}\left( (\lambda A + B) C \right) \\
& = \operatorname{trace}\left( \lambda AC + BC \right) \\
& = \operatorname{trace}\left( \lambda AC \right) + \operatorname{trace}\left( BC \right) \\
& = \lambda \operatorname{trace}\left( AC \right) + \operatorname{trace}\left( (BC) \right) \\
& = \lambda \left\langle A,C \right\rangle + \left\langle B,C \right\rangle .
\end{align*}
Therefore, it is an inner product.
-
$\left\langle A, B \right\rangle = \det (AB)$.
This is NOT an inner product. Take
\[
A = \begin{bmatrix}
1 & 0 \\
0 & 0 \\
\end{bmatrix},
\]
then
\[
\left\langle A, A \right\rangle = \det (A^2) = 0,
\]
but $A \neq 0$.
-
$\left\langle A,B \right\rangle = \operatorname{trace}\left( A \right) + \operatorname{trace}\left( B \right) $.
This is NOT an inner product. Take
\[
A = \begin{bmatrix}
-1 & 0 \\
0 & 0 \\
\end{bmatrix}.
\]
Then,
\[
\left\langle A, A \right\rangle = \operatorname{trace}\left( A \right) + \operatorname{trace}\left( A \right) = - 2 < 0.
\]