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The definiteness property of vector inner product says

$\langle x,x \rangle \geq 0$ for all $x \in \mathcal{V}$ and $\langle x,x \rangle = 0$ if and only if $x=0$

and matrix inner product is defined as $\langle A,B\rangle=TrB^TA$.

When substituting $x$ with a matrix $A \in \mathbb{R}^{n \times n}$, the first part of the statement is easy to prove. However, I am wondering how I prove $\langle A, A\rangle =0 \Rightarrow A=0$ since $\langle A,A\rangle=Tr A^TA=\sum_{i}^na_{ii}^2$, when $\langle A,A \rangle=0$, it seems that I could merely say the diagonal entries are zero and hence I could not prove $A=0$.

Could anyone figure out what I have done wrong. Thank you in advance.

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  • $(i,j)$-entry of $A^TA$ is $\sum_{k=1}^n A_{ki}A_{kj}$

  • $(i,i)$-entry of $A^TA$ is $\sum_{k=1}^n A_{ki}A_{ki}=\sum_{k=1}^n A_{ki}^2$

Hence, trace of $A^TA$ is $\sum_{i=1}^n\sum_{k=1}^n A_{ki}^2$

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Hint: The $i$th diagonal element of $A^\top A$ is not $a_{ii}^2$.

The $i$th diagonal element of $A^\top A$ is the squared Euclidean norm of the $i$th column of $A$. Thus $\text{Tr}(A^\top A)$ is the sum of the squared norms of the columns of $A$. If it is zero, then each column of $A$ must be the zero vector, and thus $A=0$.


Probably the most intuitive way to think about the matrix inner product is that $\text{Tr}(A^\top B)$ is exactly the same as the Euclidean inner product $\langle a, b \rangle$ where $a$ is the result of stacking the columns of $A$ into a vector of length $n^2$, and $b$ is defined analogously from $B$. From this interpretation all the properties that you desire follow immediately from the analogous properties of the Euclidean inner product.

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