Linear algebra: trace and dual space exercise

[mahler1]

Homework Statement .

Let $A \in \mathbb C^{m\times n}$. Prove that tr$(A^*A)=0$ if and only if $A^*A=0$ (here $0$ obviously means the zero matrix).


The attempt at a solution.

By definition of the trace of a matrix, the implication ← is obvious. I am having problems proving the other implication: first of all, I have doubts about the meaning of $A^*$, if $f:\mathbb C^n \to \mathbb C^m$ is a linear transformation and $A$ is the associated matrix to $f$, then $f^t:{(\mathbb C^m)}^* \to {(\mathbb C^n)}^*$ defined as $f^t(\phi)=\phi \circ f$ for all $\phi \in {(\mathbb C^m)}^*$, so, I suppose $A^*$ just means the matrix which represents $f^t$, am I right?

I have no idea how to prove $tr(A^*A)=0 \implies A^*A=0$. I would appreciate any suggestions.

 

[hilbert2]

Try it first with a $2\times 2$ matrix. What are the diagonal elements of $A^{*}A$ if $A=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$? Why does $Tr(A^{*}A)=0$ imply that $a,b,c$ and $d$ are all zero? (I'm supposing that the $"*"$ means hermitian conjugate). Then try to extend the solution to more general matrices.

 

[mahler1]

Try it first with a $2\times 2$ matrix. What are the diagonal elements of $A^{*}A$ if $A=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$? Why does $Tr(A^{*}A)=0$ imply that $a,b,c$ and $d$ are all zero? (I'm supposing that the $"*"$ means hermitian conjugate). Then try to extend the solution to more general matrices.

Thanks for clearing up what it means $A^*$. I want to check if this is correct:

By definition, the entry of the product matrix $(A^*A)_{ij}=\sum_{k=1}^m (a^*)_{ik}a_{kj}$, but $(a^*)_{ik}=\overline a_{ki}$, so $\sum_{k=1}^m (a^*)_{ik}a_{kj}= \sum_{k=1}^m \overline a_{ki} a_{kj}$.

Now, $tr(A^*A)=\sum_{p=1}^n (A^*A)_{pp}=\sum_{p=1}^n \sum_{k=1}^m (a^*)_{pk}a_{kp}=\sum_{p=1}^n \sum_{k=1}^m \overline a_{kp} a_{kp}=\sum_{p=1}^n \sum_{k=1}^m |a_{kp}|^2$.

Since $tr(A^*A)=0 \implies \sum_{p=1}^n \sum_{k=1}^m |a_{kp}|^2=0 \implies a_{kp}=0 \space \forall 1\leq k \leq m, 1\leq p \leq n$. From here it follows $A=0 \implies A^*A=0$.

 

[hilbert2]

That looks correct. The idea is simply that if a sum of real non-negative numbers is zero, then every term in the sum must be zero.