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Given an orthonormal basis $\{u_1,\cdots, u_n\}$ of a vector space $V$ I am asked to show that $$ \sum_{k=1}^n \|T^*u_k\|^2= \sum_{k=1}^n \|Tu_k\|^2 $$ for all $T\in \mathcal{L}(V)$ where $T^*$ represents the adjoint of $T$ and $\|u\|^2=\langle u,u \rangle$ for a positive definite hermitian inner product $\langle \cdot,\cdot \rangle.$

I know by the orthonormality of the basis we can represent $$T^*u_k = \sum_{i=1}^n \langle T^*u_k,u_i \rangle u_i $$ for each $k$ and this expression can be tinkered with by applying properties of the inner product and adjoint. I did a problem like this that was essentially a one line proof and I expect that a similar approach should work here but so far everything I've tried has expanded into a humongous mess and I get lost lines and lines of notation. A push in the right direction would be appreciated.

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  • $\begingroup$ I think is simple. Suppose $n=2$ just for simplicity. Write $Tu_k$ as you said and remember that $||Tu_k||^2 = <Tu_k,Tu_k>$ make the distributions and then use the definition of adjoint. $\endgroup$ – Rodrigo Ribeiro Nov 24 '13 at 2:10
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Another way to do it is to see $\|T^* u_k\|^2 = \sum_{i=1}^n |\langle T^*u_k,u_i \rangle|^2$, which follows from the formula you wrote. Hence $$ \sum_{k=1}^n \|T^* u_k\|^2 = \sum_{k=1}^n \sum_{i=1}^n |\langle T^*u_k,u_i \rangle|^2 $$ Or even easier: write $T$ as a matrix with respect to this orthonormal basis. Then $\sum_{k=1}^n \|T^* u_k\|^2$ is the sum of the squares of the entries of the transpose of this matrix. Now it is clear that this is no different to the sum of the squares of the entries of the matrix.

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The equality $$ \sum_{k=1}^n \|T^*u_k\|^2= \sum_{k=1}^n \|Tu_k\|^2 $$ is nothing else than $$ \mbox{Trace}(TT^*)=\mbox{Trace}(T^*T). $$ It is a particular case of the more general equality $$ \mbox{Trace}(AB)=\mbox{Trace}(BA), $$ which is proven by writing the expression and exchanging the two sums.

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You should get the answer you are looking for by using the expression for $T^*u_k$ you have given. Start with

$\langle T^*u_k, T^*u_k\rangle =\dots =\sum_{I=1}^n|\langle T^*u_k,u_i\rangle|^2$, noting that

$|z|^2=\overline{z}z$.

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Let $U =\begin{bmatrix} u_1 & \cdots u_n \end{bmatrix}$, then we have $U^*U = I$, so $U$ is unitary.

The Frobenius norm is invariant under unitary rotations and adjoints, and we have $\|A\|_F^2 = \sum_k \|A e_k\|_2^2$.

Combining gives: \begin{eqnarray} \sum_k \|T u_k\|_2^2 &=& \sum_k \|T U e_k\|_2^2 \\ &=& \|TU\|_F^2 \\ &=& \|T\|_F^2 \\ &=& \|T^*\|_F^2 \\ &=& \|T^*U\|_F^2 \\ &=& \sum_k \|T^* U e_k\|_2^2 \\ &=& \sum_k \|T^* u_k\|_2^2 \end{eqnarray}

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