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Let $V$ be some inner product vector space over $F$, let $B=\{b_1,\dots ,b_n\}$ and $C=\{c_1,\dots ,c_n\}$ be two orthonormal bases, and let $T:V\to V$ be some linear transformation. Prove or disprove: $$\sum_{i=1}^n\Vert Tb_i\Vert^2=\sum_{i=1}^n\Vert Tc_i\Vert^2$$

My feeling is that this is correct, as the length of each vector is decided by the coefficients of the orthonormal basis, whose sum should be equal for different bases, and this equality should be preserved under the transformation, but I'm sure how to prove this.

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Assume $b_i$ is the standard basis and assume $T$ has matrix $A$ in this basis. Then, $Ab_i$ is the $i-$th column of $A$, hence $\sum_{i=1}^n ||A b_i||^2$ is just the sum of the squares of the Euclidean norms of the columns of $A$. More simply said, this sum is actually the sum of the squares of the entries of $A$. This sum has a name: the square of the Frobenius norm $||A||_F$ of $A$. The article gives us

$||A||_F^2 = tr(A^*A)$

Now, let's look at the other sum: if $P$ is the change of basis matrix from $b_i $to $ c_i $, i.e $Pb_i =c_i$ then

$$\sum_{i=1}^n ||A c_i||^2 = \sum_{i=1}^n ||AP b_i||^2 = ||AP||_F^2 = tr((AP)^*AP) = tr(P^*A^*AP) = tr(A^*A) = ||A||_F^2$$ Where we used the fact that the trace doesn't change under orthonormal transformations.

Thus, the result follows.

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