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I have found this question here:

Question on why Hilbert-Schmidt operator definition is independent of the choice of basis.

But I do not understand the answer. Also I feel like my question is different as I am asking about Hilbert-Schmidt norm and not operator, am I correct?

1-If so can anyone give me a hint to the answer of my question?

2-If I am incorrect can anyone explain to me the answer in the above link in a clear way please?

EDIT:

In the solution that is given in the link above:

I do not know why he considered the given linear transformation unitary or orthogonal, could anyone explain this for me please?

EDIT2:

The statement in general is false (it is true only for orthonormal basis) ..... can anyone give me an example for showing that it is false please?

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  • $\begingroup$ Hilbert Schmidt norm is defined for Hilbert Schmidt operators so this question is already answered in the earlier post. Which part of the proof you had difficulty with? $\endgroup$ – Kabo Murphy Apr 1 at 12:30
  • $\begingroup$ @KaviRamaMurthy the general consequence of the steps of the proof is not clear for me ..... what did he do to prove the required? $\endgroup$ – Secretly Apr 1 at 12:48
  • $\begingroup$ I even do not understand where is the solution in the link mentioned @hopefully $\endgroup$ – Mathstupid Apr 1 at 15:05
  • $\begingroup$ My professor said that this statement is wrong and its correct only if we have an orthonormal basis @KaviRamaMurthy $\endgroup$ – Mathstupid Apr 2 at 18:40
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First, the proof, which is quite simple (and is true even in countably many dimensions, too, with minimal modifications, over either $\mathbb R$ or $\mathbb C$).

If $\{ e_1, \dots, e_n \}$ is an orthonormal basis, the Hilbert-Schmidt norm of $A$ is defined as

$$\| A \| ^2 = \sum _{i=1} ^n \| A e_i \| ^2 = \sum _{i=1} ^n \langle A e_i, A e_i \rangle = \sum _{i=1} ^n \langle A^t A e_i, e_i \rangle = \operatorname{Tr} (A^tA) \ ,$$

where $A^t$ denotes the transpose of $A$.

Now, if $\{ f_1, \dots, f_n \}$ is another orthonormal basis, let $M$ be the transition matrix between them, i.e. $f_i = \sum _j M_{ij} e_j$. Since it is a transition matrix between orthonormal bases, $M$ will be orthogonal, i.e. $M^t M = I_n$ or, equivalently, $\sum _{j=1} ^n (M^t)_{ij} M_{jk} = \delta_{ik}$, with $\delta_{ik}$ being Kronecker's symbol.

Then

$$\sum _{j=1} ^n \| A f_j \| ^2 = \sum _{j=1} ^n \langle A f_j, A f_j \rangle = \sum _{j=1} ^n \langle A^t A f_j, f_j \rangle = \sum _{j=1} ^n \langle A^t A \sum _{i=1} ^n M_{ji} e_i, \sum _{k=1} ^n M_{jk} e_k \rangle = \\ = \sum _{i = 1} ^n \sum _{k = 1} ^n \langle A^t A e_i, e_k \rangle \sum _{j = 1} ^n M_{ji} M_{jk} = \sum _{i = 1} ^n \sum _{k = 1} ^n \langle A^t A e_i, e_k \rangle \sum _{j = 1} ^n (M^t)_{ij} M_{jk} = \sum _{i = 1} ^n \sum _{k = 1} ^n \langle A^t A e_i, e_k \rangle \delta_{ik} = \\ \sum _{i = 1} ^n \langle A^t A e_i, e_i \rangle = \| A \| ^2 \ ,$$

which shows that, indeed, the definition of $\| A \|$ does not depend on the orthonormal basis used.

If one does not use orthonormal bases, though, the above ceases to hold. Probably the simplest illustration of this is to take $A : \mathbb R \to \mathbb R$ given by $Ax = x$. Take $e_1 = 1$ and $f_1 = 2$. The Hilbert-Schmidt norm of $A$ in the basis $\{ e_1 \}$ is $\sum _{i=1} ^1 \| A e_i \| ^2 = | A e_1 | ^2 = |e_1|^2 = 1^2 = 1$, while the norm in the basis $\{ f_1 \}$ is $|A f_1|^2 = |f_1|^2 = 4$. This happens because the basis $\{ f_1 \}$ is not orthonomal, since $\| f_1 \| ^2 = f_1 \cdot f_1 = 2 \cdot 2 = 4 \ne 1$.

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  • $\begingroup$ why in the forth line the last inequality the trace appears? what rule are you using? $\endgroup$ – Secretly Apr 7 at 15:11
  • $\begingroup$ I think $||A||^2$ instead of $||A||$only in the forth line first equality. $\endgroup$ – Secretly Apr 7 at 15:19
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    $\begingroup$ @hopefully: And I'm giving you a counterexample when $n=1$. $\endgroup$ – Alex M. Apr 7 at 15:38
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    $\begingroup$ @hopefully: Since $B e_i = \sum _j B_{ij} e_j$, we have $\langle e_i, B e_i \rangle = \langle e_i, \sum _j B_{ij} e_j \rangle = \sum _j B_{ij} \langle e_i, e_j \rangle = \sum _j B_{ij} \delta_{ij} = B_{ii}$. We have $\langle e_i, e_j \rangle = \delta_{ij}$ because the basis is orthonormal. $\endgroup$ – Alex M. Apr 7 at 16:12
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    $\begingroup$ @hopefully: This is how you define the matrix elements of a linear operator in a basis. $\endgroup$ – Alex M. Apr 7 at 17:50

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