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?


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?


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?

  • $\begingroup$ 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

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.


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

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 $\| f_1 \| ^2 = f_1 \cdot f_1 = 2 \cdot 2 = 4 \ne 1$.

  • $\begingroup$ 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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