Prove or disprove: the Hilbert-Schmidt norm is independent of the choice of basis on $\mathbb{R^n}$

I have found this question here:

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?

• 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? – Kabo Murphy Apr 1 at 12:30
• @KaviRamaMurthy the general consequence of the steps of the proof is not clear for me ..... what did he do to prove the required? – Secretly Apr 1 at 12:48
• I even do not understand where is the solution in the link mentioned @hopefully – Mathstupid Apr 1 at 15:05
• My professor said that this statement is wrong and its correct only if we have an orthonormal basis @KaviRamaMurthy – 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.

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

• why in the forth line the last inequality the trace appears? what rule are you using? – Secretly Apr 7 at 15:11
• I think $||A||^2$ instead of $||A||$only in the forth line first equality. – Secretly Apr 7 at 15:19
• @hopefully: And I'm giving you a counterexample when $n=1$. – Alex M. Apr 7 at 15:38
• @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. – Alex M. Apr 7 at 16:12
• @hopefully: This is how you define the matrix elements of a linear operator in a basis. – Alex M. Apr 7 at 17:50