Compact operators are orthogonally equivalent to a diagonal matrix?

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On Brezis's Functional Analysis, the last question of Problem 44 (near the end of the book) reads (modified to include context)

Assume that the Hilbert space $H$ is separable and $T\in\mathcal K(H)$ (but $T$ doesn't necessarily satisfy the following condition: there exists an isometry $J\colon\ker T\to\ker T^*$), the set of compact operators on $H$. Prove that there exist two orthogonal bases $(e_n)$ and $(f_n)$ of $H$ such that $$Tu=\sum_{n=1}^\infty\alpha_n(u,e_n)f_n$$ where $(\alpha_n)$ is a sequence of nonnegative real numbers and $\alpha_n\to0$ as $n\to\infty$.

Since one should assume the operator to be partial isometric instead of orthogonal/unitary, as the wiki page pointed out. Even if the operator is compact, I think it's still necessary since we can compose the one-sided shift with a compact operator corresponding to a diagonal matrix.

So I think this question is in fact false, but I doubt that I got stuck somewhere, since Brezis stressed in the parenthesis that the condition about existence of isometry isn't necessary. I want your help to clarify my misunderstandings. Thanks!