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Let us consider complex, square traceless matrix $A$. We know that it is unitarily similar to matrix which have 0 on its main diagonal, that is for $B = QAQ^H, b_{ii} = 0$.

Now, let us consider following, let A has SVD $A = U\Sigma V^H$. Let $B$ be a unitarily similar matrix to $A$ with zero diagonal. $B$ then has SVD $B =QU \Sigma V^H Q^H = QU \Sigma (QV)^H$.

Now, each $i$-th diagonal of matrix $A$ can be represented as $A_{ii} =\sigma_i \langle u_i, v_i \rangle$, analogously for $B_{ii} = \sigma_i \langle Q u_i, Q v_i\rangle$. Now because $Q$ is unitary and unitary matrices preserve inner product we can write: $$ 0 = B_{ii} = \sigma_i \langle Q u_i, Q v_i\rangle = \sigma_i \langle u_i, v_i\rangle\\ $$

By definition of singular values $s_i \geq 0$. So in cases where $s_i \neq 0, \langle u_i, v_i\rangle = 0$.

I wonder if this is correct or have I made a mistake somewhere?

Freechoice guy
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    The columns of $U$ are left singular vectors that you reference in your title i.e. consider the equation $AV = U \Sigma$ -- it is the columns of $U$ and $V$ that are singular vectors of $A$. Denote those columns $\mathbf u_k$ and $\mathbf v_k$. Then $A = \sum_{k=1}^n \sigma_k \mathbf u_k \mathbf v_k^$ which is not an inner product (actually an "outer product"). So I'd like to see proof of your assertion "Now, each $i$-th diagonal of matrix $A$ can be represented as* $A_{ii} =\sigma_i \langle u_i, v_i \rangle$," – user8675309 Oct 18 '24 at 15:17
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    In terms of really simple counter-examples, let $B=A$ be the $2\times 2$ non-identity permutation matrix. – user8675309 Oct 18 '24 at 15:25

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