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For a block diagonal matrix, we have an identity for its cholesky decomposition i.e.

$chol(Z) = chol(blockdiag(A,B,...)) = blockdiag(chol(A),chol(B),...)$

(Here, $Z = blockdiag(A,B,...)$)

I want to know whether there any such similar identities for SVD of a block diagonal matrix.

Thanks.

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Yep. If $A = U_A D_A V_A$ and $B = U_B D_B V_B$ with the $U$ and $V$ matrices unitary and the $D$ matrices diagonal, then it is easy to check that $$\operatorname{blockdiag}(A, B) = U \operatorname{blockdiag}(D_A,D_B) V$$ with $U = \operatorname{blockdiag}(U_A,U_B)$ and $V = \operatorname{blockdiag}(V_A,V_B)$ unitary.

Notice that the block SVD does not generally have its singular values in descending order, as is commonly the convention.

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