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Suppose an SVD decomposition of a $3\times3$ real invertible matrix $F = U K W$, where $ U U^T = U^TU=WW^T=W^TW=I$ and $K = $ diag$(k_1,k_2,k_3)$. Now suppose $ F = A B $ where $A$ and $B$ have SVD decompositions $A = U K_a V_a$ and $B = V_b K_b W$ where $V_a,V_b$ are also orthogonal, is there a relation between $V_a$ and $V_b$? I'd imagine that since necessarily $$ K = K_a V_a V_b K_b $$ and all $K$'s are diagonal, then the product $V_aV_b$ must also be diagonal. Does this imply that they are each other's inverse? Or can the product of two orthogonal matrices be diagonal without there being a relation between the two?

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  • $\begingroup$ As to your last question, take any pair of $2\times2$ rotation matrices whose rotation angles add up to $\pi$. There’s a relation of sorts, but they’re certainly not each other’s inverses. $\endgroup$
    – amd
    Aug 13, 2018 at 18:02

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The product of two orthogonal matrices is an orthogonal matrix. The only real orthogonal matrices that are diagonal are those whose diagonal entries are $\pm 1$. Conversely, if $A$ is an orthogonal matrix and $D$ a diagonal matrix whose diagonal entries are $\pm 1$, $B = A^{-1} D$ is an orthogonal matrix such that $AB = D$.

Of course, in your SVD case the diagonal matrices have positive diagonal elements, so then $V_a V_b = K_a^{-1} K K_b^{-1}$ must be $I$.

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  • $\begingroup$ Can you prove that the only real orthogonal matrices that are diagonal are those whose diagonal are $\pm 1$? $\endgroup$
    – TMG
    Aug 14, 2018 at 9:31
  • $\begingroup$ Yes, it's easy. If $D$ is diagonal with diagonal entries $d_1, \ldots, d_n$, then $D D^T$ is diagonal with diagonal entries $d_1^2, \ldots, d_n^2$. $\endgroup$ Aug 14, 2018 at 15:25
  • $\begingroup$ Ah awesome yeah that makes sense! Thank you! $\endgroup$
    – TMG
    Aug 14, 2018 at 17:37

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