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I have a question about Singular Value Decomposition.

Is it necessary for the orthogonal matrices used in the factorisation to have their determinants$=1$?

I ask because I attempted to decompose a matrix and it did not come out equal. I then multiplied one of the eigenvectors (one of the rows) of $V^T$ by $-1$, thus changing the sign of the determinant, and everything worked out.

Everything else seemed to be in order, so the determinant changing from $-1$ to $1$ seems to me to be the cause of the correction.

For some reason, I see a connection between this problem and the fact that we take the absolute value of the determinant in the formula for change of variables for multivariable integration.

Any explanation would be greatly appreciated.

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1 Answer 1

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No, it is not necessary for the orthogonal matrices in a Singular Value Decomposition to have determinant equal to $1$. I have just read elsewhere that it is necessary that $\mathbf{u}_i=\frac{A\mathbf{v}_i} {\sigma_i}.$ This means, in my example, that I can make everything work out by changing the matrix $U$ so that $\mathbf{u}_i=\frac{A\mathbf{v}_i} {\sigma_i}$ holds even when my $V^T$ has determinant $-1$.

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