Singular value decomposition of identity matrix

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If you run a SVD on a identity matrix, $I = U S V^T$, will the matrix $S$ also be an identity matrix?

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1 Answer

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In general, given any $m \times n$ matrix $A$ with its SVD

$$ A = U \Sigma V^T $$

the matrix $\Sigma$ is a diagonal matrix with singular values of $A$ on the diagonal entries. Furthermore, these non-zero singular values are the square roots of the eigenvalues of $AA^T$ (or $A^TA$ which has the same set of eigenvalues).

Now in your case, $I^TI = I$, and the only eigenvalue of $I$ is $1$, therefore, the singular value of $I$ is $1$ which implies that $\Sigma = I$.

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