Substitute the SVD for A and AT to show that AT A has its eigenvalues in I;TI; and AAT has its eigenvalues in I:I:T. Since a diagonal I:TI: has the same nonzeros as I:I:T, we see again that AT A and AAT have the same nonzero eigenvalues.
Substitute the SVD for A and AT to show that AT A has its eigenvalues in I;TI; and AAT has its eigenvalues in I:I:T. Since a diagonal I:TI: has the same nonzeros as I:I:T, we see again that AT A and AAT have the same nonzero eigenvalues.
Let A be an m x n matrix, with singular value decomposition A = UΣV^T, where U is an m x m orthogonal matrix, Σ is an m x n diagonal matrix with non-negative diagonal entries σ1 ≥ σ2 ≥ ... ≥ σn ≥ 0 (the singular values of A), and V is an n x n orthogonal matrix.
Then, we have:
AT A = (UΣV^T)T(UΣV^T) = VΣTU^TUΣV^T = VΣTΣV^T
Similarly,
AAT = (UΣV^T)(UΣV^T)T = UΣV^TVΣTU^T = UΣΣTU^T
Now, let's consider the eigenvalues of AT A and AAT.
Eigenvalues of AT A:
Suppose λ is an eigenvalue of AT A, and let x be the corresponding eigenvector. Then,
AT Ax = λx
Multiplying both sides by A, we get:
AAT Ax = Aλx
Now, we can substitute AAT = UΣΣTU^T and AT A = VΣTΣV^T:
UΣΣTU^T Ax = VΣTΣV^T VΣTU^T x
Since U and V are orthogonal matrices, they preserve the lengths of vectors, and thus their eigenvalues have magnitude 1. Therefore, we can multiply both sides by U^T and V to get:
ΣΣTV^T Ax = ΣTΣU^T x
Multiplying both sides by VΣ^-1 on the left and Σ^-1U on the right, we get:
VΣ^-1ΣΣTV^T Ax = Σ^-1ΣTΣU^T x
Since Σ is a diagonal matrix with non-negative diagonal entries, Σ^-1 is well-defined and has diagonal entries 1/σ1, 1/σ2, ..., 1/σn (with σi = 0 if i > r, where r is the rank of A). Thus, we have:
VΣTV^T (VΣ^-1Ax) = ΣTΣU^T (Σ^-1Ux)
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