where A^t is the transpose of A
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The least eigenvalue of A^TA is 0. The matrix A^TA is necessarily positive semi-definite. To see this, note that for any x in IRⁿ (where A is n x n)
x^T(A^TAx) = (Ax)^T(Ax) = ||Ax||² ≥ 0
where ||•|| is the standard 2-norm. Since A is singular, 0 is an eigenvalue of A. Hence 0 is an eigenvalue of A^TA--and since the eigenvalues can't be negative, this must be the smallest.
x^T(A^TAx) = (Ax)^T(Ax) = ||Ax||² ≥ 0
where ||•|| is the standard 2-norm. Since A is singular, 0 is an eigenvalue of A. Hence 0 is an eigenvalue of A^TA--and since the eigenvalues can't be negative, this must be the smallest.