The inverse of a SPD matrix is also symmetric. But what about the square root? Intuitively, I would say yes. But I'm not sure about it.
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Yes, it is. Infact it is true for any self-adjoint positive definite operator in Hilbert space by the spectral theorem. – Urgje May 13 '14 at 22:24
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An SPD matrix has a unique square root. Think of it as an analog of the scalar case: each positive real number has a positive square root. – Algebraic Pavel May 13 '14 at 22:40
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1Consider the square roots of the $2x2$ identity $I_{2}$ found by putting $\pm 1$ along the diagonal. – Disintegrating By Parts May 13 '14 at 23:33
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Each PSD matrix certainly has a symmetric square root. In fact, PSD matrices have a unique PSD square root. However, not every square root of a PSD matrix is symmetric.
For example, the $2 \times 2$ zero matrix is PSD, but $$ A = \pmatrix{0&1\\0&0} $$ Is not symmetric, even though $A^2$ is PSD.
EDIT: If, like Pavel (below) you need some positivity in your PSD matrix, you should consider $$ A = \pmatrix{1&0&0\\0&0&1\\0&0&0} $$
Ben Grossmann
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I would almost certainly think that the zero matrix is only semi-definite :) – Algebraic Pavel May 13 '14 at 22:36
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There exist both positive-definite and non-positive-definite square roots for a Symmetric Positive-Definite matrix.
If A is SPD, then A can be represented as UDU* for some unitary matrix U and diagonal matrix D, with the eigenvalues represented in D, and all eigenvalues positive.
If I let R = sqrt(D) then I claim that URU* is a square root of A. Notice:
(URU*)^2 = URU*URU* = URRU = UDU = A
To find a particular square root, I only need to worry about finding square roots of the eigenvalues of D. There will be both positive and negative values for the square roots and sometimes a mix of both.
However, a matrix is positive-definite only when all the eigenvalues are positive. You can indeed find a square root with all positive values, but you can also find some with both positive and negative eigenvalues.
Bill Province
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