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Given a real symmetric positive semi-definite matrix $A$, will there be a root $R$ which is real symmetric positive semi-definite as well? Can you comment on it's uniqueness? It will be nice if you can give proofs as well.

dineshdileep
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1 Answers1

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Consider the real Schur factorization of $A$, i.e. $A = UTU'$, where $U$ is orthogonal and $T$ is diagonal matrix with nonnegative elements $d_i$. Let $$A_{1/2} = UT_{1/2}U'$$ where $T_{1/2}$ is a real diagonal matrix with diagonal entries $d_i^{1/2}$. Then $$A_{1/2}A_{1/2} = UT_{1/2}U'UT_{1/2}U' = UT_{1/2}T_{1/2}U' = UTU' = A$$ hence, $A_{1/2}$ is a square root of $A$. Since all elements of $T_{1/2}$ are nonnegative, the matrix $A_{1/2}$ is real symmetric and semi positive definite (positive definite if $A$ is positive definite).

The square root of $A$ is not unique, since $-A_{1/2}$ is also the square root of $A$.

Pawel Kowal
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  • but, is the square root real? – dineshdileep Jul 13 '16 at 03:41
  • @dineshdileep yes, I edited my answer to make it more clear, that $A_{1/2}$ is real. – Pawel Kowal Jul 13 '16 at 04:11
  • sorry to bother more, but why is schur factorization real? I thought that $U$ can be complex in schur factorization. – dineshdileep Jul 13 '16 at 05:36
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    @dineshdileep If $A$ is an $n \times n$ matrix with real entries then $A$ can be factored into a product $UT U^T$ where $U$ is a real orthogonal matrix and $T$ is a real quasi upper triangular matrix. If $A$ is symmetric, then $T$ is diagonal. On wiki only complex Schur factorization is mentioned. Real factorization is described for example here, – Pawel Kowal Jul 13 '16 at 05:49
  • that makes sense!. +1 – dineshdileep Jul 13 '16 at 08:01