Consider a real positive semi-definite matrix $M$, it is known that there exists a unique real positive semi-definite matrix $S$ such that $S^2=M$ and there is also a unique real lower-triangular matrix $C$ with non-negative diagonal such that $CC^*=M$.

Can we conclude that $C$ and $M$ coincide in the diagonal? It seems to be true for all the matrices I have tried.

  • $\begingroup$ One of the matrices is the cholesky matrix and the other is the principal square root. $\endgroup$ – Jorge Fernández Hidalgo Jan 27 '17 at 0:54
  • $\begingroup$ Per this question, the diagonal entries of $C$ will always have the form $$ C_{kk} = \sqrt{\frac{\det M_k}{\det M_{k-1}}} $$ I don't believe that this is true for $S$. $\endgroup$ – Omnomnomnom Jan 27 '17 at 1:02
  • $\begingroup$ dang it, I just need a program to calculate the square root of a matrix in c++ , but I only need the diagonal. But I haven't been able to find a package that handles this in the "semi-definite" case ;( $\endgroup$ – Jorge Fernández Hidalgo Jan 27 '17 at 1:04
  • $\begingroup$ Armadillo can do it, but it crashes in the semi-definite case, it's a huge bummer ;'( $\endgroup$ – Jorge Fernández Hidalgo Jan 27 '17 at 1:05
  • $\begingroup$ Well, if you can find the kernel of $A$, then you can reduce the problem to finding the square root of a smaller positive definite matrix. $\endgroup$ – Omnomnomnom Jan 27 '17 at 1:09

The statement fails to hold for $$ M = \frac 13\pmatrix{1&1&1\\1&1&1\\1&1&1} \implies \\ C = \frac 1{\sqrt 3}\pmatrix{1&0&0\\1&0&0\\1&0&0}, \quad S = M $$ I'm not sure about the $2 \times 2$ case, specifically. I would think that the statement also fails for $M + \epsilon I$, with $\epsilon > 0$ sufficiently small.


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