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square root of symmetric matrix and transposition

If I have a matrix K that is symmetric and positive definite, is there a way to decompose it into matrices K = A*A'?

I had a look at, however the closest factorization is of the form K = A*S*A'.

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marked as duplicate by J. M., Asaf Karagila, t.b., Martin Sleziak, Rudy the Reindeer Dec 12 '11 at 21:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

up vote 3 down vote accepted

Yes. The Cholesky decomposition takes a symmetric(/Hermitian) positive-(semi)definite matrix $K$ and computes a lower-triangular matrix $L$ with

$$K = LL^H.$$

(When $K$ is real, $L^H = L^T$.)

This decomposition is quite common and important in numerical linear algebra, as computing the Cholesky decomposition and backsolving is one of the fastest and most robust algorithms for solving the linear system $Kx=b$. For more details see the Wikipedia page.

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I guess you are working over $\mathbb{R}$, so let $K \in M_n(\mathbb{R})$. As $K$ is symmetric, you can find a matrix $T \in GL_n(\mathbb{R})$ such that $T^{-1}KT = \text{diag}(\lambda_1,...,\lambda_n)$ and all $\lambda_i$ are real. As $K$ is positive definite, we have $\lambda_i > 0$ for all $i$. By Gram-Schmidt you may assume that $T^{-1} = T^t$. Then set $$A := \text{diag}(\sqrt{\lambda_1},...,\sqrt{\lambda_n}) \, .$$ As $A^2 = T^{-1}KT$, you have $K=TA^2T^{-1} = T A A T^t$ and if you set $S =AT^t$ you get $$S^tS = (AT^t)^tAT^t = (T^t)^tA^tAT^t = TAAT^t = K \, .$$ Here I have used $(AB)^t=B^tA^t$ and $(A^t)^t=A$ for arbitrary matrices and $A^t=A$ for a diagonal matrix.

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