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Show that if an $n × n$ matrix $A$ is positive definite, then there exists a positive definite matrix $B$ such that $A = B^tB$.

The way I set out to show this was (note that P is an orthonormal basis, thus $P inverse = P^T)$:

$A=PDP^t B=PCP^t$, Thus if we can prove $c$ has only positive values, we know $B$ has only positive eigenvalues and hence is positive definite $A=B^tB=(PCP^t)^t*PCP^t=PC^tP^t*PCP^t=PC^tCP^t$. We know that $C^tC=D$, and we know $D$ is all positive values since $A$ is positive definite, however, how does this guarantee the same about $C$? Couldn't $C_{ii}$ (diagonal elements) be +or - $sqrt(D_{ii})$, meaning that $C$ can have all negative or all positive or mixed values, thus telling us nothing about what kind of matrix $B$ is?

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We proceed with the assumption that positive definite matrices are symmetric.

First since $\boldsymbol A$ is a symmetric real matrix, $\boldsymbol A$ could be orthogonally diagonalized by some matrix $\boldsymbol T$, i.e. $\boldsymbol A = \boldsymbol T^{\mathsf T}\boldsymbol {DT}$, where $\boldsymbol T$ is orthogonal and $\boldsymbol D$ is diagonal. Since $\boldsymbol A$ is positive definite [and we denote this by $\boldsymbol A > 0$], the eigenvalues of $\boldsymbol A$ are strictly positive, hence $\boldsymbol D = \mathrm {diag}(z_j^2)_1^n [z_j \in \mathbb R^+]$. Let $\boldsymbol B = \boldsymbol T^{\mathsf T} \mathrm {diag}(z_j)_1^n \boldsymbol T$ and the decomposition is complete, since the all eigenvalues of $\boldsymbol B$ are positive hence $ \boldsymbol B > 0$ as well, and $\boldsymbol A = \boldsymbol B ^{\mathsf T} \boldsymbol B$ by direct computation.

If the assumption of symmetry was dropped, then maybe the problem would be somehow difficult.

Reply to your post: we only have to find 1 matrix whose eigenvalues are all positive, thus just take that $\boldsymbol C$ with all $+\sqrt{D_{j,j}}$ and the existence is therefore established.

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  • $\begingroup$ Thank you so much! $\endgroup$ – Paul Blart Dec 18 '17 at 17:07

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