This proof is short only because arguably the $QR$ decomposition is doing most of the heavy lifting.

Statement Suppose $B$ is a positive definite matrix. Then there exists a unique upper triangular matrix $R$ with only positive numbers on its diagonal such that

\[B = R^{\ast} R\]

where $R^{\ast}$ is the adjoint of $R$.

Proof

(Note that because $B$ is positive definite, it implies that $B$ is invertible and also that $B$ is a square matrix.)

\[\begin{aligned} B &= A^2 & \text{(Every positive definite matrix has a unique positive square root)} \\ &= A^{\ast}A & \text{(Every positive operator is self-adjoint)} \\ &= (QR)^{\ast}(QR) & \text{(Every full-rank square matrix can be QR decomposed)} \\ &= R^{\ast} Q^{\ast} QR \\ &= R^{\ast}R & \text{($Q^{\ast}Q = I$ because $Q$ is unitary)} \end{aligned}\]

Above we have used the fact that since $B$ is invertible, $A$ is also invertible and therefore full-rank.

The $R$ is unique as desired as the $QR$ decomposition is unique if $Q$ is unitary and $R$ is upper triangular with only positive entries in the diagonal.

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