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.

The primary reference used in this post is (Axler, 2024).

Bibliography
  1. Axler, S. J. (2024). Linear Algebra Done Right, 4th edition. Springer. http://linear.axler.net/