A question about positive definite matrix.

Question

$\mathbf{A}$ is a real positive definite matrix. Show that there exists an upper triangular matrix $\mathbf{B}$ such that $\mathbf{A}=\mathbf{B}\mathbf{B}^{\mathrm{T}}$, and all the numbers on the diagonal of matrix $\mathbf{B}$ are negative. Here $\mathbf{B}^{\mathrm{T}}$ denotes the transpose of $\mathbf{B}$. As we konw, a real positive definite matrix is congruent to an identity matrix. So, there exists a matrix $\mathbf{C}$ such that $\mathbf{A}=\mathbf{C}\mathbf{C}^{\mathrm{T}}$. But how to prove that numbers on the diagonal of matrix $\mathbf{C}$ are negative?

Answer 1

Consider the Cholesky decomposition of $A$:

$$A=LL^T$$

where $L$ is known to be upper triangular with $>0$ diagonal entries and take

$$C=-L$$.