Given a matrix $A$, how to find matrix $B$ such that $B^T B=A$?

Question

Is it possible to find matrix $B$ if we knew matrix $A$ and $B^T B=A$?

For example: $A$ is a symmetric matrix and $$ A = \begin{pmatrix} 1 & 2& 3 \\ 2 & 1 & 4 \\ 3 & 4 & 1 \\ \end{pmatrix}? $$

Answer 1

Hint: Cholesky decomposition.

Note that there are more than one such $B$ matrix.

Answer 2

It is possible to find the matrix $B$, as long as $A$ is a positive definite matrix (which are also always symmetric). In this case, $A$ is diagonalizable such that $A=PDP^T$, where $P$ is an orthogonal matrix (i.e., $P^T = P^{-1}$, or $P^T P = I$) and $D$ is a diagonal matrix whose entries are the eigenvalues $\lambda_i$ of $A$.

Since $A$ is positive definite, these eigenvalues $\lambda_i$ are positive. Thus, we can define a diagonal matrix $C$ whose entries are $\sqrt{\lambda_i}$, such that $D=CC$. Utilizing this trick, we can find $B$ as follows:

$A = PDP^T = PCC P^T = PCP^T P CP^T = (P^T)^TC^TP^TP C P^T = (PCP^T)^T(PCP^T) = B^TB $,

such that we find $B = P C P^T$.

Remark: In the second to last step, we utilized the fact that $(ABC)^T = C^T B^T A^T$.

Answer 3

Your matrix is indefinite,; you need a diagonal matrix in the middle that matches the counts according to Sylvester's Law of Inertia. I don't think all integers is possible. $$ \left( \begin{array}{rrr} 4 & 3 & 0 \\ 5 & 4 & 1 \\ 0 & 1 & 0 \end{array} \right) \left( \begin{array}{rrr} - \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - \frac{5}{2} \end{array} \right) \left( \begin{array}{rrr} 4 & 5 & 0 \\ 3 & 4 & 1 \\ 0 & 1 & 0 \end{array} \right) = \left( \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 3 & 4 & 1 \end{array} \right) $$