If A is positive definite (but not necessarily symmetric) can you decompose it?

Question

If A is a $2 \times 2$ matrix that is positive definite but may or may not be symmetric, does there exist another matrix B such that $A=B^TB$?

Answer 1

If that's the case, then $A^T = (B^T B)^T = B^T (B^T)^T = B^T B = A$, so $A$ is symmetric.

Answer 2

Suppose that there exists a matrix $B=\begin{bmatrix} a&b\\c&d \end{bmatrix}$, such that $$B^\top B = \begin{bmatrix} a^2+c^2 & ab+cd \\ ab+cd & b^2+d^2 \end{bmatrix} =\begin{bmatrix} X&Y \\ Y&W \end{bmatrix} =A$$ for given $X,Y,W.$ We assume the most complicated case where $Y\neq 0.$ The case where $Y = 0$ can be dealt in a similar way.

We want to check if the system with variables $a,b,c,d$: $$\left.\begin{array}{l} a^2+c^2=X\\ ab+cd=Y\\ b^2+d^2=W \end{array} \right\} $$ has a solution. Also we need to add an extra equality, since it must hold that $$\det B\cdot \det B^T = \det A \implies (\det B)^2 = \det A.$$

Thus, an extra equation which must be satisfied is $$ad-bc = \sqrt{\det A}.$$ (One could choose the equation $ad-bc = -\sqrt{\det A}$). Hence, we have a system of 4 equations with 4 variables.

With the use of software one can attain the solution for $a,b,c,d$:

$$\displaystyle\begin{cases} a= \sqrt{X-c^2}\\[2ex] b= \dfrac{aY-c\sqrt{\det A}}{X}\\[2ex] d= \dfrac{cY^2 +a Y\sqrt{\det A}}{XY} \end{cases} $$ The variable $c$ is a free variable as long as $|c|\le \sqrt{X}.$

The result is that we can always construct a $2\times 2$ matrix $B$ such that $B^\top B = A.$