There exists positive definite matrix $B$ such that $A=B^TB$ Let $A=\begin{bmatrix}
a &b \\ 
b &d 
\end{bmatrix}$ be a positive definite matrix. I want to prove or disprove that there exists a positive definite matrix $B$ satisfying $A=B^TB$.
In my understanding, such a matrix could be the upper triangular matrix $B=\begin{bmatrix}
\sqrt{a} & b/\sqrt{a} \\ 
0 & \sqrt{d-b^2/a} 
\end{bmatrix}$,
which is positive definite. But then some issues arise immediately:


*

*The need to prove that $a>0$. (This must be implied by (b), I think).$\ $

*That $d-b^2/a \ge 0$. (This one actually follows from $B$ being positive definite).


Is there an easier way to prove the existence of $B$?
 A: We use a few well-known properties of matrices:  first, $A$ is symmetric; thus there exists an orthogonal matrix $O$ such that
$D = O^TAO, \tag{1}$
where $D$ is a diagonal matrix 
$D = diag(d_1, d_2), \tag{2}$
where the $d_i$ are the eigenvalues of $A$.  Since $A$ is positive definite, the $d_i > 0$.  Set
$r_i = \sqrt d_i; \tag{3}$
set
$R = diag(r_1, r_2); \tag{4}$
then
$R^2 = D, \tag{5}$
set
$B = ORO^T; \tag{6}$
we have
$B^T = (ORO^T)^T = O^{TT} R^T O^T = ORO^T = B, \tag{7}$
whence
$B^TB = B^2 = ORO^TORO^T = OR^2O^T = ODO^T = A. \tag{8}$
NOTE: This proof easily extends to show that for any positive definite symmetric matrix $A$ there is a symmetric positive definite $B$ such that $A = B^2$.
A: A is positive definite, hence, all of its eigenvalues are positive. So we know $det(A)\gt 0$ and $tr(A)\gt 0$.
This translates to:
$$det(A)=ad-b^2\gt 0\\$$
Which implies a and d have the same sign.
$$tr(A)=a+d>0$$
Which implies, under the previous condition, that a>0.
And these are both of the conditions you required.
