Find matrix given matrix times its transpose

One answer would be the Cholesky decomposition. In general, $A$ isn't unique; there are many possible answers that would fit your criteria.

EDIT:

I think I want to mention another way to do this, namely eigendecomposition. With this decomposition you can find:

$$AA^T=Q\Lambda Q^T$$

Here $Q$ is orthogonal and has the eigenvectors in it's columns and $\Lambda$ is diagonal and has the corresponding eigenvalues. Then you can have $A=Q\sqrt\Lambda$ where you get the square root of $\Lambda$ by simply taking the square roots of its diagonal elements.

I suggested beginning with $$ \left( \begin{array}{rr} a & b \\ c & d \\ \end{array} \right) \left( \begin{array}{rr} a & c \\ b & d \\ \end{array} \right) = \left( \begin{array}{rr} 14 & 12 \\ 12 & 16 \\ \end{array} \right) $$

or $$ a^2 + b^2 = 14 \; , \; \; ac+bd = 12 \; , \; \; c^2 + d^2 = 16$$

Drawing a picture in the plane, this means I can take a vector $(a,b)$ of squared length $14,$ a vector $(c,d)$ of squared length $16,$ with their dot product $12,$ meaning the angle between them must be $ \arccos \frac{3}{\sqrt{14}}, $ approximately $36.7^\circ$

As in the diagram, I just picked $(c,d) = ( 4,0).$ Then taking $(a,b)$ in the first quadrant, we see that $a = 3,$ from which we find $b = \sqrt 5.$

Giving $$ \left( \begin{array}{rr} 3 & \sqrt 5 \\ 4 & 0 \\ \end{array} \right) \left( \begin{array}{rr} 3 & 4 \\ \sqrt 5 & 0 \\ \end{array} \right) = \left( \begin{array}{rr} 14 & 12 \\ 12 & 16 \\ \end{array} \right) $$

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