Given $A^TA$ how to find A If I am given a matrix $A^TA=\begin{bmatrix}50&-14\\-14&50\end{bmatrix}$
Is there a simple way to find A?
 A: For a geometric approach to this problem, if $A$ has column vectors $a_1$ and $a_2$, then knowing that $$A^TA = \begin{bmatrix}50 & -14 \\ -14 & 50\end{bmatrix}$$ tells us that $\langle a_1, a_1\rangle = \langle a_2, a_2\rangle = 50$ and $\langle a_1, a_2\rangle = -14$. From this, we may deduce that $\|a_1\| = \|a_2\| = \sqrt{50} = 5\sqrt2$ and $\angle(a_1, a_2) = \arccos \frac{\langle a_1, a_2\rangle}{\|a_1\| \, \|a_2\|} = \arccos -\frac{7}{25}$.
So we know the lengths of the two vectors and the angle between them, which obviously does not determine the two vectors: we may pick the direction of $a_1$ arbitrarily, and then compute $a_2$ by rotating $a_1$ clockwise or counterclockwise by $\arccos -\frac{7}{25}$. It's been pointed out that $$a_1 = \begin{bmatrix}7 \\ -1\end{bmatrix}, \quad a_2 = \begin{bmatrix}-1\\7\end{bmatrix}$$ work; we can find another example by picking $a_1$ to go in the positive $x$ direction, which gives: $$a_1 = \begin{bmatrix}5\sqrt2 \\ 0\end{bmatrix}, \quad a_2 = \begin{bmatrix}-\frac75 \sqrt2 \\ \frac{24}{5}\sqrt2\end{bmatrix}$$
A: No. The solution is not unique. For example
$$A=\begin{pmatrix}1&0\\0&1 \end{pmatrix}$$
and
$$B=\begin{pmatrix}0&1\\1&0 \end{pmatrix}$$
have
$$A^TA=B^TB=\begin{pmatrix}1&0\\0&1 \end{pmatrix}$$
More specifically, there are infinitely many solutions. What you can do is find the conditions if you assume the dimension of the matrix, e.g. if it is a square matrix. Let
$$C=\begin{pmatrix}c_{00}&c_{01}\\c_{10}&c_{11} \end{pmatrix}$$
Then
$$C^TC=\begin{pmatrix}c_{00}^2+c_{01}^2&c_{00}c_{10}+c_{01}c_{11}\\c_{00}c_{10}+c_{01}c_{11}&c_{10}^2+c_{11}^2 \end{pmatrix}$$
and you can now set each of these entries equal to the elements of your resulting matrix.
A: Given a matrix of the form $A^TA$ with $A$ a real square matrix, one example of a matrix $B$ such that $B^TB = A^TA$ is always given by taking the positive square root $B=\sqrt{A^TA}$.  Note that $A^TA$ is a real symmetric, positive semi-definite matrix, so you can orthogonally diagonalize it, $A^TA=SDS^T$, with $S^T=S^{-1}$, $D$ is a diagonal matrix with all of its diagonal entries nonnegative.  You can take $\sqrt{D}$ to be the diagonal matrix with the nonnegative square roots of the corresponding entries from $D$, and then define $B=S\sqrt{D}S^T$.  You can notice that $B^T=B$ and $B^TB=B^2=A^TA$.  
As an alternative to diagonalizing, if you find the eigenvalues $\{\lambda_1,\ldots,\lambda_n\}$, let $p(x)$ be a polynomial with real coefficients interpolating the points $\{(\lambda_1,\sqrt{\lambda_1}),\ldots,(\lambda_n,\sqrt{\lambda_n})\}$, and then take $B=p(A^TA)$.
Once you have one solution $B$, as Qiaochu commented you can get others by taking $UB$ where $U$ is any orthogonal matrix.
In your case the eigenvalues are $50\pm 14$, so the polynomial above should have $p(64)=8$ and $p(36)=6$, hence $p(x)=6+\frac{1}{14}(x-36)$ will do.  You would have $B=p(A^TA) = 6I+\frac{1}{14}\begin{bmatrix}14&-14\\-14&14\end{bmatrix}=\begin{bmatrix}7&-1\\-1&7\end{bmatrix}.$
A: Try
$$
A = 
\left(
\begin{array}{rr}
7 & -1 \\
-1 & 7
\end{array}
\right)
$$
Or
$$
A = 
\left(
\begin{array}{rr}
1 & -7 \\
-7 & 1
\end{array}
\right)
$$
