For a given $A$, find a matrix $B$ such that $B^2=A$ Find a matrix $B$ such that $B^2=A$, where
$$A = \begin{bmatrix}9 & 4 & -8 & 4\\4 & 9 & -4 & 8\\-8 &-4 & 9 & -4\\4 & 8 & -4 & 9\end{bmatrix}.$$
The matrix is symmetric and real, therefore is diagonizable with positive eigenvalues, so after the hard work of diagonizing the matrix the problem becomes trivial. My question is: there has to be an easier way to do it, but what is it?
Thanks for your time.
 A: You can try Newton's square-root-finder applied to matrices:              
$ \qquad \qquad  \displaystyle B = I \\ 
\qquad \qquad  B = (A \cdot B^{-1} + B) / 2 \qquad \qquad \text{ iterate to convergence }$
Trying it myself : after 7 iteration I've approximation to half-integers with so many digits precision that I have confidence to the rounded result:
$$
 B = \frac 12 \cdot \small \begin{bmatrix} 
 5 & 1 & -3 & 1 \\ 
 1 & 5 & -1 & 3 \\ 
 -3 & -1 & 5 & -1 \\ 
 1 & 3 & -1 & 5
 \end{bmatrix}
$$
Indeed, the final test $ B ^2 - A = 0$ holds.
A: It is easy to see that $(1,1,-1,1)$ is an eigenector with eigenvalue 25.
With a bit more work, $(1,0,1,0)$ and $(0,1,0,-1)$ are eigenvectors with eigenvalue 1.
The last one is harder to guess, but remember that it is a real symmetric matrix can diagonalized by an orthogonal matrix, which means that the last eigenvector is orthogonal to the 3 that we found. Hence, it is $(1, -1,-1, -1)$.
A: Another -simpler- way, but which is still employing the concept of diagonalization (which seems to be what you wish according to your comment at the question-box) is using SVD-decomposition. Since A is symmetric the SVD and the diagonalization coincide. SVD (and in this case diagonalization too) can be obtained by column/row-rotations only which might be seen as simpler.
