Show that $A =\left(\begin{smallmatrix}41&12\\12&34\end{smallmatrix}\right)$ is symmetric positive definite After that it asks to find a 2 x 2 matrix B such that $A=B^2$
I was thinking to find the eigenvalues. If they are positive then it means the matrix is spd. 
Then try to diagonalize the matrix and find a matrix B that has the square root of the eigenvalues of A as elements of its diagonal.
Am I in the right direction?
Thanks.
 A: Another approach is to see that $B$ can be symmetric as well, so write $$B =\left(\begin{matrix}x&y\\y&z\end{matrix}\right)\\B^2 =\left(\begin{matrix}x&y\\y&z\end{matrix}\right) ^2=\left(\begin{matrix}41&12\\12&34\end{matrix}\right)\\
x^2+y^2=41,y(x+z)=12,y^2+z^2=34$$
For which Alpha finds a mess
A: Your approach is correct, but you needs also the eigenvectors of your matrix $A$.
When you diagonalize $A$ you find $A=PDP^{-1}$ where $P$ is a matrix that has as columns the eigenvectors of $A$.
If you find the matrix $C$ such that $C^2=D$ ( as you has suggested) the you have:
$$
B^2=(PCP^{-1})^2=PCP^{-1}PCP^{-1}=PC^2P^{-1}=PDP^{-1}=A
$$
So $B$ is the square root of $A$.
A: Alternative method: since $A$ is symmetric, it can be diagonalized within $\mathbb {R}^2$. The eigenvalues $\lambda _{1,2}$ satisfy: $$\det A= \lambda _1 \lambda _2,\quad \text {tr}A=\lambda _1 + \lambda _2;$$
Note that $\det A$ and $\text {tr} A$ are stable under conjugation $A\mapsto C^{-1} A C$.
Now if $\det A\leq0$ the matrix obviously can't be positive definite. If $\det A>0$, $\lambda _1 $ and $\lambda _2$ have the same sign, which is the same sign of $\text {tr}A$. So the symmetric $2\times 2$ matrix is positive definite if and only if: $$\det A>0\quad \text{and}\quad\text {tr}A >0.$$
You can also easily solve for $\lambda _1$ and $\lambda _2$, however finding $A^{\frac{1}{2}}$ still requires some computation. 
