Find the square root of a matrix Let $A$ be the matrix
$$A = \left(\begin{array}{cc}
41 &amp; 12\\
12 &amp; 34
\end{array}\right).$$

I want to decompose it into the form of $B^2$.

I tried diagonalization , but can not move one step further.

Any thought on this? Thanks a lot!

ONE STEP FURTHER:

How to find a upper triangular $U$ such that $A = U^T U$?
This is an expansion of Arturo's comment.

The matrix has eigenvalues $50,25$, and eigenvectors $(4,3),(-3,4)$, so it eigendecomposes to $$A=\begin{pmatrix}4 &amp; -3 \\ 3 &amp; 4\end{pmatrix} \begin{pmatrix}50 &amp; 0 \\ 0 &amp; 25\end{pmatrix} \begin{pmatrix}4 &amp; -3 \\ 3 &amp; 4\end{pmatrix}^{-1}.$$

This is of the form $A=Q\Lambda Q^{-1}$. If this is $B^2$, then there will be a $B$ of the form $Q\Lambda^{1/2} Q^{-1}$ (square this to check this is formally true). A square root of a diagonal matrix is just the square roots of the diagonal entries, so we have

$$B=\begin{pmatrix}4 &amp; -3 \\ 3 &amp; 4\end{pmatrix} \begin{pmatrix}\sqrt{50} &amp; 0 \\ 0 &amp; \sqrt{25}\end{pmatrix} \begin{pmatrix}4 &amp; -3 \\ 3 &amp; 4\end{pmatrix}^{-1}$$

$$=\frac{1}{5}\begin{pmatrix}9+16\sqrt{2} &amp; -12+12\sqrt{2} \\ -12+12\sqrt{2} &amp; 16+9\sqrt{2}\end{pmatrix}.$$

Here we used $\sqrt{50}=5\sqrt{2},\sqrt{25}=5$, and a quick formula for the inverse of a $2\times 2$ matrix:

$$\begin{pmatrix}a&amp;b\\c&amp;d\end{pmatrix}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&amp;-b\\-c&amp;a\end{pmatrix}.$$

Keep in mind that matrix square roots are not unique (even up to sign), but this particular method is guaranteed to produce one example of a real matrix square root whenever $A$ has all positive eigenvalues.



Finding an upper triangular $U$ such that $A=U^TU$ is even more straightforward:

$$A=\begin{pmatrix} a&amp;0 \\ b&amp;c \end{pmatrix} \cdot  \begin{pmatrix} a&amp;b \\ 0&amp;c \end{pmatrix} $$

This is $a^2=41$ hence $a=\sqrt{41}$, $ab=12$ hence $b=\frac{12}{41}\sqrt{41}$, and $b^2+c^2=34$ hence $c=25\sqrt{\frac{2}{41}}$.

In other words,

$$U=\sqrt{41}\begin{pmatrix}1&amp;\frac{12}{41}\\0&amp;\frac{25}{41}\sqrt{2}\end{pmatrix}. $$
As traditional known the square root of any number would have two result, its cube root would have three results and so on
The matrix $$A^2 = $$
$$
\begin{pmatrix}
41 &amp; 12\\
12 &amp; 34
\end{pmatrix}
$$
Has solutions as 
Matrix $$A =$$
$$
\begin{pmatrix}
6.3254834 &amp; 0.9941125\\
0.9941125 &amp; 5.7455844
\end{pmatrix}
$$
And
Also
$$
\begin{pmatrix}
2.7254834 &amp; 5.7941125\\
5.7941125 &amp; -0.6544156
\end{pmatrix}
$$
This is a clear solution to the question, if you really checked well you'll see it.
I don't know why anyone here can block someone's comment, claiming it to be wrong