Find the square root of a matrix Let $A$ be the matrix
$$A = \left(\begin{array}{cc}
41 & 12\\
12 & 34
\end{array}\right).$$
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 $A = U^T U$?
 A: Let $\lambda$ and $\mu$ be the eigenvalues of your $2$ by $2$ real matrix $A$. (We may have $\lambda=\mu$.) Assume that $\lambda$ and $\mu$ are positive. 
If $\lambda\not=\mu$, write the equation of the secant line to the curve $y=\sqrt x$ through the points $(\lambda,\sqrt\lambda)$ and $(\mu,\sqrt\mu)$: 
$$y=\sqrt\lambda\ \ \frac{x-\mu}{\lambda-\mu}+\sqrt\mu\ \ \frac{x-\lambda}{\mu-\lambda}\quad.$$ 
The matrix you want is 
$$\sqrt\lambda\ \ \frac{A-\mu I}{\lambda-\mu}+\sqrt\mu\ \ \frac{A-\lambda I}{\mu-\lambda}\quad,$$ 
where $I$ is the identity matrix. 
If $\lambda=\mu$, write the equation of the tangent line to the curve $y=\sqrt x$ through the point $(\lambda,\sqrt\lambda)$: 
$$y=\sqrt\lambda+\frac{x-\lambda}{2\sqrt\lambda}\quad.$$ 
The matrix you want is 
$$\sqrt\lambda\ I+\frac{A-\lambda I}{2\sqrt\lambda}\quad.$$ 
Do you see why? 
Do you see how to generalize this to $n$ by $n$ matrices? 
EDIT 4. This is to just explain why this secant/tangent stuff comes into the picture. Assume to simplify that the eigenvalues $\lambda$ and $\mu$ of your two by two real matrix $A$ are real and distinct. Let $f\in\mathbb R[X]$ be a polynomial, and $s$ the unique polynomial of degree $\le1$ which agrees with $f$ at $\lambda$ and $\mu$. [Graphically, this is a secant line.] Then the characteristic polynomial $$\chi=(X-\lambda)(X-\mu)$$ will divide $f-s$. As $\chi(A)=0$ by the Cayley-Hamilton Theorem, we have $f(A)=s(A)$. But the expression $s(A)$ makes sense whenever $f$ is a (real-valued) function defined at $\lambda$ and $\mu$. Moreover, the map $f\mapsto f(A)$ is compatible with addition and multiplication. 
EDIT 1. As noticed by @Did and @user1551, there is a cute formula for the "generalized secant line" to the curve $y=\sqrt x$, by which I mean: the secant line if the points are distinct, the tangent line if they coincide. Supposing $\lambda\not=\mu$, the equation of the secant line is 
$$y=\frac{\sqrt\lambda-\sqrt\mu}{\lambda-\mu}\ \ x+
\frac{\mu\sqrt\lambda-\lambda\sqrt\mu}{\lambda-\mu}=
\frac{x+\sqrt{\lambda\mu}}{\sqrt\lambda+\sqrt\mu}\quad,$$ 
and the miracle is that the last expression makes sense even if $\lambda=\mu$. 
EDIT 2. Note that there are other solutions when $\lambda\not=\mu$. Putting 
$$E:=\frac{A-\lambda I}{\mu-\lambda}\quad,\quad 
F:=\frac{A-\mu I}{\lambda-\mu}\quad,$$ 
we get $$E^2=E,\ F^2=F,\ EF=FE=0,\ I=E+F,\ A=\mu E+\lambda F,$$ 
and thus $$(\pm\sqrt\mu\ E\pm\sqrt\lambda\ F)^2=A$$ 
for the four choices of signs. [The plus plus choice corresponds to the previous formula.] 
EDIT 3. Here is a generalization. 
Let $T$ be an $n$ by $n$ complex matrix, and 
$$p(X)=(X-\lambda_1)^{m(1)}\cdots(X-\lambda_k)^{m(k)}$$ 
its minimal polynomial (the $\lambda_i$ being distinct and the $m(i)$ positive). Let $A$ be the algebra of those functions $f(z)$ which are holomorphic in a neighborhood of the spectrum $\{ \lambda_1,\dots,\lambda_k \}$ of $T$. 
There is a unique $\mathbb C[X]$-algebra morphism from $A$ to $\mathbb C[T]=\mathbb C[X]/(p(X))$. Denote this morphism by $f(z)\mapsto f(T)$. If $f(z)$ is in $A$, then the unique representative of $f(T)$ in $\mathbb C[X]$ of degree less than $\deg p(X)$ is 
$$\sum_{i=1}^k\ \ \underset{X=\lambda_i}\heartsuit\left(
\Big(\ \underset{z=\lambda_i}\heartsuit f(z) \Big)\ \ 
\frac{(X-\lambda_i)^{m(i)}}{p(X)}\ \right)\ 
\frac{p(X)}{(X-\lambda_i)^{m(i)}}$$ 
with 
$$\underset{u=\lambda_i}\heartsuit\varphi(u):=\sum_{j=0}^{m(i)-1}\frac{\varphi^{(j)}(\lambda_i)}{j!}\ (X-\lambda_i)^j.$$ 
Moreover, the $\lambda_i$-generalized eigenspace of $T$ is contained in the $f(\lambda_i)$-generalized eigenspace of $f(T)$. 
All this follows from the Chinese Remainder Theorem, which says 
$$\frac{\mathbb C[X]}{(p(X))}=\prod_{i=1}^k\ \ 
\frac{\mathbb C[X]}{(X-\lambda_i)^{m(i)}}\quad,$$ 
and from the Taylor Formula. 
[There is an Edit 4 above.]
A: Still another explicit formula: for every nonnegative real number $\alpha$,
$$
A^\alpha=\frac{(u^\alpha-v^\alpha)A+(uv^\alpha-vu^\alpha)I}{u-v}
$$
where $u$ and $v$ are the two roots of the polynomial $\chi_A(x)=\det(xI-A)$. When $\alpha=1/2$, this yields
$$
\sqrt{A}=\frac{A+\sqrt{uv}I}{\sqrt{u}+\sqrt{v}}
$$
Note that the coefficients of this formula can be computed directly from the matrix $A$ since $t=\sqrt{uv}$ is simply $t=\sqrt{\det A}$ and $s=\sqrt{u}+\sqrt{v}$ is such that $s^2=u+v+2t$ hence 
$$
s=\sqrt{\text{tr}(A)+2\sqrt{\det(A)}}
$$
and, finally,

$$
\sqrt{A}=\frac{A+\sqrt{\det A}I}{\sqrt{\text{tr}(A)+2\sqrt{\det(A)}}}
$$

In the present case, one can also compute $$\chi_A(x)=(x-41)(x-34)-12^2=(x-25)(x-50)$$ and use the values $\sqrt{u}=5\sqrt2$ and $\sqrt{v}=5$.
A: Write $$\pmatrix{a&b\cr c&d\cr}^2=\pmatrix{41&12\cr12&34\cr}$$ multiply out the left side, set corresponding entries equal to get four equations in the four unknowns $a,b,c,d$, then see if you can work your way through the algebra to a solution. 
A: This is an expansion of Arturo's comment.
The matrix has eigenvalues $50,25$, and eigenvectors $(4,3),(-3,4)$, so it eigendecomposes to $$A=\begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix} \begin{pmatrix}50 & 0 \\ 0 & 25\end{pmatrix} \begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix}^{-1}.$$
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
$$B=\begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix} \begin{pmatrix}\sqrt{50} & 0 \\ 0 & \sqrt{25}\end{pmatrix} \begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix}^{-1}$$
$$=\frac{1}{5}\begin{pmatrix}9+16\sqrt{2} & -12+12\sqrt{2} \\ -12+12\sqrt{2} & 16+9\sqrt{2}\end{pmatrix}.$$
Here we used $\sqrt{50}=5\sqrt{2},\sqrt{25}=5$, and a quick formula for the inverse of a $2\times 2$ matrix:
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}.$$
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:
$$A=\begin{pmatrix} a&0 \\ b&c \end{pmatrix} \cdot  \begin{pmatrix} a&b \\ 0&c \end{pmatrix} $$
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,
$$U=\sqrt{41}\begin{pmatrix}1&\frac{12}{41}\\0&\frac{25}{41}\sqrt{2}\end{pmatrix}. $$
A: The second is even simpler than the first question. All you need to to look at the assumed solution, setting unknowns:
$ \qquad \qquad \small A = U^t \cdot U = \begin{pmatrix} a&0 \\ b&c \end{pmatrix} \cdot  \begin{pmatrix} a&b \\ 0&c \end{pmatrix} \qquad  $ and $ \small a \cdot a = 41$. Then you can proceed; in Pari/GP it needs something like 5 lines of code...
A: For the first part of your question, here is a solution that only works for 2-by-2 matrices, but it has the merit that no eigenvalue is needed.
Recall that in the two-dimensional case, there is a magic equation that is useful in many situations. It is $X^2-({\rm tr}X)X+(\det X)I=0$, which arises from the characteristic polynomial of a $2\times2$ matrix $X$. Now, if $X^2=A$, we have $\det X=\pm\sqrt{\det A}=r$ (say). We take the positive value for $r$. Hence
$$
(\ast):\quad ({\rm tr}X)X=X^2+rI=A+rI
$$
and $({\rm tr}X)^2 = {\rm tr}\left(({\rm tr}X)X\right) = {\rm tr}(A+rI) = {\rm tr}A + 2r$. Thus, from $(\ast)$ we obtain
$$
X = \frac{1}{\sqrt{{\rm tr}A + 2r}}(A+rI)\quad {\rm where}\quad r=\sqrt{\det A}.
$$
This method works for all 2-by-2 matrices $A$ when $\det A\ge0$ and ${\rm tr}A + 2\sqrt{\det A}>0$. In particular, it works for positive definite $A$.
For the second part of your question, as the others have pointed out, the decomposition you ask for is a Cholesky decomposition.
A: 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 $$A^2 = $$
$$
\begin{pmatrix}
41 & 12\\
12 & 34
\end{pmatrix}
$$
Has solutions as 
Matrix $$A =$$
$$
\begin{pmatrix}
6.3254834 & 0.9941125\\
0.9941125 & 5.7455844
\end{pmatrix}
$$
And
Also
$$
\begin{pmatrix}
2.7254834 & 5.7941125\\
5.7941125 & -0.6544156
\end{pmatrix}
$$
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
