Root of upper triangular matrices Suppose all of the entries on the diagonal of $P$ is nonzero. 
Is there any way to show that $P$ has a square root $A$ (i.e. $A^2=P$) by using Taylor expansion?
 A: The formulas given in the other answers don't come out of the blue. Let
$$J=\lambda I+N,\quad\lambda\ne0, \quad N\ {\rm nilpotent},$$
be a Jordan block of size $m$. Then we can write
$$J=\lambda\left(I+{N\over\lambda}\right)$$
and therefore
$$J^{1/2}=\lambda^{1/2}\left(I+{N\over\lambda}\right)^{1/2}\lambda^{1/2}=\lambda^{1/2}\sum_{k=0}^\infty{1/2\choose k}\left({N\over\lambda}\right)^k\ ,$$
whereby the series on the right only has $m$ nonzero terms.
A: I can start you off, these took a bit of work. With complex $t \neq 0,$
$$ $$
$$
\left(
\begin{array}{cc}
t & \frac{1}{2t} \\
0 & t
\end{array}
\right)^2  =
\left(
\begin{array}{cc}
t^2 & 1 \\
0 & t^2
\end{array}
\right) \; \; ,
$$
$$ $$
$$
\left(
\begin{array}{ccc}
t & \frac{1}{2t} & - \frac{1}{8 t^3} \\
0 & t &  \frac{1}{2t} \\
0 & 0 & t
\end{array}
\right)^2  =
\left(
\begin{array}{ccc}
t^2 & 1 & 0\\
0 & t^2 & 1 \\
0 & 0 & t^2
\end{array}
\right) \; \; .
$$
A: Let $f:\mathbb{C} \longrightarrow \mathbb{C}$ and suppose that $f^{(k)}(\lambda)$ is defined for $k=0,1,\dots,n-1$. It is well-known that if 
$$
J=
\begin{bmatrix}
\lambda & 1 & 0 & \cdots & \cdots & 0 \\
 & \lambda & 1 & 0 & \cdots & 0 \\
 & & \ddots & \ddots & \ddots & \vdots \\
 & & & \lambda & 1 & 0 \\
 & & & & \lambda & 1 \\
 & & & & & \lambda
\end{bmatrix} \in \textsf{M}_n(\mathbb{C}),
$$
then
$$
f(J)=
\begin{bmatrix}
f(\lambda) & f'(\lambda) & \frac{f''(\lambda)}{2} & \cdots & \cdots & \frac{f^{(n-1)}(\lambda)}{(n-1)!} \\
 & f(\lambda) & f'(\lambda) & \frac{f''(\lambda)}{2} & \cdots & \frac{f^{(n-2)}(\lambda)}{(n-2)!} \\
 & & \ddots & \ddots & \ddots & \vdots \\
 & & & f(\lambda) & f'(\lambda) & \frac{f''(\lambda)}{2} \\
 & & & & f(\lambda) & f'(\lambda) \\
 & & & & & f(\lambda)
\end{bmatrix}.
$$
For example, if $f(z) = \sqrt{z}$, and 
$$J
=
\begin{bmatrix}
\lambda & 1 & 0 \\
0 & \lambda & 1 \\
0 & 0 & \lambda 
\end{bmatrix}
$$
then the matrix
$$
B = \sqrt{J} =
\begin{bmatrix}
\sqrt{\lambda} & \frac{1}{2\sqrt{\lambda}} & -\frac{1}{8\lambda^{3/2}} \\
0 & \sqrt{\lambda} & \frac{1}{2\sqrt{\lambda}} \\
0 & 0 & \sqrt{\lambda} 
\end{bmatrix}
$$ 
satisfies $B^2=J$.
