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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?

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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.

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I can start you off, these took a bit of work. With complex $t \neq 0,$

$$ $$

$$ \left( \begin{array}{cc} t & \frac{1}{2t} \\ 0 & t \end{array} \right)^2 = \left( \begin{array}{cc} t^2 & 1 \\ 0 & t^2 \end{array} \right) \; \; , $$

$$ $$

$$ \left( \begin{array}{ccc} t & \frac{1}{2t} & - \frac{1}{8 t^3} \\ 0 & t & \frac{1}{2t} \\ 0 & 0 & t \end{array} \right)^2 = \left( \begin{array}{ccc} t^2 & 1 & 0\\ 0 & t^2 & 1 \\ 0 & 0 & t^2 \end{array} \right) \; \; . $$

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    $\begingroup$ The general term for the $k$-th off-diagonal seems to be $$\frac{1}{t^{2k-1}} \binom{1/2}{k}. $$ $\endgroup$ – Trevor Gunn Feb 21 '18 at 3:19
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    $\begingroup$ @TrevorGunn, yeah. From her wording, I am guessing her professor began with $t^2 I + N$ with $N$ the nilpotent part and took the square root as a finite power series, $t I \sqrt {I + \frac{N}{t^2}}.$ I did not go that way, I did a bunch of Jordan forms of various things $\endgroup$ – Will Jagy Feb 21 '18 at 3:24
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    $\begingroup$ @TrevorGunn if I have it right, terms will come from (-1 + sqrt( 1 + w^2 ) ) / w = 1/2*w - 1/8*w^3 + 1/16*w^5 - 5/128*w^7 + 7/256*w^9 - 21/1024*w^11 + 33/2048*w^13 + O(w^15) this being gp-pari, and w = 1/t. $\endgroup$ – Will Jagy Feb 21 '18 at 3:37
  • $\begingroup$ @TrevorGunn How do you define the $k$-th off-diagonal term? $\endgroup$ – user522841 Feb 21 '18 at 3:45
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    $\begingroup$ @VivianL. The $k$-th off-diagonal is everything $k$ steps to the right of the main diagonal: $$\begin{pmatrix} 0 & 1 & 2 & 3 & 4 \\ & 0 & 1 & 2 & 3 \\ & & 0 & 1 & 2 \\ &&& 0 &1 \\ &&&& 0 \end{pmatrix}$$ $\endgroup$ – Trevor Gunn Feb 21 '18 at 3:55
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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 $B^2=J$.

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