Consider for \(n \ge 2\) the linear space \(\mathcal M_n(\mathbb C)\) of complex matrices of dimension \(n \times n\). *Is a matrix \(T \in \mathcal M_n(\mathbb C)\) always having a square root \(S \in \mathcal M_n(\mathbb C)\), i.e. a matrix such that \(S^2=T\)?* is the question we deal with.

First, one can note that if \(T\) is similar to \(V\) with \(T = P^{-1} V P\) and \(V\) has a square root \(U\) then \(T\) also has a square root as \(V=U^2\) implies \(T=\left(P^{-1} U P\right)^2\).

### Diagonalizable matrices

Suppose that \(T\) is similar to a diagonal matrix \[

D=\begin{bmatrix}

d_1 & 0 & \dots & 0 \\

0 & d_2 & \dots & 0 \\

\vdots & \vdots & \ddots & 0 \\

0 & 0 & \dots & d_n

\end{bmatrix}\] Any complex number has two square roots, except \(0\) which has only one. Therefore, each \(d_i\) has at least one square root \(d_i^\prime\) and the matrix \[

D^\prime=\begin{bmatrix}

d_1^\prime & 0 & \dots & 0 \\

0 & d_2^\prime & \dots & 0 \\

\vdots & \vdots & \ddots & 0 \\

0 & 0 & \dots & d_n^\prime

\end{bmatrix}\] is a square root of \(D\).

### Invertible matrices

A complex matrix \(T\) is similar to a matrix \[

J=\begin{bmatrix}

J_1 & & \\

& \ddots &\\

& & J_p

\end{bmatrix}\] where each block \(J_i\) is a square matrix of the form \[

J_i=\begin{bmatrix}

\lambda_i & 1 & &\\

& \lambda_i & \ddots &\\

& & \ddots & 1\\

& & & \lambda_i

\end{bmatrix}\] \(J\) is called the Jordan normal form of \(T\). If \(T\) is supposed to be invertible, the \(\lambda_i\) are non-zero and \(J\) has a square root. For the proof, we prove that a Jordan block \[

J(\lambda)=\begin{bmatrix}

\lambda & 1 & &\\

& \lambda & \ddots &\\

& & \ddots & 1\\

& & & \lambda

\end{bmatrix}\] with \(\lambda \neq 0\) has a square root. Writing \(J(\lambda)= \lambda(I+N)\), where \(N=\frac{J(0)}{\lambda}\) and \(I\) the identity matrix, it is sufficient to prove that \(I+N\) has a square root \(K\) for any nilpotent matrix \(N\) as then \(cK\) is a square root of \(J(\lambda)\) for \(c\) a square root of \(\lambda\).

Recall the Taylor expansion of \(\sqrt{1+x}\) \[

\sqrt{1+x} = 1 + \frac{1}{2} x + \frac{\left(\frac{1}{2}\right)\left(\frac{-1}{2}\right)}{2!} x^2 + \dots + \frac{\left(\frac{1}{2}\right)\left(\frac{-1}{2}\right)\dots \left(\frac{1}{2}-(n-1)\right)}{n!} x^n + \dots\] Plug in \(N\) for \(x\), we get a finite sum as \(N\) is nilpotent. Squaring both side of the equality we get that \[

I+N=\left[1 + \frac{1}{2} N + \frac{\left(\frac{1}{2}\right)\left(\frac{-1}{2}\right)}{2!} N^2 + \dots + \frac{\left(\frac{1}{2}\right)\left(\frac{-1}{2}\right)\dots \left(\frac{1}{2}-(n-1)\right)}{n!} N^n\right]^2\] concluding the proof.

### Singular matrices

The case of singular matrices is more complex. For example the matrix \[

A=\begin{bmatrix}

1 & 0 & 0\\

0 & 0 & 0\\

0 & 0 & 0

\end{bmatrix}\] has infinitely many square roots including the matrices \[

A^\prime_t=\begin{bmatrix}

1 & 0 & 0\\

0 & 0 & t\\

0 & 0 & 0\end{bmatrix}\] for \(t \in \mathbb C\).

**However the nilpotent matrix \[
B=\begin{bmatrix}
0 & 1\\
0 & 0
\end{bmatrix}\] doesn’t have a square root**. Indeed, if \(B^\prime\) was a square root of \(B\), \(B^\prime\) would also be nilpotent and therefore be similar to \(B\) or the zero matrix using the Jordan canonical form of \(B^\prime\). However both squares of those matrices are equal to the zero matrix.