Keeping it short

If $\left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right) \; =\; \left(\begin{array}{cc}a & b \\ c & d \end{array}\right)^2 \; =\; \left(\begin{array}{cc} a^2 + bc & b(a+d) \\ c(a+d) & bc + d^2 \end{array} \right)$ Then, since $c(a+d) = 0 \neq b(a+d)$ , we deduce that $c=0$ , and hence $a^2 \; = \; 0 \hspace{1cm} b(a+d) \; = \; 1 \hspace{1cm} d^2 \; =\; 0$ Thus $a=d=0$ , which makes the condition $b(a+d)=1$ an impossibility. Thus this $2\times2$ matrix is not a square.

To within similarity, this is the only matrix which is not a square, since diagonal matrices and matrices of the form $\left( \begin{array}{cc}x & 1 \\ 0 & x \end{array} \right)$ for nonzero $x$ are all squares.

---

A more abstract proof goes like this. Suppose that $A$ is a $2\times2$ matrix with minimum polynomial $t^2$ . If $A$ was a perfect square, then $A = B^2$ for some matrix $B$ , and this means that $B^4 = 0$ . Thus the minimum polynomial of $B$ has degree at most $2$ and divides $t^4$ . If the minimum polynomial is $t$ , this means that $B=0$ . If the minimum polynomial is $t^2$ , this means that $B^2=0$ . In either case we see that $A = B^2 = 0$ . But $A=0$ does not have minimum polynomial $t^2$ . Thus no matrix with minimum polynomial $t^2$ is a square.