Prove or disprove: for every \(2\times2\) complex matrix \(A\), there is a \(2\times2\) complex matrix \(B\) such that \[ A=B^2. \]
Solution
Let \[ N\coloneqq \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}. \]
1. The matrix \(N\) is nonzero and satisfies\[ N^2=0. \]
The \((1,2)\)-entry of \(N\) is \(1\), so \(N\neq0\). Direct multiplication gives \(N^2=0\).
2. There is no \(2\times2\) complex matrix \(B\) such that\[ B^2=N. \]
Suppose that such a matrix \(B\) exists. By step 1, \[ B^4=N^2=0, \] so \(B\) is nilpotent. Hence both eigenvalues of \(B\) are zero, and its characteristic polynomial is \[ \chi_B(t)=t^2. \] The Cayley–Hamilton theorem therefore gives \[ B^2=0. \] But \(B^2=N\) and \(N\neq0\) by step 1, a contradiction.
3. The proposed statement is\[ \boxed{\text{false}}. \]
Step 2 shows that the explicit complex matrix \(N\) has no square root in \(M_2(\CC)\).
4. Q.E.D.
Step 3 gives the required disproof.