Let \(n \geq 1\) and consider the ring \(M_n(\mathbb{C})\) of \(n \times n\) matrices with coefficients in \(\mathbb{C}\). Suppose that \(A \in M_n(\mathbb{C})\) satisfies \(A^3 = A\). Let \(V = \mathbb{C}^n\), an \(n\)-dimensional vector space over \(\mathbb{C}\). Thinking of the elements of \(V\) as column vectors, consider the linear transformation \(\phi: V \to V\) defined by left multiplication by the matrix \(A\). Prove that \(V\) decomposes into a direct sum of three \(\mathbb{C}\)-linear subspaces, say \(V = U_1 \oplus U_2 \oplus U_3\), such that given \(v \in V\) with \(v = u_1 + u_2 + u_3\) where \(u_i \in U_i\), then \(\phi(v) = u_1 - u_2\).
Solution
The matrix identity \[ A^3=A \] is equivalent to \[ A(A-I)(A+I)=0. \]
1. The minimal polynomial of \(A\) divides\[ t(t-1)(t+1). \]
Let \(m_A(t)\) be the minimal polynomial of \(A\). Since \[ A^3-A=0, \] the polynomial \[ t^3-t=t(t-1)(t+1) \] annihilates \(A\). By the defining property of the minimal polynomial, \[ m_A(t)\mid t(t-1)(t+1). \]
2. The operator \(A\) is diagonalizable over \(\mathbb C\), with possible eigenvalues only \(1,-1,0\).
The polynomial \[ t(t-1)(t+1) \] splits over \(\mathbb C\) into three distinct linear factors. By 1, the minimal polynomial \(m_A\) also splits into distinct linear factors. A linear operator over a field is diagonalizable if and only if its minimal polynomial splits into distinct linear factors. Therefore \(A\) is diagonalizable. Its eigenvalues are roots of \(m_A\), hence belong to \[ \{1,-1,0\}. \]
3. Define\[ U_1:=\ker(A-I), \qquad U_2:=\ker(A+I), \qquad U_3:=\ker A. \]Then\[ V=U_1\oplus U_2\oplus U_3. \]
By 2, \(A\) is diagonalizable and its only possible eigenvalues are \(1,-1,0\). A diagonalizable operator is the direct sum of its eigenspaces. The eigenspaces corresponding to \(1,-1,0\) are exactly \[ \ker(A-I), \qquad \ker(A+I), \qquad \ker A. \] If one of these eigenvalues does not occur, the corresponding eigenspace is simply \(0\). Hence \[ V=U_1\oplus U_2\oplus U_3. \]
4. For \(v=u_1+u_2+u_3\) with \(u_i\in U_i\), one has\[ \phi(v)=u_1-u_2. \]
By definition of the three eigenspaces, \[ Au_1=u_1, \qquad Au_2=-u_2, \qquad Au_3=0. \] Since \(\phi\) is left multiplication by \(A\), \[ \phi(v) =A(u_1+u_2+u_3) =Au_1+Au_2+Au_3 =u_1-u_2. \] This is exactly the required decomposition and action formula.