Let \(A\) and \(B\) be \(n\times n\) complex matrices. Prove or disprove each of the following statements:
If \(A\) and \(B\) are diagonalizable, so is \(A+B\).
If \(A\) and \(B\) are diagonalizable, so is \(AB\).
If \(A^2=A\), then \(A\) is diagonalizable.
If \(A^2\) is diagonalizable, then \(A\) is diagonalizable.
Solution
Let \[ J\coloneqq \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}. \] The matrix \(J\) is not diagonalizable: its only eigenvalue is \(0\), so a diagonalizable matrix similar to it would have to be the zero matrix, whereas \(J\ne0\).
1. Statement 1 is false.
Take \[ A= \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}, \qquad B= \begin{pmatrix} -1&1\\ 0&1 \end{pmatrix}. \] The matrix \(A\) is diagonal. The matrix \(B\) has the two distinct eigenvalues \(-1\) and \(1\), so it is diagonalizable. However, \[ A+B = \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix} =J, \] which is not diagonalizable.
2. Statement 2 is false.
Take \[ A= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \qquad B= \begin{pmatrix} 0&1\\ 0&1 \end{pmatrix}. \] Again \(A\) is diagonal. The characteristic polynomial of \(B\) is \[ t(t-1), \] so \(B\) has distinct eigenvalues \(0\) and \(1\) and is diagonalizable. But \[ AB = \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix} =J, \] which is not diagonalizable.
3. Statement 3 is true.
If \(A^2=A\), then \[ A(A-I)=0. \] Hence the minimal polynomial of \(A\) divides \[ t(t-1). \] This polynomial splits over \(\CC\) and has no repeated root. Therefore the minimal polynomial of \(A\) also splits with no repeated root, which is equivalent to diagonalizability.
4. Statement 4 is false.
Take \(A=J\). Then \[ A^2=0, \] which is diagonalizable, while \(A=J\) itself is not diagonalizable.
5. Thus the answers are\[ \boxed{\text{false},\ \text{false},\ \text{true},\ \text{false}}. \]
This is exactly the content of steps <1>1–<1>4.
6. Q.E.D.
Step 5 answers all four parts.