Let \(A,B\) be complex \(n\times n\) matrices. Prove or disprove each assertion:
- If \(A\) and \(B\) are diagonalizable, then \(A+B\) is diagonalizable.
- If \(A\) and \(B\) are diagonalizable, then \(AB\) is diagonalizable.
- If \(A^2=A\), then \(A\) is diagonalizable.
- If \(A\) is invertible and \(A^2\) is diagonalizable, then \(A\) is diagonalizable.
Solution
1. Assertion (1) is false.
Take \[ A= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \qquad B= \begin{pmatrix} 0&1\\ 0&1 \end{pmatrix}. \] The matrix \(A\) is diagonal. The matrix \(B\) has the two distinct eigenvalues \(0\) and \(1\), so it is diagonalizable over \(\CC\). But \[ A+B = \begin{pmatrix} 1&1\\ 0&1 \end{pmatrix} \] has characteristic polynomial \((t-1)^2\) and is not the identity matrix. Its \(1\)-eigenspace is one-dimensional, so it is not diagonalizable.
2. Assertion (2) is false.
Take \[ A= \begin{pmatrix} 1&0\\ 0&2 \end{pmatrix}, \qquad B= \begin{pmatrix} 1&1\\ 0&1/2 \end{pmatrix}. \] Again \(A\) is diagonal, while \(B\) has the two distinct eigenvalues \(1\) and \(1/2\), so \(B\) is diagonalizable. However, \[ AB = \begin{pmatrix} 1&1\\ 0&1 \end{pmatrix}, \] which is the same nondiagonalizable Jordan block as in step 1.
3. Assertion (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 over \(\CC\).
4. Suppose \(A\) is invertible and \(A^2\) is diagonalizable. Then everyeigenspace of \(A^2\) is invariant under \(A\).
Let \[ E_\mu\coloneqq\ker(A^2-\mu I) \] be an eigenspace of \(A^2\). If \(v\in E_\mu\), then \[ A^2(Av) = A(A^2v) = A(\mu v) = \mu Av. \] Thus \(Av\in E_\mu\).
5. For every eigenvalue \(\mu\) of \(A^2\), the restriction\[ A|_{E_\mu} \]is diagonalizable.
Because \(A\) is invertible, \(A^2\) is invertible, so \(\mu\ne0\). On \(E_\mu\) one has \[ (A|_{E_\mu})^2=\mu I. \] Therefore the minimal polynomial of the restriction divides \[ t^2-\mu. \] Over \(\CC\), this polynomial has the two distinct roots \[ \sqrt\mu \qquad\text{and}\qquad -\sqrt\mu, \] because \(\mu\ne0\). Thus the restriction is diagonalizable.
6. Assertion (4) is true.
Since \(A^2\) is diagonalizable, \[ \CC^n = \bigoplus_{\mu\in\Spec(A^2)}E_\mu. \] By step 4 each summand is \(A\)-invariant, and by step 5 the restriction of \(A\) to each summand has a basis of eigenvectors. Combining these bases gives an eigenbasis of \(\CC^n\) for \(A\). Hence \(A\) is diagonalizable.
7. Therefore the answers are\[ \boxed{ (1)\ \text{false},\qquad (2)\ \text{false},\qquad (3)\ \text{true},\qquad (4)\ \text{true}. } \]
Steps <1>1–<1>2 give counterexamples to the first two assertions, while steps 3 and 6 prove the last two.
8. Q.E.D.
Step 7 gives the complete requested classification.