Let \(A,B\) be complex square matrices of the same size. 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 \(AB\) is diagonalizable and invertible, then \(BA\) is diagonalizable.
Solution
Let \(V\coloneqq\CC^n\) be the common underlying vector space.
1. Assertion 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 over \(\CC\). But \[ A+B = \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}. \] This matrix is nonzero and nilpotent, so its only eigenvalue is \(0\) and it cannot be diagonalizable.
2. Assertion 2 is false.
Set \[ N= \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}, \qquad A= \begin{pmatrix} 1&0\\ 1&-1 \end{pmatrix}, \qquad B= \begin{pmatrix} 0&1\\ 0&1 \end{pmatrix}. \] One has \[ A^2=I, \] so the minimal polynomial of \(A\) divides \[ t^2-1=(t-1)(t+1), \] which has distinct roots; hence \(A\) is diagonalizable. Also \[ B^2=B, \] so the minimal polynomial of \(B\) divides \[ t(t-1), \] which likewise has distinct roots; hence \(B\) is diagonalizable.
However, \[ AB=N. \] As in step 1, \(N\) is a nonzero nilpotent matrix and is not diagonalizable.
3. Assertion 3 is true.
Assume \[ A^2=A. \] For every vector \(v\), \[ v=(v-Av)+Av. \] Moreover, \[ A(v-Av)=Av-A^2v=0, \] so \(v-Av\in\ker A\), while \(Av\in\operatorname{im}A\). Thus \[ V=\ker A+\operatorname{im}A. \] If \(w\) lies in the intersection, then \(Aw=0\) because \(w\in\ker A\), while \(Aw=w\) because \(w\in\operatorname{im}A\) and \(A\) acts as the identity on its image. Hence \(w=0\), and therefore \[ V=\ker A\oplus\operatorname{im}A. \] A basis adapted to this direct sum consists of eigenvectors with eigenvalues \(0\) and \(1\), so \(A\) is diagonalizable.
4. Assertion 4 is true.
If \(AB\) is invertible, then \[ 0\neq\det(AB)=\det(A)\det(B). \] Thus both \(A\) and \(B\) are invertible. Consequently \[ BA = A^{-1}(AB)A. \] Hence \(BA\) is similar to \(AB\). Similarity preserves diagonalizability, so \(BA\) is diagonalizable whenever \(AB\) is.
5. The four answers are\[ \boxed{\text{false},\ \text{false},\ \text{true},\ \text{true}}. \]
Steps <1>1–<1>4 establish the assertions in order.
6. Q.E.D.
Step 5 records the required conclusions.