A square matrix \(A\) is nilpotent if \(A^k=0\) for some positive integer \(k\).
If \(A\) and \(B\) are nilpotent, must \(A+B\) be nilpotent? Give a proof or counterexample.
Prove that if \(A\) is nilpotent, then \(I-A\) is invertible.
Solution
1. (a) The sum of two nilpotent matrices need not be nilpotent.
Take \[ A= \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}, \qquad B= \begin{pmatrix} 0&0\\ 1&0 \end{pmatrix}. \] Then \[ A^2=0, \qquad B^2=0, \] so both \(A\) and \(B\) are nilpotent. However, \[ A+B= \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} \] and \[ (A+B)^2=I. \] Hence \[ (A+B)^{2m}=I \] for every \(m\geq1\). If \((A+B)^r=0\) for some \(r\geq1\), then \((A+B)^{2r}=0\), contradicting the displayed identity with \(m=r\). Thus \(A+B\) is not nilpotent.
2. (b) Suppose \(A^k=0\) for some \(k\geq1\), and define\[ S=I+A+A^2+\cdots+A^{k-1}. \]Then\[ (I-A)S=I. \]
Expanding and cancelling consecutive powers gives \[ \begin{aligned} (I-A)S &= (I+A+\cdots+A^{k-1}) - (A+A^2+\cdots+A^k)\\ &= I-A^k\\ &= I. \end{aligned} \]
3. Under the same hypotheses,\[ S(I-A)=I. \]
Since every term of \(S\) is a power of \(A\), it commutes with \(A\). Thus the same finite geometric-series computation gives \[ S(I-A)=I-A^k=I. \]
4. Therefore\[ \boxed{(I-A)^{-1}=I+A+A^2+\cdots+A^{k-1}}. \]
Steps 2 and 3 show that the displayed matrix is both a left and right inverse of \(I-A\).
5. Q.E.D.
Step 1 answers part (a), and step 4 proves part (b).