Let $A$ be a complex matrix $n \times n$ matrix. Suppose that $e^{A} = I + A + A^{2}$. Prove or disprove: A is the zero matrix.

Question

I know that if $A$ is diagonalizable, it must be the zero matrix. I also know that diagonalizable matrices are dense in the set of complex matrices. I don't think that helps though. Any ideas on where to start?

Answer 1

$A$ is not necessarily the zero matrix.

No shortage of such $A \ne 0$:

For suppose

$0 \ne A \in M_n(\Bbb C), \; n \ge 2, \tag 1$

but

$A^2 = 0; \tag 2$

there are plenty of such matrices; e.g., take

$A = [a_{ij}] \tag 3$

where

$a_{ij} = \delta_{i1} \delta_{jn}, \tag 4$

that is,

$a_{1n} = 1; \; a_{ij} = 0, i \ne 1 \; \text{or} \; j \ne n; \tag 5$

it is easy to see that

$A^2 = 0; \tag 6$

then for invertible

$P \in M_n(\Bbb C) \tag 7$

we also have

$P^{-1}AP \ne 0 \tag 8$

and

$(P^{-1}AP)^2 = P^{-1}APP^{-1}AP = P^{-1}A^2P = 0; \tag 9$

now for

$n \ge 2, \tag{10}$

$A^n = A^{n - 2}A^2 = 0, \tag{11}$

so

$e^A = \displaystyle \sum_0^\infty \dfrac{A^m}{m!} = I + A = I + A + A^2, \tag{11}$

by virtue of (10), (11). And, in the light of (9), the same holds for any $P^{-1}AP$ with invertible $P$.