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?
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3Let $A \neq 0$ and $A^2 = 0$. – pisco Sep 01 '18 at 18:10
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4in dimension $1$ by $1$ we also have $1.793282133..$ as the only entry – Will Jagy Sep 01 '18 at 18:13
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As well as any $n\times n$ diagonal matrix with $0$ and/or @WillJagy's number on the diagonal. So the first sentence of the answer is definitely wrong. – hmakholm left over Monica Sep 01 '18 at 18:16
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2@WillJagy If in $\mathbb{C}$, $e^z = 1+z+z^2$ has infinitely many solutions. – pisco Sep 01 '18 at 18:17
1 Answers
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$A$ 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 $P$.
Robert Lewis
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