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If $A$ is a nilpotent matrix, then how to show that the matrix $I+A$ is invertible.

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Look at this:

$(I + A)(\sum_0^n (-A)^j) = \sum_0^n (-A)^j + A\sum_0^n (-A)^j$ $ = \sum_0^n (-A)^j - (-A)\sum_0^n (-A)^j = \sum_0^n (-A)^j - \sum_0^n (-A)^{j + 1}$ $= \sum_0^n (-A)^j - \sum_1^{n + 1} (-A)^j = I + \sum_1^n (-A)^j - \sum_1^n (-A)^j - (-A)^{n + 1}$ $= I - (-A)^{n + 1}, \tag{1}$

which holds for any positive integer $n$. If $A$ is nilpotent, then $A^m = 0$ for some positive integer m. Thus, by (1),

$(I + A)(\sum_0^m (-A)^j) = I - (-A)^{m + 1} = I, \tag{2}$

showing the inverse of $I + A$ is in fact $\sum_0^m (-A)^j$. And that's how it is shown.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

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Here is a slightly different approach:

If $A$ is nilpotent, we have $A^n = 0$.

If $\det (I+A) = 0$, we have $(I+A) v = 0$ for some $v \neq 0$, that is $Av = -v$. Then $A^n v = (-1)^nv$, which contradicts $A^n = 0$.

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  • $\begingroup$ There should be some $-1$s in there. $\endgroup$
    – Slade
    Sep 26 '14 at 5:51
  • $\begingroup$ @Slade: Thanks for catching that! $\endgroup$
    – copper.hat
    Sep 26 '14 at 5:54
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Hint

Let us put $B=-A$

Compute: $$(I- B)(I+B+\cdots B^n)$$

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If $A$ is nilpotent, then $0$ is the only possible eigenvalue of $A$ (since each eigenvalue must satisfy $\lambda^k=0$ if $A^k=0$). In particular $-1$ is not an eigenvalue, i.e., $A+I$ is injective, hence invertible.

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