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As the title states, I would like to prove the following:

$A$ is a nilpotent $n \times n$ matrix ($A^k = 0$). Show that $I_n + aA$ is invertible for each $a \in \mathbb{F}$ (where $I_n$ denotes the $n\times n$ identity matrix).

I (think) I am able to show that $I_n + A$ is invertible but I am struggling to come up with an appropriate inverse to use when the scalar $a$ is introduced.

Is there an easier way to formulate an inverse than trial and error in this case?

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  • $\begingroup$ Why not use the fact that $(aA)$ is nilpotent? $\endgroup$ – ancientmathematician May 2 '17 at 8:26
  • $\begingroup$ Since $aA$ is also nilpotent you can use the same method as you used for $I+A$ $\endgroup$ – JJR May 2 '17 at 8:28
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The inverse is $B=I-aA+a^2A^2-a^3A^3+\cdots+(-1)^{n-1}a^{n-1}A^{n-1}$

because $(I+aA)B=I$ by successive cancellations.

With the restriction, of course, that $aA \neq -I$.

Have you recognized the application of the series expansion:

$$(I-X)^{-1}=I+X+X^2+\cdots +X^k+\cdots$$

which is in fact a polynomial because all powers $X^k$ for $k \geq n$ are zero ?

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Let $\lambda$ be the eigenvalue of the matrix $A$ with $x\neq 0$ as an eigenvector. Then eigenvalue of the matrix $(I+aA)$ is given by $1+a\lambda$. Now the matrix $(I+aA)$ is singular if and only if it has at least one zero eigenvalue which is imposible. Since eigenvalues of nilpotent matrices are zero.

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We usually learn at school that $$ (1-x)(1+x+\dots+x^{n-1})=1-x^n.$$

So if we are in a situation where $x^n=0$ we have that $$ (1-x)(1+x+\dots+x^{n-1})=1,$$ and know the inverse of $1-x$.

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Hint. If $(I+aA)x=0$, you may express $(aA)^kx$ as a nonzero multiple of $x$. But $A$ is nilpotent, so ...

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The most elementary prove: denote $N=-aA$, it is a nilpotent matrix. Now apply formally the series $$ \frac{1}{1-x}=1+x+x^2+x^3+\ldots $$ to the matrix $(I-N)^{-1}$ and show that due to nilpotency it may have only finite number of nonzero terms. Then show by definition that this finite sum is, in fact, the inverse of $I-N$.

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Eigenvalues of $I+aA$ are non-zero numbers. So it is invertible. It is that simple.

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