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?
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 ?
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.
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 $1-x$.
Hint. If $(I+aA)x=0$, you may express $(aA)^kx$ as a nonzero multiple of $x$. But $A$ is nilpotent, so ...
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 $I-N$.
Eigenvalues of $I+aA$ are non-zero numbers. So it is invertible. It is that simple.
