Say $A^2 = 0$. Prove that for every real number $a$, the matrix $I + a A$ is invertible

Question

Say $A^2 = 0$. Prove that for every real number $a$, the matrix $I + a A$ is invertible.

I need some help with this question.

Answer 1

Observe that:$$(I + aA)(I - aA) = I - a^2A^2 = I$$

Answer 2

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

Answer 3

Let's have $v\in\ker(I+aA)$

Then $(I+aA)v=v+aAv=0$

We multiply by $A$ again to get $Av+aA^2v=Av+0=0$

But then $v+aAv=v+0=0$

So $\ker(I+aA)=\{0\}$

Answer 4

Suppose we are given a nilpotent matrix $\mathrm A \in \mathbb R^{n \times n}$ such that $\rm A^2 = O_n$. Let its Jordan decomposition be $\rm A = P J P^{-1}$. Since $\rm A^2 = O_n$, we have $\rm J^2 = O_n$, which allows us to conclude that all the eigenvalues of $\rm A$ are zero. Let $\gamma \in \mathbb R$. Hence, all the eigenvalues of $\mathrm I_n + \gamma \mathrm A$ are $1$ and, thus, $\mathrm I_n + \gamma \mathrm A$ is invertible.