Can $A^3=0$ imply $|I+A|=0$?

Question

Suppose $A$ is a non-zero matrix such that $A^3=0$. Prove the following assertions or provide counter examples:-

$(1) A^2$ is a zero matrix $(2) A+A^2$ can have zero trace $(3) A-A^2$ can have zero trace $(4) I+A$ is singular.

My Attempt:- I know if $A^3=0$ then $A^2=0$ can be true (though not always). I have no idea whether $tr(A+A^2)=0$ or $tr(A-A^2)=0$ is possible or not if $A^3=0$. But when I looked closely at $|I+A|$ then I found that $$|I+A|=0$$ For $2\times2$ matrix, we have $$\Rightarrow |A|+tr(A)+1=0 $$ $$\Rightarrow \lambda_1\lambda_2+\lambda_1+\lambda_2+1=0 $$ where $\lambda_1$ and $\lambda_2$ are the two eigenvalues of $A$ $$\Rightarrow \lambda_1(\lambda_2+1)+1.(\lambda_2+1)=0 $$ $$\Rightarrow (\lambda_2+1)(\lambda_1+1)=0 $$ $$\Rightarrow \lambda_2=-1, \lambda_1=-1 \tag1$$ But we have $$A^3=0$$ $$\Rightarrow |A^3|=0$$ $$\Rightarrow |A|^3=0$$ $$\Rightarrow |A|=0$$ So, either $\lambda_1=0$ or $\lambda_2=0$ (or both may be zero) which contradicts with equation $(1)$. So, $I+A$ is non singular.

Am I Correct ?

Answer 1

You're on the right track.

Your first sentence about option (1) is correct, it can be ruled out.
Your thoughts about $2\times2$ matrices and option (4) are also correct.
However, here we might deal with bigger matrices. Especially because for a $2\times2$ matrix $A$, we do have $A^3=0 \implies A^2=0$.

It's generally a good practice to think about examples. A typical example for $A^3=0$ is $$\pmatrix{0&1&0\\0&0&1\\0&0&0}$$ This example rules out options (2) and (3), so we're indeed only left with (4).

For this, can you find an inverse for $I+A$, knowing $A^3=0$?

Answer 2

Notice that $$(I+A)(I-A+A^2)=I,$$ hence $I+A$ is nonsingular.

For $(2), (3)$, try the matrix $\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}$

Answer 3

$(a)$ can be true, set $A = \pmatrix{ 0 & 1 \\ 0 & 0}$.

$A^3 = 0$ implies that $\sigma(A) = \{0\}$, because the minimal polynomial of $A$ divides $x^3$.

This in turn means that $\sigma(A^k) = \{0\}$ so $\operatorname{Tr} A^k = 0$ for all $k \in \mathbb{N}$. Therefore $(b)$ and $(c)$ are always true:

$$\operatorname{Tr}(A \pm A^2) = \operatorname{Tr} A \pm \operatorname{Tr} A^2 = 0$$

Also, $\sigma(I + A) = \{1\}$ so $\det(I + A) = 1 \ne 0$, so $(d)$ is certainly false.