Q:

A question based on nilpotent real matrix

This particular question was asked in a masters exam for which I am preparing.

Let $A$ be an $n\times n$ nilpotent real matrix .Define $e^{A}=I_{n} +A+\frac{1}{2!}A^{2}+\frac{1}{3!}A^{3}+\dots$.

Then choose correct statements from below:

There exists a basis of $\mathbb{R}^n$ such that $e^{A}$ is upper triangular.
There exist $B$, $P$ belonging to $GL_{n}{(\mathbb{R})}$ such that $B=Pe^{A}P^{-1}$ and $\operatorname{trace}(B)=0$.
There exists a basis of $\mathbb{R}^{n} $ such that $A$ is lower-triangular.

Unfortunately, I have no clue on how this problem can be approached. My linear algebra is weak although I am working on it.
I would really appreciate if anyone can help me with any one of the options 1,2 and 3.

A:

The first assertion holds: if $A$ is a $n\times n$ real matrix such that all of its eigenvalues are real, then $A$ is similar to an upper triangular matrix (this can be proved by induction). Since your matrix is nilpotent, its only eigenvalue is $0$. So, it is similar to a strictly upper triangular matrix. The same argument proves the third assertion.
The second assertion is false. Note that$$\operatorname{tr}(B)=\operatorname{tr}\left(e^A\right)=n.$$

