I am trying to solve the following entrance exam problem.
Let $š \in š_4(\mathbb{R})$ be such that $P^4$ is the zero matrix, but $P^3$ is a nonzero matrix.
Then which one of the following is FALSE?
(A) For every nonzero vector $š£ \in \mathbb{R}^4$, the subset $\{š£, Pš£, P^2š£, P^3š£\}$ of the real vector space $\mathbb{R}^4$ is linearly independent.
(B) The rank of $P^š$ is $4 ā š$ for every $š \in \{1,2,3,4\}$.
(C) $0$ is an eigenvalue of $P$.
(D) If $š \in š_4(\mathbb{R})$ is such that $Q^4$ is the zero matrix, but $Q^3$ is a nonzero matrix, then there exists a nonsingular matrix $S \in š_4(\mathbb{R}$ such that $š^{ā1}šš = š$.
Note: $š_š (\mathbb{R})$ = the real vector space of all $š \times š$ matrices with entries in $\mathbb{R}$.
My attempt: After numerous trials and errors, I successfully constructed a matrix denoted $P$ given by $\begin{pmatrix}
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0\\
\end{pmatrix}$
However, I am currently facing a significant roadblock in solving this problem. I am struggling to comprehend the next steps to proceed forward.
Thank you very much for your assistance
Best Answer
Option $A$ Looks Obviously false... I can give a counter example $$T:\mathbb{R^4} \rightarrow \mathbb{R^4}: T(e_1)=0,T(e_2)=e_1,T(e_3)=e_2, T(e_4)=e_3$$ Where $\{e_i:1\leq i\leq 4 \}$ is the standard basis of $\mathbb{R^4}$ . Do you Know what is the matrix of T..? It is the same matrix given by you..
then we check A is false... $ $\begin{pmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ \end{pmatrix}$ $
Now take any one of $e_1,e_2,e_3$ let's take $e_1$. It is clear that $e_1 \neq 0$ then $$\{Te_1,T^2e_1,T^3e_1,T^4e_1\}=\{e_1,0,0,0\}$$ and hence it is obvious that it is not linearly independent.
For option D check Prove that all nxn nilpotent matrices of order n are similar.
OR
Proof that all $n\times n$ matrices that are nilpotent of order $n$ are similar.