Let $V$ be a vector space over R such that $\dim(V) =3$ and let $T : V → V$ be a
linear transformation which satisfies $T^3 = 0$ and $T^2 \ne 0$. Prove:
(i) $\dim(\ker(T)) = 1$
In my proof I claim that $$T^2x = 0, \forall x \in V$$
Does this imply that $$Tx = 0, \forall x \in V$$
Beacuse $$T^2 \neq 0$$
Best Answer
Your title and seems to ask a different question than your problem statement. In general, $ker(T) \neq Ker(T^2)$. This is easily shown with a counter-example. Consider the nilpotent matrix $A$:
\begin{bmatrix} 0 & 2 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}
$A^2$ = \begin{bmatrix} 0 & 0 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} $A^3 =$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} So all of the parameters of your problem are met. However, the kernel of $A$ is the span of:\begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} and the kernel $A^2$ is the span of: \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} AND \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix}
This is made more clear by considering the fact that the kernel is the same as the set of eigenvectors with associated eigenvalue $\lambda = 0$. The eigenvalues for a nilpotent matrix must be $0$ for all eigenvectors, therefore the kernel of a nilpotent matrix is comprised of all eigenvectors. This can be easily proven:
$Av = \lambda v$
$A^2v = \lambda^2 v$
$A^3v = \lambda^3 v$
but since $A$ is nilpotent of degree $3$, $A^3v = 0$, and so $\lambda^3v = 0$, which is only (non-trivially) zero if $\lambda = 0$. For the same reason, it is clear that if $v$ is an eigenvector of $A$, it is also an eigenvector of $A^2$. However, this argument does not run both ways. You cannot say that an eigenvector of $A^2$ is necessarily an eigenvector of $A$.
The statement that $T^2(x) = 0$ implies $T(x)=0$ is false, as nilpotent transformations are not isomorphic. I.e., there can be no inverse transformation for:
$T^{-1}(T^2(x)) = T^{-1}(0) = 0 = T(x)$.
To prove the question asked in your problem statement, you need to consider the fact that, for an $n$ dimensional nilpotent matrix, $dim(ker(T^k)) < dim(ker(T^{k+1}))$ for $k \le n$. In this problem, $dim(ker(T^3)) = 3$, as any $3$-dimensional vector mapped by $T^3$ is the zero vector. This consequently forces $dim(ker(T)) = 1$