Consider the $0$ matrix. It has one eigenvalue: $0$, and the dimension of its eigenspace is $n$, since it sends everything to $0$. If I have $n$ distinct eigenvalues, one of them zero, then the eigenspace for $0$ must have dimension $1$, hence the rank is $n−1$.
So is my conclusion right:
If the a linear transformation T has only one eigenvalue ${\lambda }$= $0$, then dim(ker($T$)) = $1$?
Best Answer
If $0$ is an eigenvalue for the linear transformation $T:V\rightarrow V$, then by the definitions of eigenspace and kernel you have
$$V_0=\{v\in V| T(v)=0v=0\}=\text{ker}T.$$
If you have only one eigenvalue, which is $0$ the dimension of $\text{ker}T$ is equal to the dimension of $V_0$.
For instance: consider the endomorphism, whose associated matrix respect to the canonical base is:
\begin{pmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{pmatrix}
the the only eigenvalue is $0$, but the dimension of the kernel is $2$.