1. ## Advance linear transformation exercise

Let $V$ be a finite-dimensional vector space and let $T$ be a linear operator on $V$.
Suppose that rank $(T^2)$ = rank $(T)$,

Prove that the range and null space of $T$ are disjoint therefore have only the zero vector in common.

THANKS

2. Originally Posted by osodud

Let $V$ be a finite-dimensional vector space and let $T$ be a linear operator on $V$.
Suppose that rank $(T^2)$ = rank $(T)$,

Prove that the range and null space of $T$ are disjoint therefore have only the zero vector in common.

THANKS
(1) by the rank-nullity theorem $\text{nullity}(T)=\text{nullity}(T^2).$ we also know that the null space of $T$ is a subspace of the null space of $T^2.$ so the null space of $T$ is equal to the null space of $T^2.$

now if $v=T(u)$ is in the null space of $T,$ then $T^2(u)=T(v)=0$ and thus $v=T(u)=0$ by (1).

3. Originally Posted by osodud
Let $V$ be a finite-dimensional vector space and let $T$ be a linear operator on $V$.
Suppose that rank $(T^2)$ = rank $(T)$,
Prove that the range and null space of $T$ are disjoint therefore have only the zero vector in common.
1. Prove null-space of $T$= null space of $T^2$, under the given conditions. (This can be done by showing they have same dimension).
2. Using 1 above establish that if $x\in Null(T)$ AND $x \in Range(T)$ implies $x=0$. As $x \in Range(T)$ there is a $y \in V$ such that $T(y)=x$. What can you say about $T^2(y)$ and hence $T(y)$ and hence $x$ from this? That should provide you the prove.