Let be $F$ a fied, $V$ a finite dimensional vector space over $F$, and a lineal operator $T:V \rightarrow V$ distinct to zero and not invertible. Prove that there exists $G$ and $H$ lineal operations in $V$ distinct to zero sush that: $T \circ G = 0$ and $H \circ T = 0$
I think that I can use the fact that a lineal operator $S$ is invertible if and only if $S$ is biyective.
Then if we suppose that there don't exists $G$ and $H$ such that $T \circ G = 0$ and $H \circ T = 0$, then $Ker(T)={0}$ so that implies that T is inyective. By other side, we have that $Im(T)=V$ so T is suprayective. Then T is biyective, but that's a contradiction, so we conclude that there exists $G$ and $H$ with the desired properties.
Am I correct? Or how can I prove it? I would really appreciate your help!
By assumption $T$ is not injective so the kernel of $T$ is some subspace $W\not=\{0\}$ of $V$. There exists a basis $\{e_1,...,e_n\}$ of the whole space $V$ such that the first $k$ elements are a basis of $W$.
Can you from there construct a linear map $G:V\rightarrow V$ which is the identity in $W$ and such that $T\circ G=0$? It is enough to decide what $G$ does on the basis $\{e_1,...,e_n\}$ and extend by linearity.
The other assertion works similarly, $T$ is not surjective so its range is a proper subspace $W\not=V$. Construct an appropriate basis and then a linear map $H$ such that $H\circ T=0$.