Prove that the set of linear transformations is a subspace of $Hom(V,V)$

Question

Let $T: V \to V$ be a linear transformation over n-dimensional vector space. Prove that the set of linear transformations $X: V \to V$ such that $T \circ X=0$ is a linear subspace of $Hom(V,V)$ and calculate its dimension.

Any help or guidance would be appreciated

Answer 1

One way to show that a space is a linear subspace is to express it as the null space of some linear operator.

Define $\phi:\operatorname{Hom}(V,V) \to \operatorname{Hom}(V,V)$ by $\phi(X) = T \circ X$.

Then $\{X | T \circ X = 0 \} = \ker \phi$.

Note that $X \in \ker \phi$ iff ${\cal R}X \subset \ker T$. If we write $X = [c_1 \cdots c_n ]$, then we see that $X \in \ker \phi$ iff $c_k \in \ker T$ for all $k$. Hence $\dim \ker \phi = n(\dim \ker T)$.

Answer 2

If $T$ is the null function, then your space is $\hom(V,V)$, whose dimension is $n^2$.

Otherwise, let $v_1,\ldots,v_k$ be a basis of $\ker T$. Let $v_{k+1},\ldots,v_n\in V$ be such that $B=\{v_1,\ldots,v_n\}$ be a basis of $V$.Then the elements of your space are those $f\in\hom(V,V)$ such that$$(\forall i\in\{1,\ldots,n\}):f(v_i)\in\langle v_1,\ldots, v_k\rangle.$$In other words, the last $n-k$ lines the matrix $[f]_B$ are null. So, the dimension is $nk$.

Answer 3

Let $A$ be a ring.

You're aware that for $A$-modules $M$ and $N$ that $\text{Hom}_A(M,N)$ is a group, and when $A$ is commutative, that this is also an $A$-module.

Here you are dealing with $M=N$ both equal to the free $A$-module $V\cong A^n$ where $A=K$ is a field, which is of course commutative.

So $\text{Hom}(V,V)$ has $K$-vector space structure. Of course, even better, you even have ring structure via composition. Let $X\in\text{End}(V)$, and consider those $A\in \text{End}(V)$ such that $AX=0$. You wish to treat the endomorphism algebra as a right $\text{End}_A(V)$-module, and you are considering the right annihilator $\text{ann}_{\text{End}_A(V)}(\{T\})$, which is of course a right submodule as desired.