Prove that kernel of a transformation is a subset of the composite transformation.

Question

Let T:L→L Be a linear transformation. Prove that ker( T )⊆ker( T∘T ) .

I tried classical ker( T )⊆L and postulated that since composite transformations keep their properties such as being injection, surjection or bijection, this implication is true

( ker( T )⊆L∧ker( T∘T )⊆L )⇒ker( T )⊆ker( T∘T ) .

But it doesn't seem right.

Answer 1

Let $v\in \ker(T).$ Then, by definition of kernel, $T(v)=0_L$.

Applying $T$ to the null vector of $L$, we have that

$$T(0_L)=T(T(v))=\left(T\circ T\right)(v)$$

On the other hand,

$$T(0_L)=T(0_L+0_L)=T(0_L)+T(0_L)$$

Then, $T(0_L)=0_L$, i.e, $\left(T\circ T\right)(v) = 0_L$. Therefore, $v \in \ker(T\circ T)$. By the arbitrariness of $ v $, the result follows