91Ó°ÊÓ

Consider two linear spaces Vand W. A function Tfrom Vto Wis called linear transformation if

1. T(f+g) =T (f) + T (g)
2. T (kf) = kT (f)

for all elements f and g of Vand for all scalars.

If Vis finite dimensional, then

$\mathbf{dim}\mathbf{}\mathbf{\left(}\mathbf{v}\mathbf{\right)}\mathbf{=}\mathbf{rank}\mathbf{}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{nullity}\mathbf{}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{dim}\mathbf{}\mathbf{\left(}\mathbf{imT}\mathbf{\right)}\mathbf{+}\mathbf{dim}\mathbf{}\mathbf{(}\mathbf{ker}\mathbf{}\mathbf{T}\mathbf{)}$

Given that the composition of two linear transformations $L∘T:V→U$.

Then, $ker(L∘T)\underline{⊂}V$.

Thus, we can restrict T to $\left(L∘T\right)$.

Let $S={\overline{)T}}_{ker(L∘T)}$.

Let $v∈ker\left(T\right)$, then $T\left(v\right)=0$.

Also $\left(T\right)\underline{⊂}ker\left(L∘T\right)$,

Since T (w) =0, this implies that L ( T (w) )= 0.

This implies $v∈ker\left(L∘T\right)$.

Hence, S (v) = T (v) = 0.

So, $v∈ker\left(S\right)$then, S (v) = 0.

This implies T (v) =S (v) = 0.

Thus, $v∈ker\left(T\right)$.

Hence, ker(T) = ker (S).

$$\begin{gathered} \mathrm{Let}lm\left(S\right)\underline{⊂}ker\left(L\right)\mathrm{and}w∈lm\left(S\right)\mathrm{then}\mathrm{there}\mathrm{exists}v∈ker(L∘T) \\ \mathrm{such}\mathrm{that}w=S\left(v\right)=T\left(v\right) \end{gathered}$$

Now, w = S (v)= T (v) implies that L ( T (v) ) = 0 , so $T\left(v\right)∈ker\left(L\right)$.

Hence, $w=S\left(v\right)=T\left(v\right)∈ker\left(L\right)$.

Thus, $lm\left(S\right)\underline{⊂}ker\left(L\right)$.

Here $ker\left(T\right)=ker\left(S\right)$impliesker (S) is finite dimensional since it is given that $ker\left(T\right)$is finite dimensional. Also, $lm\left(S\right)\underline{⊂}ker\left(L\right)$implies lm (S) is finite dimensional since it is given that ker (L) is finite dimensional.

Then by rank-nullity theorem, $\left(L∘T\right)$is finite dimensional.

Also, dim (ker(LoT) = dim (ker (S)) + dim (lm (S)) which implies

dim (ker (LoT)) = dim (ker (T) + dim (lm(S) ) Since, ker (T) = ker (S) .

This implies ( ker (LoT)) = dim (ker (T)) + dim (ker (L)) as lm (S) = ker (L) .