91Ó°ÊÓ

Consider two linear spacesand. A functionfromtois called linear transformation if

(i)$\mathbf{T}\mathbf{(}\mathbf{f}\mathbf{+}\mathbf{g}\mathbf{)}\mathbf{=}\mathbf{T}\mathbf{\left(}\mathbf{f}\mathbf{\right)}\mathbf{+}\mathbf{T}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

(ii)$\mathbf{}\mathbf{T}\mathbf{\left(}\mathbf{kf}\mathbf{\right)}\mathbf{=}\mathbf{kT}\mathbf{\left(}\mathbf{f}\mathbf{\right)}$

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

If Vis finite dimensional, then

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

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

Then, $ker(L∘T)⊆V$.

Thus, we can restrictT to $ker(L∘T)$.

Let $S=T{\backslash }_{ker\left(L∘T\right)}$.

Let $v∈\mathrm{ke}r\left(T\right)$, then $T\left(v\right)=0$.

Also,$\left(T\right)⊆\mathrm{ke}r\left(\mathrm{LoT}\right)$

Since $T\left(w\right)=0$, this implies that $L\left(T\left(w\right)\right)=0$.

This implies $V∈\mathrm{ke}r\left(\mathrm{Lo}T\right)$.

Hence,$\mathrm{S}\left(V\right)=T\left(V\right)=0$.

So,$v∈\mathrm{ke}r\left(S\right)$then,$S\left(v\right)=0$.

This implies $T\left(V\right)=S\left(V\right)=0$.

Thus, $v∈\mathrm{ke}r\left(T\right)$.

Hence,$\mathrm{ke}r\left(T\right)=\mathrm{ke}r\left(\mathrm{S}\right)$.

Let $\mathrm{Im}\left(\mathrm{S}\right)⊆\mathrm{ke}r\left(L\right)$and $w∈\mathrm{Im}\left(S\right)$then there exists $v∈\mathrm{ke}r(L∘T)$such that $w=S\left(v\right)=T\left(v\right)$.

Now,$w=S\left(v\right)=T\left(v\right)$implies that $L\left(T\left(v\right)\right)=0$, so $T\left(v\right)∈\mathrm{ke}r\left(L\right)$.

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

Thus, $\mathrm{Im}\left(\mathrm{S}\right)⊆\ker \left(\mathrm{L}\right)$.

Here $\mathrm{ke}r\left(T\right)=\mathrm{ke}r\left(\mathrm{S}\right)$implies $\mathrm{ke}r\left(S\right)$is finite dimensional since it is given that $\mathrm{ke}r\left(T\right)$is finite dimensional. Also,$\mathrm{Im}\left(\mathrm{S}\right)⊆\ker \left(\mathrm{L}\right)$implies is finite dimensional since it is given that $\mathrm{ke}r\left(L\right)$is finite dimensional.

Then by rank-nullity theorem,$\mathrm{ke}r(L∘T)$is finite dimensional.