91Ó°ÊÓ

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

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

(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

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

In finite dimensional linear space V, $dim\left(V\right)=n$, all of the subspaces have dimension $≤n$.

Let $x∈ker\left(T\right)⇒t\left(x\right)=\mathrm{θ}$.

Then for every $x,y∈ker\left(T\right)$and for every $\mathrm{Î\pm },\mathrm{β}∈F$.

We have,$\mathrm{Î\pm }x+\mathrm{β}y∈ker\left(T\right)$

$\mathrm{Î\pm }t\left(x\right)+\mathrm{β}t\left(y\right)=\mathrm{Î\pm }\mathrm{θ}+\mathrm{β}\mathrm{θ}=\mathrm{θ}$

This implies,$\mathrm{ke}r\left(T\right)$ is a subspace of V.

Since,$\mathrm{ke}r\left(T\right)$is a subspace of V it is possible to find its basis.

Let$\mathrm{ke}r\left(T\right)$ has dimension a, wherea<n

$B\left(ker\right(T\left)\right)=\left\{v_1,v_{2,},...,v_a\right\}$

.

Then the basis ofV is $B\left(V\right)=\left\{v_1,v_{2,},),...,v_a\right\}∪\left\{v_{a+1},...,v_n\right\}$.

If the vectors$\left\{v_{a+1},...,v_n\right\}$are linearly independent and if$\left\{t\left(v_{a+1}\right),...,t\left(v_n\right)\right\}$form a basis for image $\left(T\right)$.

$B\left(t\right(V\left)\right)=Im\left(V\right)=\left\{\mathrm{θ},\mathrm{θ},...,\mathrm{θ},t\left(v_{a+1}\right),...,t\left(v_n\right)\right\}$

So,${\mathrm{Î\pm }}_{a+1}t\left(v_{a+1}\right),...,{\mathrm{Î\pm }}_{n}t\left(v_n\right)=\mathrm{θ}⇒a_i=0$for every $i∈\left\{a+1,...,n\right\}$.

By definition, If the image of T is finite dimensional, then$\dim \left(imT\right)$ is called the rank of T, and if the kernel of T is finite dimensional, then $\dim \left(\mathrm{ke}rT\right)$ is the nullity of T.

If V is finite dimensional, then the rank-nullity theorem holds.

$dim\left(V\right)=rank\left(T\right)+nullity\left(T\right)=dim\left(imT\right)+dim(kerT)$.

Thus, we get

$$\begin{gathered} t\left({\mathrm{Î\pm }}_{a+1}{v}_{a+1}+...+{\mathrm{Î\pm }}_n{v}_n\right)=\mathrm{θ} \\ ⇒{\mathrm{Î\pm }}_{a+1}{v}_{a+1}+...{\mathrm{Î\pm }}_n{v}_n∈ker\left(t\right) \\ {\mathrm{Î\pm }}_{a+1}{v}_{a+1}=\mathrm{θ},v_i∈B\left(V\right) \\ ⇒{\mathrm{Î\pm }}_I=0,âˆ\P Äi∈\left\{a+1,...,n\right\} \end{gathered}$$

$B\left(Im\right(V\left)\right)=\left\{t\left(v_{a+1}\right),...,t\left(v_n\right)\right\}$

$z∈V⇒z=\underset{i=1}{\overset{n}{∑}}{\mathrm{γ}}_i{v}_i$

$z-t\left(z\right)∈ker\left(t\right)$

$t\left(z\right)∈Im\left(t\right)⇒t\left(z\right)=\underset{i=a+1}{\overset{n}{∑}}{\mathrm{γ}}_i{v}_i$

$$\begin{gathered} \underset{i=1}{\overset{n}{∑}}{\mathrm{γ}}_i{v}_i-\underset{i=a+1}{\overset{n}{∑}}=\underset{i=1}{\overset{a}{∑}}{\mathrm{γ}}_i{v}_i+\underset{i=a+1}{\overset{n}{∑}}{\mathrm{γ}}_i{v}_i-\underset{i=a+1}{\overset{n}{∑}}{\mathrm{γ}}_i{v}_i \\ =\underset{i=1}{\overset{n}{∑}}{\mathrm{γ}}_i{v}_i∈ker\left(t\right) \end{gathered}$$

Thus,$dim\left(ker\right(t))+dim\left(Im\right(t))=a+(n-a)=n=dim\left(V\right)$or it can be said that the elements$u_1,…u_r,v_1,…v_n$ spanV.