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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

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

Let be a transformation from V to W.

$$\begin{gathered} dim\left(ker\right(t\left)\right)=a鈮�+鈭�鈬�B\left(ker\right(t\left)\right)=\left\{v_1,...,v_a\right\} \\ dim\left(Im\right(t\left)\right)=b鈮�+鈭�鈬�B\left(Im\right(t\left)\right)=\left\{v_1,...,v_b\right\} \end{gathered}$$

To proveV is finite-dimensional let us consider the contradiction thatV is not finite-dimensional linear space.

Let $v鈭�V$,

$t\left(v\right)鈭�Im\left(t\right)鈬�t\left(v\right)=\underset{I=1}{\overset{b}{鈭�}}{纬}_i{v}_i$

Since,$Im\left(t\right)=\left\{v_1,...,v_b\right\}$there exists $r鈭�V$,$t\left(r_i\right)=V_i$in $\mathrm{Im}\left(t\right)$ for all i in $\left\{1,...,b\right\}$.

$t\left(V\right)=\underset{i=1}{\overset{b}{鈭�}}{纬}_1{v}_i=\underset{i=1}{\overset{b}{鈭�}}{纬}_{1}t\left(r_i\right)=t\underset{i=1}{\overset{b}{鈭�}}{纬}_i\left(r_i\right)$

This implies,$V=\underset{i=1}{\overset{b}{鈭�}}{纬}_i{r}_i$.

Let $v鈭�\mathrm{ke}r\left(t\right)$, then this implies $t\left(v\right)=\underset{i=1}{\overset{a}{鈭�}}{未}_i{v}_i$.

Both the times we were able to express as linear combination of finite number vectors from V.

Therefore, our assumptionV is non-finite dimensional linear space is wrong.

Since,$v鈭�V$was any vector, this follows for any other vector.

$V=\underset{i=1}{\overset{b}{鈭�}}{纬}_I{r}_i$or $V=\underset{i=1}{\overset{a}{鈭�}}{未}_i{r}_i$and those are linearly independent.

Thus,$dim\left(V\right)=a+b$.

This implies, $dim\left(V\right)=dim\left(ker\right(t))+dim\left(Im\right(t))$.