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

By cancellation law, $0 = T(0)$.

$T(v_1 - v_2) = T(v_1 + (-v_2))$.

$x, y \in N(T) \implies T(x) = 0, T(y) = 0$.
$T(x + y) = T(x) + T(y) = 0 + 0 = 0.$
$\therefore x + y \in N(T)$.

$x \in N(T) \implies T(x) = 0, c \in F$.
$T(cx) = cT(x) = c0 = 0$
$\therefore cx \in N(T)$.

$N(T) \leq V$.

$w_1, w_2 \in R(T) \implies \exists x_1, x_2 \in V, s.t. T(x_1) = w_1, T(x_2) = w_2$.
$w_1 + w_2 = T(x_1) + T(x_2) = T(x_1 + x_2)$. $\therefore (w_1 + w_2) \in R(T)$.

$w \in R(T) \implies \exists x \in V s.t. T(x) = w$.
$cw = cT(x) = T(cx)$.
$\therefore cw \in R(T)$.

Let $\dim N(T) = k, N(T) \leq V$, and $\beta = \set{x_1, x_2, …, x_{k}}$ be a basis of $N(T)$.

Since $N(T) \leq V, \beta$ can be extended to $\gamma $, a basis of $V$. i.e. $\gamma = \beta \cup \set{y_1, y_2, …, y_{n-k}} = \set{x_1, …, x_k, y_1, …, y_{n-k}}$, where $n = \dim V$.

Need to prove $S = \set{T(v_{k + 1}), T(v_{k + 2}), …, T(v_{n})}$ is a basis for $R(T)$.

First we prove that $S $ generates $R(T)$.

Let $w \in R(T)$, then $\exists x \in v$ s.t. $T(x) = w$.

Since $\beta$ is a basis of $V, \exists c_1, c_2, …, c_{n} \in F$, s.t. $x = \sum c_ix_i$.

Then

$$T(x) = T(\sum_{i=1}^{n} c_ix_i) = \sum_{i=1}^{n} c_i T(x_i)$$

Since $\forall i = 1, 2, …, k, T(v_i) = 0$,

$$\begin{align*} R(T) &= \spa{T(x_i), …, T(x_k), T(y_1), … , T(y_{n-k})} \br &= \spa{T(y_1), …, T(y_{n-k})} \end{align*}$$

$\therefore R(T) \subseteq \set{T(y_1), T(y_2), …, T(y_{ n-k })}$.

Next we need to prove this set is linearly independent.

Consider the equation, $a_1T(y_1) + … + a_{n-k}T(y_{n-k}) = 0$

$\implies T(a_1y_1 + a_2y_2 + … + a_{n-k} y_{n-k}) = 0$

$\implies \sum_{i=1}^{n} a_iy_i \in N(T)$

Since $\beta = \set{x_1, x_2, …, x_{k}}$ is a basis of $N(T)$,

$\exists b_1, b_2, …, b_{k} \in F$ s.t.

$$\sum_{i=1}^{n-k} a_iy_i = b_1x_1 + b_2x_2 + … + b_{k} x_{k}$$

$-b_1x_1 -b_2x_2 … -b_{k} x_{k} + a_1y_1 + a_2y_2 + … + a_{n-k} y_{n-k}= 0$(*)

Since $\set{x_1, x_2, …, x_{k}, y_1, y_2, …, y_{n-k}}$ is a basis of $V$, it is a linearly independent set, in particular, for equation (*) we get

$$-b_1 = -b_2 = -b_k = a_1 =a_2 =…= a_{n-k} = 0$$

$\therefore \set{T(y_1), T(y_2), …, T(y_{ n-k })}$ is a linearly independent set.

In particular, it forms a basis of $R(T)$.

$\therefore \rank{T} = \dim R(T) = n-k$

Recall that

$$\begin{align*} \nullity{T} = \dim N(T) = k \br \dim V = n \end{align*}$$

$\therefore \rank{T} + \nullity{T} = \dim V$

$(1) \implies (2)$

Assume $T$ is$\oto$.

$\implies N(T) = {0}$

$\implies \nullity{T} = \dim N(T) = 0$

Now by the dimension theorem, $\nullity{T} + \rank{T} = \dim V$

$\implies \rank{T} = \dim V$

$\dim R(T) = \dim W$

$\because \dim V = \dim W$

$\therefore R(T) \subseteq W$

We have, $R(T) = W$

FC. $R(T) = W ,T$ is onto

$(ii) \implies (iii)$

Assume that $T$ is onto, i.e. $R(T) = W$

i.e. $\rank{T} = \dim W = \dim V$

$(iii) \implies (i)$

Assume that $\rank{T} = \dim V$

$\nullity{T} + \rank{T} = \dim V$

Analysis:

$\beta = \set{x_1, x_2, …, x_{n}} \subseteq V \to W \supseteq \gamma = \set{w_1, w_2, …, w_{n}}$

Want to define: $T: V \to W,$ s.t. $T(x_1) = w_1, T(x_2) = w_2, T(x_3) = w_3$

T: $\beta \subseteq V \to \gamma \subseteq W$

You want to extend $T$ to $V \to W$

Define a linear transformation $\R^3$ to $\R^2$ whose image is the line $y = x$ in $\R^2$.

$y = x$

$w = \spa{\set{(1, 1)}}$

$T: \R^3 \to \R^2$

$\beta = \set{e_1, e_2, e_{3}}$

Define $T(e_1) = T(e_2) = T(e_3)$

$R(T) = \spa{T(e_1), T(e_2), T(e_3)} = \spa{\set{(1, 1)}}$

$T(a, b, c) = T(a_1e_1 + a_2e_2 + a_{3} e_{3})$

$= aT(e_1) + bT(e_2) + cT(e_3) = (a+b+c)(1,1)$

Since $x \in V$ be an arbitrary vector. We need to know $T(x)=?$.

Since $\set{x_1, x_2, …, x_{n}}$ is a basis, _ $\exists$ unique scalars $a_1, a_2, …, a_{n}$ s.t. $x = \sum_{i=1}^{n} a_i x_i$

We define $T(x)$ as $T(x) := \sum_{i=1}^{n} a_i w_i$(*)

$T$ is linear transformation.

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It follows from the uniqueness of the previous theorem.