Mathematical Immaturity

2.7 One-to-one linear transformations

$\quad$ In this section, $V$ and $W$ denote linear spaces with the same scalars, and $T: V \to W$ denotes a linear transformation in $\mathscr{L}(V,W).$ The linearity property of $T$ allows us to express the one-to-one property in several equivalent forms.

$\quad$ Theorem 2.10.$\quad$ Let $T:V \to W$ be a linear transformation in $\mathscr{L}(V,W).$ Then, the following statements are equivalent.
$\quad$ (a) $\quad$ $T$ is one-to-one on $V.$
$\quad$ (b) $\quad$ $T$ is invertible and its inverse $T^{-1}:T(V) \to V$ is linear.
$\quad$ (c) $\quad$ For all $x$ in $V,$ $T(x) = O$ implies $x = O.$ That is, the null space $N(T)$ contains only the zero element of $V.$

$\quad$Proof.$\quad$ We will prove that (a) implies (b), (b) implies (c), and (c) implies (a). First, suppose that (a) is true; that is, assume $T$ is one-to-one on $V.$ Then, we know from Theorem 2.9 that $T$ has a left inverse if and only if $T$ is one-to-one on $V.$ But by definition, this means that $T$ is invertible with the unique left inverse denoted by $T^{-1}.$ To prove that $T^{-1}$ is linear, let $u = T(x)$ and $v = T(y)$ be elements of $T(V)$ and let $a$ and $b$ be scalars in $\mathscr{L}(V,W).$ Because $T$ is linear, we get \begin{align*} T^{-1}(au + bv) &= T^{-1}[aT(x) + bT(y)] \\ &= T^{-1}T(ax + by) \\ &= ax + by \end{align*} But since $u = T(x)$ and $v = T(y),$ it follows that $ax + by= aT^{-1}(u) + bT^{-1}(v).$ Thus, $T^{-1}$ is linear, which means (a) implies (b).

$\quad$ Now, suppose $T$ is invertible and its inverse $T^{-1}: T(V) \to V$ is linear. Then, let $x$ be an element of $V$ such that $T(x) = O.$ Taking the left inverse, we get $$T^{-1}T(x) = T^{-1}(O) = x.$$ But because $T(V)$ is a subspace of the linear space $W,$ from Theorem 1.3 (a), we can write $O = 0w$ for any $w$ in $T(V).$ And because $T^{-1}$ is a linear transformation with values in the linear space $V$, we have \begin{align*} x &= T^{-1}(O) \\ &= T^{-1}(0w) \\ &= 0T^{-1}(w) \\ &= O \end{align*} Hence, if $T$ is invertible with its inverse $T^{-1}: T(V) \to T$ being linear, then $T(x) = O$ implies $x = O$ for all $x$ in $V.$ In other words, (b) implies (c).

$\quad$ Finally, assume that the null space of $T$ contains only the zero element of $V.$ That is, for all $x$ in $V,$ $T(x) = O$ implies $x = O.$ Then, let $u$ and $v$ be elements of $V$ such that $T(u) = T(v).$ Because $T$ is a linear transformation, we have $T(u) - T(v) = T(u - v) = O.$ But by our assumption, this implies that $u - v = O,$ and that $u = v.$ Thus, $T$ is one-to-one on $V.$ As such, we have shown that (a) implies (b), (b) implies (c), and (c) implies (a). This completes the proof.

$\quad$ When $V$ is finite-dimensional, the one-to-one property can be formulated in terms of independence and dimensionality:

$\quad$ Theorem 2.11. $\quad$ Let $T: V \to W$ be a linear transformation in $\mathscr{L}(V, W)$ and assume that $V$ is finite-dimensional, say $\dim V = n.$ Then, the following statements are equivalent.
$\quad$ (a) $\quad$ $T$ is one-to-one on $V.$
$\quad$ (b) $\quad$ If $e_1, \dots, e_p$ are independent elements in $V,$ then $T(e_1), \dots, T(e_p)$ are independent elements in $T(V).$
$\quad$ (c) $\quad$ $\dim T(V) = n.$
$\quad$ (d) $\quad$ If $\{e_1, \dots, e_p\}$ is a basis for $V,$ then $\{T(e_1), \dots, T(e_p)\}$ is a basis for $T(V).$

$\quad$ Proof. $\quad$ We will prove that (a) implies (b), (b) implies (c), (c) implies (d), and (d) implies (a). First, assume that $T$ is one-to-one on $V.$ Now, suppose we have \begin{align*} \sum_{i=1}^p a_iT(e_i) &= O \end{align*} where $e_1, \dots, e_p$ are independent elements of $V$ and $a_1, \dots, a_p$ are scalars shared by $V$ and $W.$ Then, by linearity, we can rewrite the above sum as \begin{align*} T\left(\sum_{i=1}^p a_ie_i\right) &= O \end{align*} But as we proved in Theorem 2.10, this means that $\sum_{i=1}^p a_ie_i = O.$ And since $e_1, \dots, e_p$ are independent, this implies that $a_1 = \dots = a_p = 0,$ and thus, that $T(e_1), \dots, T(e_p)$ are independent. Hence, (a) implies (b).

$\quad$ Now, assume that $\{e_1, \dots, e_n\}$ are independent elements of $V.$ By assumption, $T(e_1), \dots, T(e_n)$ are independent elements of $T(V).$ From Theorem 1.7 we know that these elements are a subset of a basis for $T(V).$ Hence, $\dim T(V) \geq n.$ But from the Nullity Plus Rank Theorem, we have $\dim T(V) \leq n.$ Hence, $\dim T(V) = n,$ which means (b) implies (c).

$\quad$ Next, assume that $\dim T(V) = n.$ Then, let $\{e_1, \dots, e_p\}$ be a basis for $V.$ (Note that in order for this set to be a basis for $V,$ $p$ must be equal to $n.$) Because $\{e_1, \dots, e_p\}$ is a basis for $V,$ by definition it is an independent set of elements in $V.$ Thus, if (b) holds, then $\{T(e_1), \dots, T(e_p)\}$ is a set of $n$ independent elements in $T(V).$ But from Theorem 1.7, we recall that if $\dim T(V) = n,$ then $\{T(e_1), \dots, T(e_p)\}$ is a basis for $T(V).$ Thus, we have shown that if (b) holds, then (c) implies (d).

$\quad$ Finally, assume that (d) holds. That is, if $\{e_1, \dots, e_p\}$ is a basis for $V,$ then $\{T(e_1), \dots, T(e_p)\}$ is a basis for $T(V).$ We will now prove that this implies $T$ is one-to-one on $V.$ Let $T(x) = O$ for any $x$ in $V.$ We can rewrite $T(x)$ as the linear combination \begin{align*} T(x) &= T\left(\sum_{i=1}^n c_ie_i\right) \\ &= \sum_{i=1}^n c_iT(e_i) \end{align*} But if (d) holds, then $T(x) = O$ implies $c_1 = \dots = c_n = 0,$ hence \begin{align*} x &= \sum_{i=1}^n c_ie_i = O \end{align*} From Theorem 2.10, we recall that this implies $T$ is one-to-one on $V.$ Thus, we have shown that (d) implies (a), completing the proof.