📂Linear Algebra

Necessary and Sufficient Conditions for Linear Transformations to be Surjective and Injective

Theorem 1¹

The following two propositions are equivalent concerning a linear transformation $T: V \to W$.

• $T$ is one-to-one.
• $N(T) = \text{ker}(T) = \left\{ \mathbf{0} \right\}$

Explanation

This means that understanding the kernel of $T$ is a method to determine whether $T$ is one-to-one or not. According to the theorem, a linear transformation being one-to-one is equivalent to the following condition.

$$ \mathbf{x} \ne \mathbf{0} \implies T(\mathbf{x}) \ne \mathbf{0} $$

Proof

• $(\implies)$

  Let’s assume that $T$ is one-to-one. Since $T$ is a linear transformation, the following holds.

  $$ T(\mathbf{0})=\mathbf{0} $$

  However, assuming that $T$ is one-to-one, the element of $V$ that satisfies the above equation is uniquely $\mathbf{0}$. Therefore, the following holds.

  $$ \text{ker}(T) = \left\{ \mathbf{0} \right\} $$

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• $(\impliedby)$

  Let’s assume $\text{ker}(T) = \left\{ \mathbf{0} \right\}$. Let’s say $\mathbf{u}, \mathbf{v} \in V$ are distinct vectors. Therefore, $\mathbf{u} - \mathbf{v} \ne \mathbf{0}$. Then, by assumption, the following holds.

  $$ T(\mathbf{u}) - T(\mathbf{v}) = T(\mathbf{u} - \mathbf{v}) \ne \mathbf{0} $$

  Thus, we obtain the following.

  $$ \mathbf{u} \ne \mathbf{v} \implies T(\mathbf{u}) - T(\mathbf{v}) $$

  Therefore, $T$ is one-to-one.

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Theorem 2

Concerning a linear transformation $T: V \to V$, if $V$ is finite-dimensional, the following propositions are equivalent.

• $T$ is one-to-one.
• $N(T) = \text{ker}(T) = \left\{ \mathbf{0} \right\}$
• $T$ is onto. In other words, the range of $T$ is the same as $V$. $R(T)=V$

Explanation

This is a special case when $V$ is finite-dimensional and $W=V$, as proved in Theorem 1. The first two propositions being equivalent was proved in Theorem 1, so we will prove the equivalence of the first and third propositions.

Proof²

Let $S = \left\{ \mathbf{v}_{1}, \dots, \mathbf{v}_{n} \right\}$ be the basis of $V$. Then, since every $T(\mathbf{v})$ can be represented by a linear transformation of $T(\mathbf{v}_{i})$, we can know that the set $Q$ generates $R(T)$.

$$ Q = \left\{ T(\mathbf{v}_{1}),\dots T(\mathbf{v}_{n}) \right\} $$

Here, since the number of elements in $Q$ is $\dim(V)=n$, $Q$ being linearly independent is equivalent to $Q$ generating $V$. But since $Q$ generates $R(T)$, showing that $Q$ is linearly independent is the same as showing $R(T)=V$. Therefore, the proof shifts to proving the following.

$T$ is one-to-one $\iff$ $Q$ is linearly independent

• $(\implies)$

  Let’s assume that $T$ is one-to-one. And let’s assume the constants $c_{i}$ satisfy the following equation.

  $$ \begin{equation} c_{1}T(\mathbf{v}_{1}) + c_{2}T(\mathbf{v}_{2}) + \cdots + c_{n}T(\mathbf{v}_{n}) = \mathbf{0} \label{1} \end{equation} $$

  Then, since $T$ is a linear transformation, the following holds.

  $$ T (c_{1}\mathbf{v}_{1} + \cdots + c_{n}\mathbf{v}_{n}) = \mathbf{0} $$

  Since we assumed $T$ is one-to-one, by Theorem 1, the constants satisfying the above equation are only $\mathbf{0}$.

  $$ c_{1}\mathbf{v}_{1} + \cdots + c_{n}\mathbf{v}_{n} = \mathbf{0} $$

  But since $\mathbf{v}_{i}$ are the elements of the basis, the constants that satisfy the above equation are only $0$.

  $$ c_{1}=c_{2}=\cdots=c_{n}=0 $$

  Thus, since the constants that satisfy $\eqref{1}$ are only $0$, $Q$ is linearly independent.

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• $(\impliedby)$

  Assume that $Q$ is linearly independent. And let’s assume the constants $c_{i}$ satisfy the following equation.

  $$ c_{1}T(\mathbf{v}_{1}) + c_{2}T(\mathbf{v}_{2}) + \cdots + c_{n}T(\mathbf{v}_{n}) = \mathbf{0} $$

  Then, by assumption, we know that the constants that satisfy the above equation are only $0$.

  $$ \begin{equation} c_{1}=c_{2}=\cdots=c_{n}=0 \label{2} \end{equation} $$

  But since $T$ is a linear transformation, the following holds.

  $$ T(c_{1}\mathbf{v}_{1} + \cdots + c_{n}\mathbf{v}_{n}) = c_{1}T(\mathbf{v}_{1}) + c_{2}T(\mathbf{v}_{2}) + \cdots + c_{n}T(\mathbf{v}_{n}) = \mathbf{0} $$

  Then, by $\eqref{2}$, only $\mathbf{0}$ satisfies the above equation.

  $$ T(\mathbf{0}) = T(0\mathbf{v}_{1} + \cdots + 0\mathbf{v}_{n}) = \mathbf{0} $$

  Therefore, by Theorem 1 $T$ is one-to-one.

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