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4.F Injective, surjective, and bijective transformations

The following definition is used throughout mathematics, and applies to any function, not just linear transformations.

Definition 4.31:

Let $T:V\to W$ be a function. $T$ is called injective if for any two elements $\vec{x},\vec{y}\in V$ we have that: if $T(\vec{x})=T(\vec{y})$ then $\vec{x}=\vec{y}$ .

The following Theorem shows that for linear transformations, injective is the same as having trivial kernel.

Theorem 4.32.

Let $T:V\to W$ be a linear transformation. The following statements are equivalent.

i.

$\ker T=\{\vec{0}\}.$
ii.

$T$ is injective.

Proof.

First we prove that (i) implies (ii). Assume $\ker T=\{\vec{0}\}$ . Take two vectors $\vec{x},\vec{y}\in V$ such that $T(\vec{x})=T(\vec{y})$ . Then, by linearity, $T(\vec{x}-\vec{y})=T(\vec{x})-T(\vec{y})=\vec{0}$ . Therefore, by the definition of the kernel, $\vec{x}-\vec{y}\in\ker T$ . But we assumed the only vector in the kernel is zero, and so $\vec{x}-\vec{y}=\vec{0}$ . Hence (i) implies (ii).

For the other direction, assume that (ii) is true. If $T(\vec{x})=\vec{0}$ , then $T(\vec{x})=T(\vec{0})$ . So, by (ii) $\vec{x}=\vec{0}$ . Therefore, $\ker T=\{\vec{0}\}$ . So (ii) implies (i). ∎

[Aside: The word “injective” is synonymous with “one-to-one”. I recommend thinking of injective functions as those which map the domain in to the codomain (no two elements map to the same element).]

For the following exercises you are expected to use Theorem 4.32.

Exercise 4.33:

Let $A=\begin{bmatrix}1&2&3\\ 4&5&6\end{bmatrix}$ . Prove that $A$ defines a non-injective linear transformation, whilst $A^{T}$ defines an injective linear transformation.

Exercise 4.34:

Write down 3 of your own linear transformations which are injective, and 3 which are not injective.

[End of Exercise]

Definition 4.35:

A linear transformation $T:V\to W$ is surjective if $\operatorname{im}\nolimits T=W$ .

[Aside: I recommend remembering surjective, because the french word “sur” means “onto”; and for such a linear transformation, for each vector $\vec{w}\in W$ there is a vector in $V$ which maps on to $\vec{w}$ .]

In the following examples, one can use Theorem 4.16 to justify whether or not $\operatorname{im}\nolimits T$ equals the codomain.

Example 4.36.

Here are 3 surjective linear transformations from ${\mathbb{R}}^{n}\to{\mathbb{R}}^{m}$ :

$A:=\begin{bmatrix}1&0&2\\ 0&1&3\end{bmatrix}$
$T(x,y)=x-y$
$A:=\begin{bmatrix}0&0&3\\ 2&0&0\\ 0&1&0\end{bmatrix}$

Example 4.37.

Here are 3 non-surjective linear transformations:

$A:=\begin{bmatrix}1&0&2\\ 2&0&4\end{bmatrix}$
$T(x,y)=(x-y,y-x)$
$A:=\begin{bmatrix}3&-3&0\\ 0&2&-2\\ 1&0&-1\end{bmatrix}$

Definition 4.38:

If a linear transformation $T:V\to W$ is both injective and surjective, then it is called bijective.

Theorem 4.39.

Let $T:V\to W$ be a linear transformation between finite dimensional vector spaces. Let $\mathcal{B}$ and $\mathcal{C}$ be any two bases of $V$ and $W$ , respectively. The following conditions are equivalent to each other:

i.

$T$ is bijective,
ii.

The matrix ${}_{\mathcal{C}}[T]_{\mathcal{B}}$ is invertible.

Moreover, when these are satified, the inverse transformation of $T$ is the linear transformation associated to the inverse matrix $(_{\mathcal{C}}[T]_{\mathcal{B}})^{-1}$ .

Recall from Theorem 1.12 that a matrix $A$ is invertible if and only if $\det A\neq 0$ .

[Aside: By a set-theoretic result from MATH112, any function is bijective if and only if it has an “inverse function”.]

Example 4.40.

Let $T:\mathcal{P}_{3}({\mathbb{R}})\to\operatorname{M}_{{2}}({{\mathbb{R}}})$ be the function defined by

T(a+bx+cx^{2}+dx^{3})=\begin{bmatrix}a&b\\ c&d\end{bmatrix},

for any $a,b,c,d\in{\mathbb{R}}$ . Prove, using Theorem 4.39 that $T$ defines a bijective linear transformation.

Solution: To apply that Theorem, we first need to check that $T$ is a linear transformation. Let $\vec{v}=a_{1}+b_{1}x+c_{1}x^{2}+d_{1}x^{3}$ and $\vec{w}=a_{2}+b_{2}x+c_{2}x^{2}+d_{2}x^{3}$ , and $\alpha\in{\mathbb{R}}$ . Then:

	$\displaystyle T(\alpha\vec{v}+\vec{w})$	$\displaystyle=T((\alpha a_{1}+a_{2})+(\alpha b_{1}+b_{2})x+(\alpha c_{1}+c_{2}% )x^{2}+(\alpha d_{1}+d_{2})x^{3})$
		$\displaystyle=\begin{bmatrix}\alpha a_{1}+a_{2}&\alpha b_{1}+b_{2}\\ \alpha c_{1}+c_{2}&\alpha d_{1}+d_{2}\end{bmatrix}=\alpha\begin{bmatrix}a_{1}&% b_{1}\\ c_{1}&d_{1}\end{bmatrix}+\begin{bmatrix}a_{2}&b_{2}\\ c_{2}&d_{2}\end{bmatrix}=\alpha T(\vec{v})+T(\vec{w}).$

This simultaneously verifies 1 and 2.

Next, to apply the Theorem, we need to choose a basis for the domain and the codomain. Let us use the standard ones:

\mathcal{B}=(1,x,x^{2},x^{3})

is the standard basis for $\mathcal{P}_{3}({\mathbb{R}})$ ,

\mathcal{C}=(\begin{bmatrix}1&0\\ 0&0\end{bmatrix},\begin{bmatrix}0&1\\ 0&0\end{bmatrix},\begin{bmatrix}0&0\\ 1&0\end{bmatrix},\begin{bmatrix}0&0\\ 0&1\end{bmatrix})

is the standard basis for $\operatorname{M}_{{2}}({{\mathbb{R}}})$ .

Then we compute that ${}_{\mathcal{C}}[T]_{\mathcal{B}}=\begin{bmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix}$ , which is the identity matrix, and obviously invertible. Therefore, $T$ is a bijective linear transformation.

Exercise 4.41:

Prove that the linear transformation ${\mathbb{R}}^{2}\to{\mathbb{R}}^{2}$ defined by

T(x,y)=(2x-y,x+3y)

is bijective, and find its inverse.

Exercise 4.42:

Write down your own examples of

i.

3 linear transformations which are injective but not surjective,
ii.

3 linear transformations which are surjective but not injective,
iii.

3 linear transformations which are neither injective nor surjective.

[End of Exercise]

Theorem 4.43.

Assume $T:V\to W$ is a bijective linear transformation between vector spaces over a field $F$ . If $\mathcal{B}=(\vec{x_{1}},\cdots,\vec{x_{n}})$ is a basis for $V$ , then $\mathcal{C}:=(T(\vec{x_{1}}),\cdots,T(\vec{x_{n}}))$ is a basis for $W$ .

Proof.

Since $T$ is bijective, it is surjective. So for any $\vec{y}\in W$ , there is an $\vec{x}\in V$ such that $T(\vec{x})=\vec{y}$ . Since $\mathcal{B}$ spans $V$ , there are scalars $\alpha_{i}\in F$ such that

\vec{y}=T(\sum_{i=1}^{n}\alpha_{i}\vec{x_{i}})=\sum_{i=1}^{n}\alpha_{i}T(\vec{% x_{i}}),

where the last equality follows since $T$ is linear.

Therefore, $\mathcal{C}$ spans $W$ .

To prove $\mathcal{C}$ is linearly independent, assume we have scalars $\beta_{i}\in F$ as follows:

\vec{0}=\sum_{i=1}^{n}\beta_{i}T(\vec{x_{i}})=T(\sum_{i=1}^{n}\beta_{i}\vec{x_% {i}}),

, where the last eqality follows since $T$ is linear.

Since $T$ is bijective, it is injective, and therefore $\ker T=\{\vec{0}\}$ . In particular, $\sum_{i=1}^{n}\beta_{i}\vec{x_{i}}=\vec{0}$ . Since $\mathcal{B}$ is linearly independent, $\beta_{i}=0$ for every $i=1,\cdots,n$ . Hence $\mathcal{C}$ is linearly independent. So $\mathcal{C}$ is basis for $W$ . ∎

Theorem 4.44.

Let $T:V\to W$ be a linear transformation between (finite-dimensional) vector spaces over $F$ . If $\dim V=\dim W$ , then the following are equivalent:

i.

$T$ is injective,
ii.

$T$ is surjective.

Proof.

First we prove (i) implies (ii). Assume $T$ is injective. Then by Theorem 4.32, we have $\ker T=\{\vec{0}\}$ . So by the Dimension Theorem 4.24, this implies $\dim\operatorname{im}\nolimits T=\dim V=\dim W$ . But since $\operatorname{im}\nolimits T\subset W$ , if we choose a basis for $\operatorname{im}\nolimits T$ then it must also be a basis for $W$ , and hence $\operatorname{im}\nolimits T=W$ .

To prove (ii) implies (i), assume that $T$ is surjective. So $\dim(\operatorname{im}\nolimits T)=\dim W=\dim V$ , which by the Dimension Theorem, implies that $\dim\ker T=0$ , and hence $\ker T=\{\vec{0}\}$ . By Theorem 4.32, this implies $T$ is injective. ∎

Definition 4.45:

If there exists a bijective linear transformation $T:V\to W$ , then $V$ and $W$ are said to be isomorphic.

Theorem 4.46.

Let $V$ and $W$ be finite dimensional vector spaces over the same field $F$ . Then $V$ and $W$ are isomorphic if and only if $\dim V=\dim W$ .

Proof.

If a bijective linear transformation exsits, by Theorem 4.43 the dimensions must be equal. Conversely, if the dimensions are equal, when we choose a basis for each one, they must be of the same size. So define the linear transformation associated to the identity matrix using these basis, and this must be a bijective linear transformation. ∎

Exercise 4.47:

Find a bijective linear transformation between the vector spaces $\mathcal{P}_{8}({\mathbb{R}})$ and $\operatorname{M}_{{3}}({{\mathbb{R}}})$ over ${\mathbb{R}}$ .

[End of Exercise]

[Aside: Theorem 4.46 shows that, in linear algebra, the concept of isomorphism is “uninteresting” since it is equivalent to the dimensions being the same. The reason we introduce the terminology here is due to its wide usage in other mathematical disciplines, as a way of describing when two different mathematical objects are “the same” (i.e. isomorphic), in a precisely defined sense. For example, ${\mathbb{R}}$ and ${\mathbb{R}}^{2}$ are isomorphic as sets (because there is a set-theoretic bijection between them), but they are not isomorphic as vector spaces (since their dimensions are different). Isomorphisms of groups and of rings will be studied in MATH225. Those are both abstract mathematical concepts which are defined using axioms, like we have done for fields and vector spaces.]