Definition 1.2.1.
The kernel or nullspace of \(T\) is defined as
\begin{equation*}
\ker(T) = \Null(T) =
T^{-1}(\0_W) = \{v \in V\mid T(v) = \0_W\}.
\end{equation*}
Section 1.2 Measuring injectivity and surjectivity
Subsection 1.2.1 Injective and surjective linear maps: assessment and implications.
Given a linear map \(T:V \to W\) (between vector spaces \(V,W\)), we know the function-theoretic definitions of injective and surjective. Let’s first give an alternate characterization of these primitives, and then explore how linearity informs and refines our knowledge.
Given a function \(f:X\to Y\) between sets \(X\) and \(Y\text{,}\) and an element \(y \in Y\text{,}\) the inverse image of \(y\) is the set of elements of \(X\) which map onto \(y\) via \(f\text{,}\) that is
\begin{equation*}
f^{-1}(y) = \{ x\in X\mid f(x) =y\}.
\end{equation*}
Thus an equivalent way in which to say that a function \(f\) is surjective is if for every \(y \in
Y\text{,}\) the inverse image, \(f^{-1}(y)\) is non-empty, and an equivalent way to say that a function is injective is to say for every \(y \in Y\text{,}\) the inverse image, \(f^{-1}(y)\) is either empty or consists of a single element.
For a linear map \(T:V\to W\text{,}\) the inverse image of the \(\0_W\) in \(W\) plays a special role and is given name recognition:
One recalls that since \(T(\0_V) = \0_W\text{,}\) we always have \(\0_V \in
\ker(T)\text{,}\) and indeed the kernel (null space) is a subspace of \(V\text{.}\)
Now using that for a linear map \(T\text{,}\) \(T(v) = T(v')\) if and only if \(T(v-v') = \0_W\text{,}\) one easily deduces the familiar proposition below. Since it should be clear from context, we shall henceforth simply write \(\0\text{,}\) leaving to the reader to understand the space to which we are referring.
Proposition 1.2.2.
A linear map \(T:V\to W\) is injective if and only if \(\ker(T) = \{\0\}.\)
The significance of this proposition is that rather than checking that \(T^{-1}(w)\) consists of at most one element for every \(w\in W\) (as for a generic function), for linear maps it is enough to check for the single element \(w=\0\text{.}\) The kernel also says something about the image of a linear map. Suppose \(T(v_0) = w\text{.}\) Then \(T(v) =
w\) if and only if \(v = v_0+k\text{,}\) where \(k\in
\ker(T)\text{.}\) Said another way
\begin{equation}
T^{-1}(w) = \{v_0+k \mid k \in \ker(T)\} = v_0 + \ker(T).\tag{1.2.1}
\end{equation}
Now that we have reminded ourselves of the definitions and basic properties, we explore how bases dovetail with the notion of injective and surjective linear maps.
Proposition 1.2.3. Linear maps and bases.
Let \(T: V\to W\) be a linear map between vector spaces and suppose that \(V\) is finite-dimensional with basis \(\cB=\{v_1, \dots, v_n\}.\) Then
- \(T\) is injective if and only if \(\{T(v_1), \dots, T(v_n)\}\) is a linearly independent subset of \(W\text{.}\)
- \(T\) is surjective if and only if \(\{T(v_1), \dots, T(v_n)\}\) is a spanning set for \(W\text{.}\)
Proof of (1).
First suppose that \(T\) is injective and to proceed by contradiction that \(\{T(v_1), \dots, T(v_n)\}\) is linearly dependent. Then there exist scalars \(a_1, \dots,
a_n\) not all zero, so that
\begin{equation*}
a_1T(v_1) + \cdots + a_nT(v_n) = \0.
\end{equation*}
By (1.1.1)
\begin{equation*}
T(a_1v_1+
\cdots+a_n v_n) = a_1T(v_1) + \cdots + a_r T(v_r) =\0,
\end{equation*}
which means that \((a_1v_1+
\cdots+a_n v_n)\in \ker(T).\) Since \(\cB=\{v_1, \dots,
v_n\}\) is a linearly independent set and the \(a_i\)’s are not all zero, we conclude \(\ker(T) \ne \{\0\}\) which contradicts that \(T\) is injective.
Conversely suppose that \(\{T(v_1), \dots, T(v_n)\}\) is a linearly independent subset of \(W\text{,}\) but that \(T\) is not injective. Then \(\ker(T) \ne \{\0\},\) and since \(\{v_1, \dots, v_n\}\) is a basis for \(V\text{,}\) there exist scalars \(a_1, \dots, a_n\) not all zero so that \(a_1v_1 + \cdots + a_n v_n \in
\ker(T).\) But this in turn says that
\begin{equation*}
\0 = T(a_1v_1 + \cdots + a_n v_n) = a_1T(v_1) +
\cdots + a_n T(v_n),
\end{equation*}
(again by Proposition 1.1.5) showing that \(\{T(v_1), \dots, T(v_n)\}\) is linearly dependent, a contradiction.
Proof of (2).
First suppose that \(\{T(v_1), \dots,
T(v_n)\}\) is a spanning set for \(W\text{.}\) Since \(T(V),\) the image of \(T\text{,}\) is a subspace of \(W\text{,}\) and \(\{T(v_1), \dots,
T(v_n)\}\subset T(V)\)
\begin{equation*}
W = \Span\{ T(v_1), \dots,
T(v_n)\} \subseteq T(V),
\end{equation*}
so \(T\) is surjective.
Conversely if \(T\) is surjective, then \(T(V) =
W.\) But with a very slight generalization of Proposition 1.1.5, we see that
\begin{equation*}
W=T(V) = T(\Span\{v_1, \dots, v_n\}) = \Span\{T(v_1),
\dots, T(v_n)\},
\end{equation*}
showing that \((v_1),
\dots, T(v_n)\}\) is a spanning set for \(W.\)
Example 1.2.4. Some easy-to-check isomorphisms.
- For an integer \(n \ge 1\text{,}\) the vector spaces \(V= F^{n+1}\) and \(W=P_n(F)\) are isomorphic. One bijective linear map which demonstrates this is \(T: V\to W\) given by \(T(a_0, \dots, a_n) = a_0 + a_1 x+ \cdots + a_n x^n\) where we have written the element \((a_0,\dots, a_n)\in F^{n+1}\) as a row vector for typographical simplicity.
- A more explicit example is that \(F^6\) is isomorphic to \(M_{2\times3}(F)\) via \(T(a_1, \dots, a_6) = \begin{bmatrix}a_1\amp a_2\amp a_3\\a_4\amp a_5\amp a_6\end{bmatrix}\text{.}\)
Subsection 1.2.2 Notions connected to isomorphism
There are many important concepts related to isomorphism. Taking a top-down approach, one of the most important theorems in the classification of vector spaces applies to finite-dimensional vector spaces. The classification theorem is
Theorem 1.2.5. Classification theorem for finite-dimensional vector spaces.
Two finite-dimensional vector spaces \(V\) and \(W\) defined over the same field \(F\) are isomorphic if and only if \(\dim V = \dim
W.\)
The proof of this theorem (often stated succinctly as “map a basis to a basis”) captures a great deal about the dynamics of linear algebra including how to define a map known to be linear and how to determine whether it is injective or surjective. Try to write the proof on your own.
Proof.
First let’s suppose that \(\dim V = \dim W\text{.}\) That means that any bases for the two spaces have the same cardinality. Let \(\{v_1, \dots, v_n\}\) be a basis for \(V\text{,}\) and \(\{w_1, \dots, w_n\}\) be a basis for \(W\text{.}\) By Theorem 1.1.6, there is a unique linear map which takes \(T(v_i) = w_i\text{,}\) for \(i=1, \dots, n.\) By Proposition 1.2.3, it follows that \(T\) is both injective and surjective, hence an isomorphism.
Conversely, suppose that \(T:V\to W\) is an isomorphism and the \(\{v_1, \dots, v_n\}\) is a basis for \(V.\) Once again by Proposition 1.2.3, it follows that \(\{T(v_1), \dots, T(v_n)\}\) is a basis for \(W\text{,}\) and since the cardinality of any basis determines the dimension of the space, we have \(\dim V =
\dim W.\)
