This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
In this post, we define functions between vector spaces, that is, linear maps.
Linear Maps
Since a vector space is essentially a set, a function between two vector spaces \(V,W\) exists simply as a function between sets. However, unlike an ordinary set, a vector space has an addition of elements and a scalar multiplication by elements of \(\mathbb{K}\) defined on it, so we are only interested in those functions between vector spaces (as sets) that preserve these operations.
Definition 1 Let two \(\mathbb{K}\)-vector spaces \(V,W\) be given. A function \(L:V\rightarrow W\) is called a linear map if
- for any \(\alpha\in\mathbb{K}\) and \(v\in V\), \(L(\alpha v)=\alpha L(v)\), and
- for any \(v_1,v_2\in V\), \(L(v_1+v_2)=L(v_1)+L(v_2)\).
In the special case where \(V=W\), we call these linear operators. The following propositions are almost immediate from the definition.
Proposition 2 For two \(\mathbb{K}\)-vector spaces \(V,W\) and a linear map \(L:V\rightarrow W\),
- \(L(0)=0\).
- For any \(v\in V\), \(L(-v)=-L(v)\).
- For any \(u,v\in V\), \(L(u-v)=L(u)-L(v)\).
Proof
Since a linear map preserves scalar multiplication, the first and second claims follow respectively from §Vector Spaces, ⁋Proposition 2 and §Vector Spaces, ⁋Corollary 3. Now, using the fact that a linear map preserves addition of vectors and the second claim, we have
\[L(u-v)=L\bigl(u+(-v)\bigr)=L(u)+L(-v)=L(u)+\bigl(-L(v)\bigr)=L(u)-L(v)\]so the third claim also holds.
Proposition 3 For two \(\mathbb{K}\)-vector spaces \(V,W\) and a linear map \(L:V\rightarrow W\), and scalars \(\alpha_1,\ldots,\alpha_n\) and vectors \(v_1,\ldots, v_n\) in \(V\),
\[L\left(\sum_{i=1}^k\alpha_i v_i\right)=\sum_{i=1}^kL(\alpha_iv_i)\]holds.
Proof
This is obvious by induction on \(k\).
Just as the composition of functions is a function, the composition of linear maps is also a linear map. Moreover, as we will verify later, if a linear map has an inverse, then its inverse is automatically a linear map.
Proposition 4 For three \(\mathbb{K}\)-vector spaces \(U,V,W\) and linear maps \(L_1:U\rightarrow V\), \(L_2:V\rightarrow W\), the composition \(L_2\circ L_1:U\rightarrow W\) is linear.
Proof
For any \(\alpha\in\mathbb{K}\), \(u\in U\),
\[(L_2\circ L_1)(\alpha u)=L_2(L_1(\alpha u))=L_2(\alpha L_1(u))=\alpha(L_2(L_1(u)))=\alpha(L_2\circ L_1)(u)\]Similarly, one can show that \((L_2\circ L_1)(u_1+u_2)=(L_2\circ L_1)(u_1)+(L_2\circ L_1)(u_2)\) holds for vectors.
Kernel and Image of a Linear Map
Now let us define the following.
Definition 5 For two \(\mathbb{K}\)-vector spaces \(V,W\) and a linear map \(L:V\rightarrow W\),
- if \(v_1=v_2\) whenever \(L(v_1)=L(v_2)\), then \(L\) is said to be injective.
- if for every \(w\in W\) there exists \(v\in L\) satisfying \(L(v)=w\), then \(L\) is said to be surjective.
In general, the definitions above are almost the only tools available when dealing with injective or surjective functions. However, when the objects under consideration are not mere sets but are equipped with some operations, as in the present situation, algebraic tools can also be used.
Definition 6 For two \(\mathbb{K}\)-vector spaces \(V,W\) and a linear map \(L:V\rightarrow W\), the kernel \(\ker L\) of \(L\) is the set defined by the formula
\[\ker L=\{v\in V\mid L(v)=0\}\]Also, the image \(\im L\) of \(L\) is the set defined by the formula
\[\im L=\{w\in W\mid L(v)=w\text{ for some $v\in V$}\}\]The following can be checked without difficulty.
Proposition 7 For two \(\mathbb{K}\)-vector spaces \(V,W\) and a linear map \(L:V\rightarrow W\), we have \(\ker L\leq V\) and \(\im L\leq W\).
Proof
First, \(\ker L\) is a subspace of \(V\). For any \(\alpha\in\mathbb{K}\), \(v\in\ker L\),
\[L(\alpha v)=\alpha L(v)=\alpha\cdot 0=0\]and similarly for any \(v_1\), \(v_2\in \ker L\),
\[L(v_1+v_2)=L(v_1)+L(v_2)=0+0=0\]so \(\alpha v\in\ker L\) and \(v_1+v_2\in\ker L\).
Likewise, \(\im L\) is a subspace of \(W\). Take arbitrary \(w,w_1,w_2\in W\) and \(\alpha\in\mathbb{K}\); then by definition there exist \(v,v_1,v_2\in V\) satisfying
\[L(v)=w,\quad L(v_1)=w_1,\quad L(v_2)=w_2\]and therefore
\[\alpha w=\alpha L(v)=L(\alpha v)\in\im L\]and
\[w_1+w_2=L(v_1)+L(v_2)=L(v_1+v_2)\in \im L\]We can now determine whether \(L\) is injective or surjective by means of \(\ker L\) and \(\im L\).
Proposition 8 For two \(\mathbb{K}\)-vector spaces \(V,W\) and a linear map \(L:V\rightarrow W\),
- \(L\) is injective if and only if \(\ker L=\{0\}\), and
- \(L\) is surjective if and only if \(\im L=W\).
Proof
The second claim is a tautology.
If \(L\) is injective, then the \(v\) satisfying \(L(v)=0\) must be unique, and by Proposition 2, \(0\) satisfies this equation, so we must have \(\ker L=\{0\}\). Thus, for the first claim it suffices to show
\[\ker L=\{0\}\implies\text{$L$ is injective}\]
Assume \(v_1,v_2\in V\) are given with \(L(v_1)=L(v_2)\). Then by Proposition 3,
\[0=L(v_1)-L(v_2)=L(v_1-v_2)\]so \(v_1-v_2\in\ker L\). Since \(\ker L=\{0\}\), we have \(v_1-v_2=0\), and therefore \(L\) is injective.
Speaking loosely, the smaller \(\ker L\) is, the closer \(L\) is to being injective, and the larger \(\im L\) is, the closer \(L\) is to being surjective.
Corollary 9 Let two \(\mathbb{K}\)-vector spaces \(V,W\) and a linear map \(L:V\rightarrow W\) be given.
- If \(L\) is injective, then for any linearly independent subset \(S\subset V\), the image \(L(S)\) is also linearly independent in \(W\).
- If \(L\) is surjective, then for any \(S\subset V\) satisfying \(\langle S\rangle=V\), the image \(L(S)\) also satisfies \(\span L(S)=W\).
Proof
-
For elements \(L(x_1),\ldots, L(x_k)\) of \(L(S)\), if
\[\sum_{i=1}^k\alpha_i L(x_i)=0\]then by Proposition 3,
\[0=L\left(\sum_{i=1}^k\alpha_ix_i\right)\]so by the preceding proposition we must have \(\sum\alpha_ix_i=0\). Since \(S\) is a linearly independent subset, \(\alpha_i=0\) holds for every \(i\).
-
Let any \(w\in W\) be given. Then since \(\im L=W\), there exists a suitable \(v\in V\) with \(L(v)=w\). On the other hand, since \(\langle S\rangle=V\), we can express \(v\) as a linear combination of elements of \(S\):
\[v=\sum_{i=1}^n\alpha_ix_i\]Applying \(L\) to both sides and using Proposition 3,
\[w=L(v)=L\left(\sum_{i=1}^n\alpha_ix_i\right)=\sum_{i=1}^n\alpha_i L(x_i)\]Thus every \(w\in W\) can be expressed as a linear combination of elements of \(L(S)\).
In fact, the converses of the two statements in the above corollary also hold, and their proofs are not difficult, but we omit them since we will have no occasion to use them.
Examples of Linear Maps
Example 10 For arbitrary \(\mathbb{K}\)-vector spaces \(V\) and \(W\), the map \(L:V\rightarrow W\) defined by the formula
\[L(v)=0\text{ for all $v\in V$}\]is linear. In this case, \(\im L=\{0\}\) and \(\ker L=V\).
The function defined in the above example is itself sometimes denoted by \(0\). This notation may cause confusion with the additive identity \(0\), but this function is actually the identity element in an appropriate vector space. The proof of this is not difficult, but we postpone it.
Example 11 Let an arbitrary \(\mathbb{K}\)-vector space \(V\) and a subspace \(W\leq V\) be given. The map \(\iota:W\rightarrow V\) defined by the formula
\[\iota(w)=w\text{ for all $w\in W$}\]is a linear map. In this case, \(\im\iota=W\) and \(\ker \iota=\{0\}\). That is, \(\iota\) is injective.
In the above example, in the special case where \(W=V\), the map \(\iota\) becomes the identity function \(\id_V\). ([Set Theory] §Operations Between Functions, ⁋Example 3)
Example 12 Consider arbitrary \(\mathbb{K}\)-vector spaces \(V\), \(W\), and their product \(V\times W\). Then the map \(\pr_1:V\times W\rightarrow V\) defined by the formula
\[\pr_1((v,w))=v\]is a linear map. We can easily check that \(\im\pr_1=V\) and
\[\ker \pr_1=\{(0,w)\mid w\in W\}\]Of course, \(\pr_2:V\times W\rightarrow W\) can be defined similarly, and this can be extended to ordered \(n\)-tuples. In particular, for Euclidean space \(\mathbb{K}^n\),
\[\pr_i((a_1,\ldots, a_n))=a_i\]defines a linear map \(\pr_i:\mathbb{K}^n\rightarrow \mathbb{K}\).
\(\pr\) is the initial letter of projection, and is sometimes written simply as \(p\) or \(\pi\).
Example 13 Let the function \(D:\mathbb{K}[\x]\rightarrow \mathbb{K}[\x]\) on \(\mathbb{K}[\x]\) be defined by the formula
\[D\left(\sum_{i=0}^\infty a_i\x^i\right)=\sum_{i=1}^\infty ia_i\x^{i-1}\](Here \((a_i)\) is finitely supported.) Then \(D\) is linear, and \(\im D= \mathbb{K}[\x]\). Also, \(\ker D\) is the set of all constant polynomials.
The last example is not merely a simple example; it can be thought of as the prototype of the isomorphism that we will examine in the next post.
Example 14 Let an arbitrary \(n\)-dimensional \(\mathbb{K}\)-vector space \(V\) be given, and let \(\mathcal{B}=\{x_1,\ldots, x_n\}\) be a basis of \(V\). That is, for any \(v\in V\), there always exist scalars \(v_1,\ldots, v_n\) such that
\[v=\sum_{i=1}^n v_i x_i\]and these scalars are uniquely determined. Therefore, we can define a function \(L:V\rightarrow \mathbb{K}^n\) by the formula \(v\mapsto (v_1,v_2,\ldots, v_n)\in\mathbb{K}^n\).
Then \(L\) is linear. For any \(v,w\in V\), writing
\[v=\sum_{i=1}^n v_i x_i,\quad w=\sum_{i=1}^n w_i x_i\]for any \(\alpha\in\mathbb{K}\) we have
\[\alpha L(v)=\alpha(v_1,v_2,\ldots,v_n)=(\alpha v_1,\alpha v_2,\ldots,\alpha v_n)\]while
\[\alpha v=\alpha\sum_{i=1}^nv_i x_i=\sum_{i=1}^n\alpha v_i x_i\]so the value of \(L(\alpha v)\) is the same as that of \(\alpha L(v)\). Similarly, comparing the values of \(L(v)+L(w)\) and \(L(v+w)\),
\[L(v)+L(w)=(v_1+w_1,v_2+w_2,\ldots,v_n+w_n)=L(v+w)\]holds.
\(\ker L\) becomes \(\{0\}\) because \(\mathcal{B}\) is linearly independent. On the other hand, for any \((\alpha_1,\ldots,\alpha_n)\in\mathbb{K}^n\) the linear combination
\[\sum_{i=1}^n\alpha_i x_i\]obviously belongs to \(V\), so \(L\) is injective.
References
[Goc] M.S. Gockenbach, Finite-dimensional linear algebra, Discrete Mathematics and its applications, Taylor&Francis, 2011.
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