Space of Linear Maps

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Extension by linearity

Theorem 1 (Extension by linearity) Let \(V\) be an arbitrary \(\mathbb{K}\)-vector space with basis \(\mathcal{B}\). For any other \(\mathbb{K}\)-vector space \(W\), whenever a function \(g:\mathcal{B}\rightarrow W\) is given, there exists a unique linear map \(G:V\rightarrow W\) such that \(g=G\circ\iota\).

Here \(\iota:\mathcal{B}\rightarrow V\) denotes the inclusion \(\mathcal{B}\hookrightarrow V\).

Proof

For a given function \(g\), it is clear that a linear map \(G\) satisfying the condition must be unique. Indeed, if \(G':V\rightarrow W\) is another linear map satisfying the given condition, then for any \(v\in V\), writing

\[v=\sum_{x\in \mathcal{B}}v_xx\]

we have

\[\begin{aligned}(G-G')\left(\sum_{x\in \mathcal{B}}v_xx\right)&=\sum_{x\in\mathcal{B}}v_x(G-G')(x)=\sum_{x\in\mathcal{B}}v_x(G-G')(\iota(x))\\&=\sum_{x\in\mathcal{B}}v_x(G\circ \iota-G'\circ\iota)(x)=\sum_{x\in\mathcal{B}}v_x(g-g)(x)=0\end{aligned}\]

as claimed.

Now we must actually construct \(G\). Naturally, for any \(v=\sum_{x\in\mathcal{B}}v_xx\), we define

\[G(v)=\sum_{x\in\mathcal{B}} v_xg(x)\]

Since the expression of \(v\) as a linear combination of elements of \(\mathcal{B}\) is unique, \(G\) is well-defined, and one can easily verify that \(G\) is a linear map.

That is, we can always find \(G:V\rightarrow W\) making the following diagram commute.

Conversely, given any linear map \(G:V\rightarrow W\), we may restrict it to \(\mathcal{B}\) to define a function \(g=G\circ\iota\), and by the uniqueness part of the theorem above, \(G\) is the only linear map satisfying this equation. Thus there is a bijection between the following two sets:

\[\{\text{functions from $\mathcal{B}$ to $W$}\}\longleftrightarrow\{\text{linear maps from $V$ to $W$}\}\]

In other words, a linear map from \(V\) to \(W\) is completely determined by how it acts on the basis \(\mathcal{B}\), and if \(V\) is finite-dimensional, this means that the linear map is determined solely by its values at finitely many elements.

In particular, assume that the codomain \(W\) is also a finite-dimensional \(\mathbb{K}\)-vector space, and fix a basis \(\mathcal{B}=\{x_1,\ldots, x_n\}\) of \(V\) and a basis \(\mathcal{C}=\{y_1,\ldots,y_m\}\) of \(W\). Then by the preceding argument, a linear map \(L\) from \(V\) to \(W\) is completely determined by the \(n\) vectors in \(W\)

\[L(x_1),L(x_2)\ldots, L(x_n)\]

and as elements of \(W\), these can again be expressed as linear combinations of elements of \(\mathcal{C}\)

\[\begin{aligned}L(x_1)&=\alpha_{11}y_1+\alpha_{21}y_2+\cdots+\alpha_{m1}y_m\\L(x_2)&=\alpha_{12}y_1+\alpha_{22}y_2+\cdots+\alpha_{m2}y_m\\&\phantom{a}\vdots\\L(x_n)&=\alpha_{1n}y_1+\alpha_{2n}y_2+\cdots+\alpha_{mn}y_m\end{aligned}\tag{1}\]

Then the value of \(L\) at an arbitrary element of \(V\) can be expressed as a linear combination with respect to the basis \(\mathcal{C}\), using the scalars \(\alpha_{i,j}\) together with the scalars \(v_1,\ldots, v_n\) appearing in the coordinate expression \(v=\sum_{i=1}^n v_ix_i\).

\[\begin{aligned}L(v_1x_1)&=\alpha_{11}v_1y_1+\alpha_{21}v_1y_2+\cdots+\alpha_{m1}v_1y_m\\L(v_2x_2)&=\alpha_{12}v_2y_1+\alpha_{22}v_2y_2+\cdots+\alpha_{m2}v_2y_m\\&\phantom{a}\vdots\\L(v_nx_n)&=\alpha_{1n}v_ny_1+\alpha_{2n}v_ny_2+\cdots+\alpha_{mn}v_ny_m\end{aligned}\]

Thus, adding the corresponding sides, the left-hand side becomes

\[L(v_1x_1)+L(v_2x_2)+\cdots+L(v_nx_n)=L(v_1x_1+v_2x_2+\cdots+v_nx_n)=L(v)\]

and the right-hand side becomes

\[(\alpha_{11}v_1+\alpha_{12}v_2+\cdots+\alpha_{1n}v_n)y_1+(\alpha_{21}v_1+\alpha_{22}v_2+\cdots+\alpha_{2n}v_n)y_2+\cdots+(\alpha_{m1}v_1+\alpha_{m2}v_2+\cdots+\alpha_{mn}v_n)y_m\]

Therefore \(L\) can be understood as the correspondence

\[v=\sum_{i=1}^n v_ix_i\quad\mapsto\quad \sum_{j=1}^m\left(\sum_{i=1}^n\alpha_{ji}v_i\right)y_j=L(v)\]

Using the theorem above, we can prove the following proposition corresponding to §Retraction and Section, ⁋Proposition 1.

Corollary 2 Let \(V,W\) be two \(\mathbb{K}\)-vector spaces and let \(L:V\rightarrow W\) be a linear map.

1. If \(L\) is injective, there exists a linear map \(R:W\rightarrow V\) such that \(R\circ L=\id_V\).
2. If \(L\) is surjective, there exists a linear map \(S:W\rightarrow V\) such that \(L\circ S=\id_W\).

Proof

1. First suppose \(L\) is injective, and choose a basis \(x_1,\ldots,x_n\) of \(V\). Then \(L(x_1),\ldots, L(x_n)\) are linearly independent, and therefore we can find a basis \(\mathcal{B}\) of \(W\) containing them. Now define a function \(r:\mathcal{B}\rightarrow V\) by

   \[r(v)=\begin{cases}x_i&\text{if $v=L(x_i)$}\\0&\text{otherwise}\end{cases}\]

   and apply Theorem 1 (Extension by linearity) to obtain a linear map, which we call \(R\). Then for any element \(x_i\) of the basis \(\{x_1,\ldots,x_n\}\) of \(V\), we have \((R\circ L)(x_i)=x_i\), and hence by the uniqueness part of Theorem 1, \(R\circ L=\id_V\).

2. Suppose \(L\) is surjective, and choose a basis \(x_1,\ldots,x_n\) of \(V\). Then \(L(x_1),\ldots, L(x_n)\) span \(W\), so we can select some of these vectors to form a basis \(\mathcal{B}\) of \(W\). Without loss of generality, let \(\mathcal{B}=\{L(x_1),\ldots, L(x_m)\}\) (\(m\leq n\)). Define a function \(s:\mathcal{B}\rightarrow V\) by

   \[s(v)=x_k\qquad v=L(x_k)\]

   and apply Theorem 1 (Extension by linearity) to obtain a linear map, which we call \(S\). Now for any element \(L(x_k)\) of the basis \(\mathcal{B}\) of \(W\), we have \((L\circ S)(L(x_k))=L(x_k)\), so again by the uniqueness part of Theorem 1, \(L\circ S=\id_W\).

Space of Linear Maps

Lemma 3 Let \(V\) and \(W\) be \(\mathbb{K}\)-vector spaces. If \(L,L_1,L_2\) are linear maps from \(V\) to \(W\) and \(\alpha\in\mathbb{K}\), then

\[L_1+L_2:v\mapsto L_1(v)+L_2(v),\qquad \alpha L:v\mapsto \alpha L(v)\]

are also linear.

Proof

Let \(v, v_1,v_2\in V\) and \(\alpha\in\mathbb{K}\). Then

\[\begin{aligned} (L_1+L_2)(v_1+v_2)&=L_1(v_1+v_2)+L_2(v_1+v_2)\\ &=L_1(v_1)+L_1(v_2)+L_2(v_1)+L_2(v_2)\\ &=L_1(v_1)+L_2(v_1)+L_1(v_2)+L_2(v_2)\\ &=(L_1+L_2)(v_1)+(L_1+L_2)(v_2) \end{aligned}\]

and

\[\begin{aligned} (L_1+L_2)(\alpha v)&=L_1(\alpha v)+L_2(\alpha v)=\alpha L_1(v)+\alpha L_2(v)\\ &=\alpha(L_1(v)+L_2(v))\\ &=\alpha (L_1+L_2)(v). \end{aligned}\]

Therefore \(L_1+L_2\) is a linear map. The second claim can be shown similarly.

Thus we may make the following definition.

Definition 4 For two \(\mathbb{K}\)-vector spaces \(V,W\), the \(\mathbb{K}\)-vector space obtained by equipping the set of linear maps from \(V\) to \(W\) with the operations from Lemma 3 is denoted \(\Hom_\mathbb{K}(V,W)\), or simply \(\Hom(V,W)\) when the field \(\mathbb{K}\) is clear from context.

In particular, when \(W=\mathbb{K}\), we call \(\Hom(V,\mathbb{K})\) the dual space of \(V\) and denote it by \(V^\ast\). The elements of \(V^\ast\) are called linear functionals.

The zero vector in the vector space \(\Hom(V,W)\) is the function \(0\) sending every element to 0. (§Linear Maps, ⁋Example 10) When referring to this function, we shall call it the zero function for convenience.

Suppose both spaces \(V,W\) are finite-dimensional, and let \(\mathcal{B}=\{x_1,\ldots, x_n\}\) and \(\mathcal{C}=\{y_1,\ldots, y_m\}\) be bases of \(V\) and \(W\), respectively. Consider the \(mn\) functions from \(\mathcal{B}\) to \(W\)

\[f_i^j(x)=\begin{cases}y_j&\text{if $x=x_i$}\\0&\text{otherwise}\end{cases}\]

That is, \(f_i^j\) is the function sending only \(x_i\) to \(y_j\) and everything else to 0. Then by Theorem 1 (Extension by linearity), there exists a unique linear map \(B_i^j\) such that \(f_i^j=B_i^j\circ\iota\).

Proposition 5 Let \(V,W\) be two finite-dimensional \(\mathbb{K}\)-vector spaces with bases \(\{x_1,\ldots,x_n\}\) and \(\{y_1,\ldots,y_m\}\), respectively. Then \(\Hom(V,W)\) is an \(mn\)-dimensional vector space, and the \(mn\) linear maps \(B_i^j\) defined above form a basis of \(\Hom(V,W)\).

Proof

It suffices to prove the claim about the basis.

First, the \(B_i^j\) are linearly independent. For scalars \(\alpha_{11},\ldots,\alpha_{mn}\), suppose

\[\alpha_{11}B_1^1+\alpha_{12}B_2^1+\cdots+\alpha_{mn}B_n^m=0\]

That is, both sides are the zero function from \(V\) to \(W\), and therefore for any \(v\in V\) the equation

\[\alpha_{11}B_1^1(v)+\alpha_{12}B_2^1(v)+\cdots+\alpha_{mn}B_n^m(v)=0\]

holds. In particular, this equation also holds for \(v=x_1,\ldots, x_n\), and then

\[\alpha_{11}B_1^1(x_k)+\alpha_{12}B_2^1(x_k)+\cdots+\alpha_{mn}B_n^m(x_k)=0\]

But by the definition of \(B_i^j\), the value \(B_i^j(x_k)\) is \(y_j\) only when \(i=k\), so the above equation becomes

\[\alpha_{1k}y_1+\alpha_{2k}y_2+\cdots+\alpha_{mk}y_k=0\]

Now \(y_1,\ldots,y_k\) are linearly independent, so \(\alpha_{1k},\ldots,\alpha_{mk}\) are all \(0\). Since \(k\) was arbitrary, \(\alpha_{11},\ldots,\alpha_{mn}\) are all 0, and the \(B_i^j\) are linearly independent.

On the other hand, these \(B_i^j\) span \(\Hom(V,W)\). Given an arbitrary \(L\in\Hom(V,W)\), we can find scalars \(\alpha_{11},\ldots,\alpha_{mn}\) satisfying equation (1) from the introduction. Now the map defined by

\[L'(v)=\sum_{i,j}\alpha_{ji}B_i^j(v)\]

is a linear map. Moreover, substituting \(v=x_k\) gives

\[L'(x_k)=\sum_{i,j}\alpha_{ji}B_i^j(x_k)=\sum_{j=1}^m\alpha_{jk}B_k^j(x_k)=\sum_{j=1}^m\alpha_{jk}y_j=L(x_k)\]

Hence by the uniqueness part of Theorem 1 (Extension by linearity), we have \(L'=L\).

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References

[Lee] 이인석, 선형대수와 군, 서울대학교 출판문화원, 2005.

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