Isomorphisms

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 mathematics, after defining certain objects we usually proceed to group and classify them according to whether they are the same. For example, when dealing with sets, two sets \(A,B\) of the same size are regarded as the same, which by definition means there exists a bijection between \(A\) and \(B\).

Of course, we cannot simply carry this over to vector spaces. If we were to regard two vector spaces of the same set-theoretic size as the same, then by [Set Theory] §Natural Numbers and Infinite Sets, ⁋Corollary 15, all finite-dimensional vector spaces over an infinite field \(\mathbb{K}\) would have to be regarded as the same. Moreover, since functions in general do not preserve the addition and scalar multiplication of a vector space, this is clearly inappropriate for studying vector spaces.

Isomorphic Vector Spaces

Therefore we define as follows.

Definition 1 Let two \(\mathbb{K}\)-vector spaces \(V,W\) and a linear map \(L:V\rightarrow W\) be given. Then \(L\) is called an isomorphism if there exists another linear map \(L':W\rightarrow V\) such that \(L\circ L'=\id_W\) and \(L'\circ L=\id_V\). When such an isomorphism between \(V\) and \(W\) exists, we say that \(V,W\) are isomorphic and write \(V\cong W\).

By definition, any isomorphism is a bijection between the two sets \(V,W\). Moreover, the converse also holds by the following lemma.

Lemma 2 Let \(L:V\rightarrow W\) be an isomorphism between two \(\mathbb{K}\)-vector spaces \(V\), \(W\). Then the inverse function \(L^{-1}\) exists, and \(L^{-1}\) is linear.

Proof

The existence of \(L^{-1}\) is a result from set theory, and in this case \(L\circ L^{-1}=\id_W\) and \(L^{-1}\circ L=\id_V\).

Thus it suffices to show that \(L^{-1}\) is linear. First, for any \(\alpha\in\mathbb{K}\) and \(w\in W\), we must show that \(L^{-1}(\alpha w)=\alpha L^{-1}(w)\). For any \(w\in W\) there exists a unique \(v\in V\) such that \(L(v)=w\), and then \(L(\alpha v)=\alpha L(v)=\alpha w\). Now

\[L^{-1}(\alpha w)=L^{-1}(L(\alpha v))=\alpha v=\alpha L^{-1}(w).\]

Similarly, \(L^{-1}(w_1+w_2)=L^{-1}(w_1)+L^{-1}(w_2)\) can also be shown.

The following proposition involves the same set-theoretic issue that was briefly mentioned after [Set Theory] §Cardinals, ⁋Definition 1. Namely, it is uncertain whether the collection of all $$\mathbb{K}$$-vector spaces is actually a set, but we will pass over this without further comment.

Proposition 3 The relation \(\cong\) of Definition 1 is an equivalence relation.

Proof

We must show that the relation \(\cong\) is reflexive, symmetric, and transitive.

1. First, it is obvious that \(V\cong V\) for any \(\mathbb{K}\)-vector space \(V\), because \(\id_V:V\rightarrow V\) is an isomorphism from \(V\) to \(V\).
2. By the preceding Lemma 2, it is obvious that \(\cong\) is symmetric.
3. Finally, suppose \(U\cong V\) and \(V\cong W\). Then there exist two isomorphisms \(L_1:U\rightarrow V\), \(L_2: V\rightarrow W\) such that

Although this proposition may seem obvious, it proves the less obvious part of the task of classifying all finite-dimensional \(\mathbb{K}\)-vector spaces.

Corollary 4 Any two \(n\)-dimensional \(\mathbb{K}\)-vector spaces are always isomorphic.

Proof

[§Linear Maps, ⁋Example 14] means that any \(n\)-dimensional \(\mathbb{K}\)-vector space \(V\) satisfies \(V\cong \mathbb{K}^n\). For another \(n\)-dimensional \(\mathbb{K}\)-vector space \(W\) we also have \(W\cong \mathbb{K}^n\), so from the fact that \(\cong\) is an equivalence relation we know that \(V\cong W\).

Of course the converse also holds, and therefore we see that the only invariant determining the structure of a finite-dimensional vector space is its dimension.

Proposition 5 Let two isomorphic \(\mathbb{K}\)-vector spaces \(V,W\) and an isomorphism \(L:V\rightarrow W\) be given. If \(\mathcal{B}\) is a basis of \(V\), then \(L(\mathcal{B})\) is also a basis of \(V\).

Proof

[§Linear Maps, ⁋Corollary 9].

Rank-Nullity Theorem

Meanwhile, for a given linear map \(L\), the spaces \(\ker L\) and \(\im L\) measure how far \(L\) is from being injective and surjective, respectively. Since we have seen above that dimension is the only invariant determining a vector space, it is enough to look at the dimensions of \(\ker L\) and \(\im L\) rather than the spaces themselves.

Definition 6 Let two \(\mathbb{K}\)-vector spaces \(V,W\) and a linear map \(L:V\rightarrow W\) be given. Then

1. \(\dim\ker L\) is called the nullity of \(L\), and is denoted by \(\nullity L\).
2. \(\dim\im L\) is called the rank of \(L\), and is denoted by \(\rank L\).

The following two statements are hardly worth giving new numbers.

1. \(L\) is injective if and only if \(\nullity L=0\).
2. \(L\) is surjective if and only if \(\rank L=\dim W\).

Moreover, the following theorem holds.

Theorem 7 (Rank-nullity theorem) Let two \(\mathbb{K}\)-vector spaces \(V,W\) and a linear map \(L:V\rightarrow W\) be given. Then the equation

\[\rank L+\nullity L=\dim V\]

always holds.

Proof

For convenience, write \(\dim V=n\) and \(\nullity L=k\). The following two cases are obvious.

1. If \(n=k\), then \(\ker L\) is a subspace having the same dimension as \(V\), so \(\ker L=V\). Therefore \(L=0\), and since \(\im L=0\) we have \(\rank L=0\), so the theorem holds.
2. Similarly, if \(k=0\) then \(\ker L=0\), so \(L\) is injective. Hence, restricting the codomain of \(L\) from \(W\) to \(\im L\), we obtain that \(L\) is a bijective linear map between \(V\) and \(\im L\). Therefore \(\dim V=\dim\im L=\rank L\).

Now it suffices to consider the case \(0 < k < n\). Let \(\left\{x_1,x_2,\ldots,x_k\right\}\) be a basis of \(\ker L\). Since this set is a linearly independent subset of \(V\), we can extend it to a basis \(\left\{x_1,x_2,\ldots,x_k,x_{k+1},\ldots,x_n\right\}\) of \(V\). Then we can show that the set \(\left\{L(x_{k+1}),L(x_{k+2}),\ldots,L(x_n)\right\}\) is a basis of \(\im L\) as follows.

First, this set is linearly independent: if

\[\alpha_{k+1}L(x_{k+1})+\alpha_{k+2}L(x_{k+2})+\cdots+\alpha_nL(x_n)=0\]

holds, then by linearity \(L(\sum_{i=k+1}^n \alpha_i x_i)=0\), so \(\sum_{i=k+1}^n\alpha_ix_i\in\ker L\), and therefore for some \(\alpha_1\), \(\alpha_2\), \(\ldots\), \(\alpha_k\) we have

\[\sum_{i=k+1}^n\alpha_ix_i=\alpha_1x_1+\alpha_2x_2+\cdots+\alpha_kx_k\]

or equivalently

\[\alpha_1x_1+\alpha_2x_2+\cdots+\alpha_kx_k-\alpha_{k+1}x_{k+1}-\cdots-\alpha_nx_n=0.\]

Since \(\left\{x_1,x_2,\ldots,x_k,x_{k+1},\ldots,x_n\right\}\) is linearly independent, we must have \(\alpha_1=\alpha_2=\cdots=\alpha_n=0\), and in particular \(\alpha_{k+1}=\alpha_{k+2}=\cdots=\alpha_n=0\).

Also, this set spans \(\im L\). Let any \(w\in \im L\) be given. Then there exists \(v\in V\) with \(L(v)=w\). Writing \(v=\sum_{i=1}^n \alpha_ix_i\), we have

\[u=L\left(\sum_{i=1}^n\alpha_ix_i\right)=L\left(\sum_{i=1}^k\alpha_ix_i\right)+L\left(\sum_{i=k+1}^n\alpha_i x_i\right)=\sum_{i=k+1}^n\alpha_i L(x_i)\]

because \(\sum_{i=1}^k\alpha_ix_i\in\ker L\).

From the above, \(\rank L=\dim\im L=n-k=\dim V-\nullity L\), so the equation of the theorem holds.

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References

[Goc] M.S. Gockenbach, Finite-dimensional linear algebra, Discrete Mathematics and its applications, Taylor&Francis, 2011.

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