K

Grad. student in mathematics

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.

The spaces studied in linear algebra are vector spaces, which generalize the coordinate spaces learned in high school. A vector space is, as the name suggests, a space of vectors, and addition between these vectors is well-defined. Also, we can multiply any vector by a constant to obtain another vector; when dealing with vector spaces, these constants are called scalars. The set of scalars must form a field, which we define shortly.

Fields

Definition 1 A set \(G\) together with a binary operation \(+:G\times G\rightarrow G\) defined on \(G\) is called an abelian group if the following conditions are satisfied.

  1. \(+\) is associative. That is, for any \(a,b,c\in G\), the equality \((a+b)+c=a+(b+c)\) holds.
  2. There exists an identity element for \(+\). That is, there exists some \(0\in G\) such that for any \(a\in G\), \(a+0=0+a=a\) holds.
  3. Every element has an inverse with respect to \(+\). That is, for any given \(a\in G\), there exists \(-a\in G\) such that \(a+(-a)=(-a)+a=0\) holds.
  4. \(+\) is commutative. That is, for any \(a,b\in G\), \(a+b=b+a\) holds.

If the last condition does not hold, then \(G\) is called a group.

Anyway, we can prove a simple proposition using the definition.

Proposition 2 Let \(G\) be an abelian group. Then

  1. The identity element of \(G\) is unique.
  2. For any \(a\in G\), the inverse \(-a\) of \(a\) is also unique.
  3. If \(a+b=a+c\), then \(b=c\).
Proof

  1. Suppose \(0'\) is another element satisfying the second condition of Definition 1. Then applying the second condition to \(a=0\) and the identity element \(0'\), we obtain

    \[0+0'=0'+0=0\]

    On the other hand, applying the second condition to \(a=0'\) and the identity element \(0\), we likewise obtain

    \[0+0'=0'+0=0'\]

    Therefore \(0=0'\), so the identity element is unique.

  2. The proof is similar to the first one. Suppose \((-a)'\) is another element satisfying the third condition of Definition 1. Then

    \[(-a)=(-a)+0=(-a)+(a+(-a)')=((-a)+a)+(-a)'=0+(-a)'=(-a)'\]

    so \((-a)=(-a)'\) holds.

  3. Simply add \((-a)\) to both sides.

In an abelian group, thanks to the condition \(a+b=b+a\) in the fourth property, it is sufficient to think of the identity and inverse conditions as the single equations \(a+0=a\) and \(a+(-a)=0\), respectively. However, we did not use such logic in the proof above, and therefore the uniqueness of the identity and inverse holds for general groups as well.

The following corollary follows immediately from the uniqueness of the identity and inverse, and it also holds for general groups.

Corollary 3 Let \((G,+)\) be an abelian group and let \(a,b\) be arbitrary elements. Then the following hold.

  1. $-(-a)=a$,
  2. $-(a+b)=-a+(-b)=-a-b$.
Proof

  1. We must show that the inverse \(-(-a)\) of \(-a\) equals \(a\). Since inverses are unique, if for some \(x\in G\) the equality

    \[(-a)+x=x+(-a)=0\]

    holds, then \(x\) must be \(-(-a)\). But when \(x=a\), the above equality holds because \(-a\) is the inverse of \(a\). Therefore \(a=-(-a)\).

  2. As in the previous proof, it suffices to show that both \(x=-a+(-b)\) and \(-a-b\) satisfy

    \[(a+b)+x=x+(a+b)=0\]

    For example, if \(x=-a+(-b)\), then

    \[\begin{aligned}(a+b)+x&=(a+b)+((-a)+(-b))=(b+a)+((-a)+(-b))=b+(a+(-a))+(-b)=b+(-b)\\&=0\end{aligned}\]

    and similarly, or by commutativity, one can also show that \(x+(a+b)=0\).

In Definition 1 we assumed that the operation on \(G\) is addition, but in fact it need not be addition. Even if the operation on \(G\) is written as multiplication, if \(G\) satisfies all the conditions from 1 to 4, then \(G\) is still called an abelian group. Of course, in this case it would be natural to write the identity element as \(1\) instead of \(0\), and the inverse as \(a^{-1}\) instead of \(-a\).

Example 4 The set of real numbers \(\mathbb{R}\) is an abelian group under addition, and the same is true for the set of complex numbers \(\mathbb{C}\).

Multiplication is also defined on \(\mathbb{R}\) and \(\mathbb{C}\), but these are not abelian groups under multiplication. For \(0\in\mathbb{R}\) (or \(\mathbb{C}\)), there is no real (or complex) number \(a\) satisfying \(0a=a0=1\). Instead, \(\mathbb{R}^\times=\mathbb{R}\setminus\{0\}\) (or \(\mathbb{C}^\times=\mathbb{C}\setminus\{0\}\)) is an abelian group under multiplication.

We define the situation in the above example as follows.

Definition 5 Suppose two binary operations \(+\) and \(\times\) are defined on a set \(\mathbb{K}\). Then \(\mathbb{K}\) is called a field if the following three conditions are satisfied:

  1. \(\mathbb{K}\) is an abelian group under \(+\).
  2. \(\mathbb{K}^\times=\mathbb{K}\setminus\{0\}\) is an abelian group under \(\times\).

  3. \(\times\) is distributive over \(+\). That is, for any \(a,b,c\in\mathbb{K}\),

    \[a\times (b+c)=(a\times b)+(a\times c)\]

    holds.

In the above definition, \(0\) and \(1\) must be distinct elements, because \(1\in\mathbb{K}^\times=\mathbb{K}\setminus\{0\}\).

Ever since high school we have preferred not to explicitly write the operation \(\times\), but rather to abbreviate it as \(a\cdot b\), and in fact we prefer even more to omit the operation entirely and write \(ab\). Henceforth we follow this convention and write \(ab\) instead of \(a\times b\).

Meanwhile, applying Proposition 2 to \(\mathbb{K}\) and \(\mathbb{K}^\times\) respectively, one can easily verify that the following hold:

  1. \(0\) and \(1\) are unique.
  2. For any \(a\in\mathbb{K}\), \(-a\) is unique, and moreover if \(a\neq 0\) then \(a^{-1}\) also exists uniquely.
  3. If \(a+b=a+c\), then \(b=c\). Also, if \(a\neq 0\) and \(ab=ac\), then \(b=c\).

Proposition 5 For any element \(a\in\mathbb{K}\) of a field \(\mathbb{K}\), the equalities \(0a=0\) and \((-1)a=-a\) hold.

Proof

First we need to carefully unpack what the proposition means: \(0a=0\) means that multiplying \(0\) and \(a\) yields the additive identity \(0\), and \((-1)a=-a\) means that multiplying \((-1)\) and \(a\) yields the inverse of \(a\). To do this, just as we proved several properties right after Proposition 2, we can use the uniqueness of inverses and the identity. To prove the first equality, we would need to show that \(0a+b=b+0a=b\) holds for arbitrary \(b\), but there seems to be no obvious way to simplify \(0a+b\). We need a different approach. Let us use the third statement of Proposition 2. If we can show that \(0a+0a=0a\), then since \(0a=0a+0\), from \(0a+0a=0a+0\) we obtain \(0a=0\). Thus it suffices to show that \(0a+0a=0a\), which is immediate from

\[0a+0a=(0+0)a=0a\]

Once this is proved, the second part is even easier. We need to show that \((-1)a\) satisfies the condition for the inverse of \(a\), and

\[(-1)a+a=(-1)a+1a=((-1)+1)a=0a=0\]

so the proof is complete.

References

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

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