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Let $\mathbb A$ be an algebra of dimension $k$ over the field $\mathbb F$. It is true that $\mathbb A$ is isomorphic to a subalgebra of the matrix algebra $M_k(\mathbb F)?$

DonAntonio
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zacarias
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    No, take zero multiplication in $\mathbb{A}$ – Norbert Aug 25 '12 at 15:57
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    @Norbert: if you don't require your subalgebras to share identities, then there's no problem. Formally adjoin an identity to $\mathbb{A}$ and then proceed with Sean Eberhard's answer. – Qiaochu Yuan Aug 25 '12 at 17:49

2 Answers2

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That depends: Do you require an algebra to have an identity? If no, go to Norbert's comment. If yes, continue.

As a vector space, $\mathbb{A}\cong\mathbb{F}^k$ (choose a basis). The left-multiplication action of $\mathbb{A}$ on itself therefore defines a homomorphism of algebras $\mathbb{A}\to\text{End}(\mathbb{F}^k) \cong M_k(\mathbb{F})$. This map is injective: if $a$ maps to the zero endomorphism, then $ax=0$ for all $x\in\mathbb{A}$, in particular for $x=1$.

Sean Eberhard
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  • but, aren't all subalgebras of an associative algebra associative? – James S. Cook Aug 25 '12 at 19:43
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    Often "algebra" means an associative algebra. If you mean something more general, you have to specify that. – Noah Snyder Aug 25 '12 at 20:34
  • @SeanEberhard The line "Take zero-multiplication in $\mathbb{A}$" (@Norbert), I assume means that "take the example of an algebra $\mathbb{A}$ that contains an $x$ such that $xa = 0, \forall a \in \mathbb{A}$". – Arpita Korwar Apr 25 '14 at 03:21
  • @ArpitaKorwar The comment meant "let $\mathbb{A}$ be any algebra over $F$ with multiplication defined by $xy=0$ for all $x$ and $y$". – Sean Eberhard Apr 25 '14 at 10:11
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Yes, if the algebra $\mathbb{A}$ is associative then as Sean Eberhard points out the left-multiplication maps give linear transformations on the algebra $\mathbb{A}$ whose matrices form a subalgebra of $M_k(\mathcal{F})$ which is isomorphic to the given associative algebra.

Take $\mathbb{A} = \mathbb{F}^k$ which multiplication $*$. Let us examine where the associativity is necessary. Suppose $T: \mathbb{A} \rightarrow \mathbb{A}$ is left multiplication by $A \in \mathbb{A}$ this means $T_A(v) = A*v$. Now, suppose $T_B(v) = B*v$ is another such left-multiplication map. We should like $A \mapsto T_A$ to define an isomorphism. Consider $T_A \,{\scriptstyle \stackrel{\circ}{}}\, T_B$. Observe,

$$ (T_A \,{\scriptstyle \stackrel{\circ}{}}\, T_B)(v) = T_A(T_B(v)) = T_A(B*v)=A*(B*v) $$

Yet, $T_{A*B}(v) = (A*B)*v$. If $*$ is not an associative operation there is no reason that $T_A \,{\scriptstyle \stackrel{\circ}{}}\, T_B = T_{A *B}$ should hold true.

It may still be possible to find an isomorphism to some subset $S$ of $M_k(\mathcal{F})$ if we give the subset an operation other matrix multiplication. This means $S$ is not strictly speaking a subalgebra of $M_k(\mathcal{F})$ since the multiplication on $S$ is not inherited from the matrix multiplication operation on $M_k(\mathcal{F})$.

Some authors use different notation for the same point-set to indicate a different choice of multiplication. For example, $M(\mathbb{F})=\mathbb{F}^{n \times n}$ with matrix multiplication whereas $gl_n(\mathbb{F})= \mathbb{F}^{n \times n}$ with the commutator-bracket multiplication $[A,B] = AB-BA$. $gl_n(\mathbb{F})$ is not an associative algebra, the departure from associativity is quantified by the Jacobi identity $[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0$. For a finite dimensional Lie algebra of dimension $k$ it turns out that you can find an isomorphic copy of the algebra in $gl_n(\mathbb{F})$ where $n \geq k$. Equality may not be possible. See Ado's Theorem; http://en.wikipedia.org/wiki/Ado%27s_theorem . This is also known for super Lie algebras.

I tend to think that if we have a finite dimensional algebra then it is possible to embedd it in matrices of sufficiently large order, even if it is nonassociative. But, the operation need not be simple matrix multiplication.

James S. Cook
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