Algebra over a Field

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 $M_k(\mathbb F)?$

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No, take zero multiplication in $\mathbb{A}$ – Norbert Aug 25 '12 at 15:57
@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 Answers

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.

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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 $x=1$.

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but, aren't all subalgebras of an associative algebra associative? – James S. Cook Aug 25 '12 at 19:43
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 3:21
@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