Is there any finite dimensional algebra that is not isomorphic to some algebra of matrices? Suppose that $\mathcal A$ is a finite dimensional algebra. Is it true that there always exist some isomorphism
\begin{equation}
\phi:\mathcal{A}\rightarrow C\subset M_{K\times K},
\end{equation}
where $C$ is some subalgebra of the algebra of $K\times K$ matrices? If not, what is the requirement for an algebra to be isomorphic to some algebra of matrices?
I study physics, and in general, physicists are usually interested in some algebra $\mathcal{A}$ of operators over some vector space $V$, and if $V$ is finite dimensional, the statement is obviously true. However, I do not know how to prove it for any finite dimensional algebra, and neither think of a counterexample.
 A: Every finite-dimensional algebra is indeed representable as subalgebra of some full matrix algebra.
I’m assuming a field $K$, over which your algebra $A$ is finite-dimensional. We have the “regular representation” of $A$, by associating $b\in A$ to the $K$-linear map that will take any $z\in A$ to $bz$. That is, $b$ gives us $\mu_b:A\to A$, by $\mu_b(z)=bz$. We need to verify that $\mu_b+\mu_{b'}=\mu_{b+b'}$, $\mu_b\circ\mu_{b'}=\mu_{bb'}$. Furthermore, we have to verify that for $\lambda\in K$, we get $\lambda\mu_b=\mu_{\lambda b}$.
The first requirement is easily verified; the other two need to be looked at closely in light of the fact that $A$ may be noncommutative.
Well, $(\mu_b\circ\mu_{b'})(z)=\mu_b\bigl(\mu_{b'}(z)\bigr)=
\mu_b(b'z)=bb'z$. On the other hand, $\mu_{bb'}(z)=bb'z$ as well, so the requirement on multiplicativity in $A$ corresponding to composition of linear
mappings is satisfied.
Finally, $\lambda\mu_b(z)=\lambda(bz)=(\lambda b)z=\mu_{\lambda b}(z)$, as desired. We need to check injectivity as well, that no $b\in A$ yields the zero-map of $A$ into itself. And here it seems to me that we need the assumption that $A$ have a unit $1_A$ from which the nonzero $b$ gives $\mu_b(1_A)=b\ne0$, so that $\mu_b\ne0$. (I don’t want to consider algebras without unit elements; if yours are not so nice as this, I think you’ll have to puzzle the matter out yourself.)
You see that I’ve constructed an injection of $A$ into the ring of $K$-linear self-maps of the $K$-vector space $A$. This last is an $M_{K\times K}$, as I hope you are aware.
A: In general, the answer is no. The octonions are not isomorphic to any matrix algebra because they're not associative.
If the algebra is associative then the answer is yes.
A: If finite dimension algebra A (with dim(A)= n) has an identity, then A is isomorphic with subalgebra of $\frak{B}$(A)(the algebra of bounded linear transformation on A, that is isomorphic with matrix algebras of $M_{n×n}$) under left regular representation and if A has not an identity, then left regular representarion is not necessarily injective.but if $\textit{A}_{l}$(A)=$0$ or A be an annihilator algebra then left regular representation is injective and A is isomorphic with subalgebra of matrix algebra $M_{n×n}$
and $\textit{A}_{l}$(A)={$x\in A|xA=0$}
