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Is it true that the ring of all $n\times n$ matrices with entries in $\mathbb R$ $M(n;\mathbb R)$ is isomorphic to direct sum of $\mathbb R$ I.e $(\mathbb R \times\mathbb R\times \ldots \times \mathbb R)$ $n^2$ times?

I tried to prove it by assuming that the multiplication between any two elements in the direct sum is as matrices (components instead of entries) to prove the ring homomorphism

is this true?

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Ah, no, it's not true. Multiplication is totally different. It's true though for the underlying Abelian group of addition. – Berci Apr 1 '13 at 17:44
$M(n;\mathbb{R})$ is noncommutative, but the RHS is commutative. – i707107 Apr 1 '13 at 18:11
I changed the multiplication in the direct sum to be as in matrices , so the direct sum ring is not commutative – Bbbh Apr 1 '13 at 18:19
I am just thinking to prove the isomorphism, because in some books , they are writing matrices with direct sum and I have tried to understand why? so I am looking for the correspondence – Bbbh Apr 1 '13 at 18:27

As a vector space, $M_n(\mathbb{R})$ is indeed isomorphic to $\bigoplus_{k=1}^{n^2} \mathbb{R}$. In fact, any two vector spaces with the same dimension are isomorphic. However, the operation in a vector space is addition. Multiplication is not defined for a vector space. There is notion of multiplcation in $\bigoplus_{k=1}^{n^2}\mathbb{R}$.

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But the OP did not ask about the direct sum. He asked about the direct product, which has a canonical structure as a ring. – Tobias Kildetoft Apr 1 '13 at 17:53
@TobiasKildetoft The OP mentions the "direct sum" twice, and does not mention the direct product. – Fly by Night Apr 1 '13 at 17:55
Can I define the multiplication between any two ordered paired other than the usual multiplication – Bbbh Apr 1 '13 at 17:56
What is the difference between finite direct sum and finite direct product – Bbbh Apr 1 '13 at 17:58
@Jmath None at all. The difference only kicks in with infinite direct sums and infinite direct products: the infinite direct sum is the unital subring/subspace (depending on context) of the infinite direct product consisting of elements with only finitely many non-zero entries. – Branimir Ćaćić Apr 1 '13 at 18:59

That $M_n(\mathbb{R}) \cong \mathbb{R}^{n^2}$ as vector spaces follows almost immediately from simply exhibiting a basis for $M_n(\mathbb{R})$.

Now, since $\mathbb{R}$ is a ring, so too is $\mathbb{R}^{n^2}$ viewed as the $n^2$-fold product of $\mathbb{R}$ as a ring with itself. In particular, for $a$, $b \in \mathbb{R}^{n^2}$, $a \cdot b$ is defined entrywise by $(a \cdot b)_k = a_k b_k$ for all $k$, and $1 \in \mathbb{R}^{n^2}$ is defined by $1_k = 1$ for all $k$. However, $M_n(\mathbb{R})$ is non-commutative as a ring, whilst $\mathbb{R}^{n^2}$ is commutative as a ring, so $M_n(\mathbb{R})$ and $\mathbb{R}^{n^2}$ can't possibly be isomorphic as rings. Another obstruction to the two being isomorphic as rings is that $M_n(\mathbb{R})$ has no proper ideals, whilst each of the $n^2$ copies of $\mathbb{R}$ forming $\mathbb{R}^{n^2}$ yields a proper ideal of $\mathbb{R}^{n^2}$.

Of course, if you treat $\mathbb{R}^{n^2}$ as a vector space, you can use a vector space isomorphism $T : M_n(\mathbb{R}) \cong \mathbb{R}^{n^2}$ to translate the ring structure on $M_n(\mathbb{R})$ into a new ring structure on $\mathbb{R}^{n^2}$ by $a \cdot_T b := T(T^{-1}(a)T^{-1}(b))$ for $a$, $b \in \mathbb{R}^{n^2}$; then $T : M_n(\mathbb{R}) \cong \mathbb{R}^{n^2}$ becomes an isomorphism of rings entirely by construction.

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