Is it true that $M(n;\mathbb R)$ is isomorphic to direct sum of $\mathbb R$? 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?
 A: 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 $\bigoplus_{k=1}^{n^2}\mathbb{R}$.
A: 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.
