The idea of a vector space can be extended to include objects that you would not initially consider to be ordinary vectors. Matrix spaces. Consider the set ${\mathbb{M}}_{2x3}(\mathbb{R})$${\mathbb{M}}_{2x3}(\mathbb{R})$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}) of 2 by 3 matrices with real entries. This set is closed under addition, since the sum of a pair of 2 by 3 matrices is again a 2 by 3 matrix, and when such a matrix is multiplied by a real scalar, the resulting matrix is in the set also. Since ${\mathbb{M}}_{2x3}(\mathbb{R})$${\mathbb{M}}_{2x3}(\mathbb{R})$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}), with the usual algebraic operations, is closed under addition and scalar multiplication, it is a real Euclidean vector space. The objects in the vector space, the vectors, are now matrices. This proof is trivial and it is left as an exercise for the reader.

Since ${\mathbb{M}}_{2x3}(\mathbb{R})$${\mathbb{M}}_{2x3}(\mathbb{R})$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}) is a vector space, what is its dimension? First, note that any 2 by 3 matrix is a unique linear combination of the following six matrices:

Therefore, they span ${\mathbb{M}}_{2x3}(\mathbb{R})$${\mathbb{M}}_{2x3}(\mathbb{R})$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}).

Furthermore, these elements of the vector space are linearly independent: none of these matrices is a linear combination of the others. (ie, the only way $k_1{E}_1+k_2{E}_2+k_3{E}_3+k_4{E}_4+k_5{E}_5+k_6{E}_6$$k_1{E}_1+k_2{E}_2+k_3{E}_3+k_4{E}_4+k_5{E}_5+k_6{E}_6$k_(1)E_(1)+k_(2)E_(2)+k_(3)E_(3)+k_(4)E_(4)+k_(5)E_(5)+k_(6)E_(6)k_1 E_1 + k_2 E_2 + k_3 E_3 + k_4 E_4 + k_5 E_5 + k_6 E_6 will give the 2 by 3 zero matrix is if each scalar coefficient, $k_i$$k_i$k_(i)k_i , in this combination is zero) These six elemments therefore form a basis for ${\mathbb{M}}_{2x3}(\mathbb{R})$${\mathbb{M}}_{2x3}(\mathbb{R})$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}), $$\therefore \dim ({\mathbb{M}}_{2x3}(\mathbb{R}))=6$$$$\therefore \dim ({\mathbb{M}}_{2x3}(\mathbb{R}))=6$$:.dim(M_(2x3)(R))=6\therefore\dim(\mathbb{M}_{2x3}(\mathbb{R}))=6

If the entries in a given 2 by 3 matrix are written out in a single row (or column), the result is a vector in ${\mathbb{R}}^6$${\mathbb{R}}^6$R^(6)\mathbb{R}^6. For example,

The rule here is simple: Given a 2 by 3 matrix, form a 6-vector by writing the entries in the first row of the matrix followed by the entries in the second row. Then, to every matrix in ${\mathbb{M}}_{2x3}(\mathbb{R})$${\mathbb{M}}_{2x3}(\mathbb{R})$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}) there corresponds a unique vector in ${\mathbb{R}}^6$${\mathbb{R}}^6$R^(6)\mathbb{R}^6, and vice versa. This one-to-one correspondence:

is compatible with the vector space operations of addition and scalar multiplication. This means that

The conclusion is that the spaces ${\mathbb{M}}_{2x3}(\mathbb{R})$${\mathbb{M}}_{2x3}(\mathbb{R})$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}) and ${\mathbb{R}}^6$${\mathbb{R}}^6$R^(6)\mathbb{R}^6 are structurally identical, that is, isomorphic, a fact which is denoted ${\mathbb{M}}_{2x3}(\mathbb{R})\cong {\mathbb{R}}^6$${\mathbb{M}}_{2x3}(\mathbb{R})\cong {\mathbb{R}}^6$M_(2x3)(R)~=R^(6)\mathbb{M}_{2x3}(\mathbb{R}) \cong \mathbb{R}^6.

One consequence of this structural identity is that each basis of the vector space $E_i$$E_i$E_(i)E_i given above for ${\mathbb{M}}_{2x3}(\mathbb{R})$${\mathbb{M}}_{2x3}(\mathbb{R})$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}) corresponds to the standard basis vector $e_i$$e_i$e_(i)e_i for ${\mathbb{R}}^6$${\mathbb{R}}^6$R^(6)\mathbb{R}^6.

The only real difference between the spaces ${\mathbb{R}}^6$${\mathbb{R}}^6$R^(6)\mathbb{R}^6 and ${\mathbb{M}}_{2x3}(\mathbb{R})$${\mathbb{M}}_{2x3}(\mathbb{R})$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}) is in the notation: The six entries denoting an element in ${\mathbb{R}}^6$${\mathbb{R}}^6$R^(6)\mathbb{R}^6 are written as a single row (or column), while the six entries denoting an element in ${\mathbb{M}}_{2x3}(\mathbb{R})$${\mathbb{M}}_{2x3}(\mathbb{R})$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}) are written in two rows of three entries each. This example can be generalized further.