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# The set of matrices as a vector space

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}\left(\mathbb{R}\right)$${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}) of 2 by 3 matrices with real entries. This set is closed under addition, since the sum of a pair of 2 by 3 matrices is again a 2 by 3 matrix, and when such a matrix is multiplied by a real scalar, the resulting matrix is in the set also. Since ${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$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}\left(\mathbb{R}\right)$${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}) is a vector space, what is its dimension? First, note that any 2 by 3 matrix is a unique linear combination of the following six matrices:
$\begin{array}{lll}{E}_{1}=\left(\begin{array}{lll}1& 0& 0\\ 0& 0& 0\end{array}\right)& {E}_{2}=\left(\begin{array}{lll}0& 1& 0\\ 0& 0& 0\end{array}\right)& {E}_{3}=\left(\begin{array}{lll}0& 0& 1\\ 0& 0& 0\end{array}\right)\\ \\ {E}_{4}=\left(\begin{array}{lll}0& 0& 0\\ 1& 0& 0\end{array}\right)& {E}_{5}=\left(\begin{array}{lll}0& 0& 0\\ 0& 1& 0\end{array}\right)& {E}_{6}=\left(\begin{array}{lll}0& 0& 0\\ 0& 0& 1\end{array}\right)\end{array}$$\begin{array}{l}{E}_{1}=\left(\begin{array}{lll}1& 0& 0\\ 0& 0& 0\end{array}\right) {E}_{2}=\left(\begin{array}{lll}0& 1& 0\\ 0& 0& 0\end{array}\right) {E}_{3}=\left(\begin{array}{lll}0& 0& 1\\ 0& 0& 0\end{array}\right)\\ \\ {E}_{4}=\left(\begin{array}{lll}0& 0& 0\\ 1& 0& 0\end{array}\right) {E}_{5}=\left(\begin{array}{lll}0& 0& 0\\ 0& 1& 0\end{array}\right) {E}_{6}=\left(\begin{array}{lll}0& 0& 0\\ 0& 0& 1\end{array}\right)\end{array}${:[E_(1)=([1,0,0],[0,0,0]),E_(2)=([0,1,0],[0,0,0]),E_(3)=([0,0,1],[0,0,0])],[],[E_(4)=([0,0,0],[1,0,0]),E_(5)=([0,0,0],[0,1,0]),E_(6)=([0,0,0],[0,0,1])]:}\begin{array}{lll} E_{1}=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right) & E_{2}=\left(\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right) & E_{3}=\left(\begin{array}{lll} 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right) \\ \\ E_{4}=\left(\begin{array}{lll} 0 & 0 & 0 \\ 1 & 0 & 0 \end{array}\right) & E_{5}=\left(\begin{array}{lll} 0 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right) & E_{6}=\left(\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right) \end{array}
Therefore, they span ${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}).
Furthermore, these elements of the vector space are linearly independent: none of these matrices is a linear combination of the others. (ie, the only way ${k}_{1}{E}_{1}+{k}_{2}{E}_{2}+{k}_{3}{E}_{3}+{k}_{4}{E}_{4}+{k}_{5}{E}_{5}+{k}_{6}{E}_{6}$${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}$${k}_{i}$k_(i)k_i , in this combination is zero) These six elemments therefore form a basis for ${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}), $\therefore \mathrm{dim}\left({\mathbb{M}}_{2x3}\left(\mathbb{R}\right)\right)=6$$\therefore \mathrm{dim}\left({\mathbb{M}}_{2x3}\left(\mathbb{R}\right)\right)=6$:.dim(M_(2x3)(R))=6\therefore\dim(\mathbb{M}_{2x3}(\mathbb{R}))=6
If the entries in a given 2 by 3 matrix are written out in a single row (or column), the result is a vector in ${\mathbb{R}}^{6}$${\mathbb{R}}^{6}$R^(6)\mathbb{R}^6. For example,
$\left[\begin{array}{rrr}1& 3& -4\\ -2& -1& 0\end{array}\right]\stackrel{\text{gives}}{⟶}\left(1,3,-4,-2,-1,0\right)$[[1,3,-4],[-2,-1,0]]longrightarrow^(" gives ")(1,3,-4,-2,-1,0)\left[\begin{array}{rrr} 1 & 3 & -4 \\ -2 & -1 & 0 \end{array}\right] \stackrel{\text { gives }}{\longrightarrow}(1,3,-4,-2,-1,0)
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}\left(\mathbb{R}\right)$${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}) there corresponds a unique vector in ${\mathbb{R}}^{6}$${\mathbb{R}}^{6}$R^(6)\mathbb{R}^6, and vice versa. This one-to-one correspondence:
$\begin{array}{c}{\mathbb{M}}_{2x3}\left(\mathbb{R}\right)\stackrel{\phi }{⟶}{\mathbb{R}}^{6}\\ \\ \left[\begin{array}{lll}a& b& c\\ d& e& f\end{array}\right]\begin{array}{c}\stackrel{\phi }{⟶}\\ \stackrel{{\phi }^{-1}}{⟵}\end{array}\left(a,b,c,d,e,f\right)\end{array}$$\begin{array}{c}{\mathbb{M}}_{2x3}\left(\mathbb{R}\right)\stackrel{\phi }{⟶}{\mathbb{R}}^{6}\\ \\ \left[\begin{array}{lll}a& b& c\\ d& e& f\end{array}\right]\begin{array}{c}\stackrel{\phi }{⟶}\\ \stackrel{{\phi }^{-1}}{⟵}\end{array}\left(a,b,c,d,e,f\right)\end{array}${:[M_(2x3)(R)longrightarrow^(varphi)R^(6)],[],[[[a,b,c],[d,e,f]]{:[longrightarrow^(varphi)],[longleftarrow^(varphi^(-1))]:}(a","b","c","d","e","f)]:}\begin{array}{c} \mathbb{M}_{2x3}(\mathbb{R}) \stackrel{\varphi}{\longrightarrow}\mathbb{R}^6 \\ \\ \left[\begin{array}{lll} a & b & c \\ d & e & f \end{array}\right] \begin{array}{c} \stackrel{\varphi}{\longrightarrow}\\ {\stackrel{\varphi^{-1}}{\longleftarrow}} \end{array}(a, b, c, d, e, f) \end{array}
is compatible with the vector space operations of addition and scalar multiplication. This means that
$\phi \left(A+B\right)=\phi \left(A\right)+\phi \left(B\right)$$\phi \left(A+B\right)=\phi \left(A\right)+\phi \left(B\right)$varphi(A+B)=varphi(A)+varphi(B)\varphi(A+B)=\varphi(A)+ \varphi(B)
$\phi \left(k×A\right)=k×\phi \left(A\right)$$\phi \left(k×A\right)=k×\phi \left(A\right)$varphi(k xx A)=k xx varphi(A)\varphi(k \times A)=k\times \varphi(A)
The conclusion is that the spaces ${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}) and ${\mathbb{R}}^{6}$${\mathbb{R}}^{6}$R^(6)\mathbb{R}^6 are structurally identical, that is, isomorphic, a fact which is denoted ${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)\cong {\mathbb{R}}^{6}$${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)\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}$${E}_{i}$E_(i)E_i given above for ${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$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}$${\mathbb{R}}^{6}$R^(6)\mathbb{R}^6 and ${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}) is in the notation: The six entries denoting an element in ${\mathbb{R}}^{6}$${\mathbb{R}}^{6}$R^(6)\mathbb{R}^6 are written as a single row (or column), while the six entries denoting an element in ${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$${\mathbb{M}}_{2x3}\left(\mathbb{R}\right)$M_(2x3)(R)\mathbb{M}_{2x3}(\mathbb{R}) are written in two rows of three entries each. This example can be generalized further.

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