# Why isthe dimension of the vector space $M_{3\times4}$ is $3\times 4$?

I understood the concept of dimension of a vector space, which is the number of vectors in all basis of a certain vector space $V$. I understand that, for example, the $dim(\mathbb{R}^2)=2$ or $dim(\mathbb{R}^3)=3$, but demonstrate the correctness of these statements is more complicated.

For example, why the dimension of a general vector space $M_{3\times4}$ matrices is $3\times 4=12$?

Apparently, all matrices that form the standard basis for a vector space $M_{3\times4}$ matrices are special matrices: they have all zeros, except in the positions $ij$, where $i$ refers to the rows and $j$ to the columns, where they have $1$.

For example:

$\left[\begin{matrix}1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{matrix}\right]$

Would be one of the matrices of the standard basis of the vector space $M_{3x4}$, where $i=1$ and $j=1$, right?

What do I need to do show to demonstrate it is indeed $12$?

Let $E_{ij}\in M_{3 \times 4}$ be the matrix with $1$ at the $ij$-entry and zeroes everywhere. For example: $$E_{12} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 &0 & 0 & 0\end{pmatrix}.$$

Let $A = (a_{ij}) \in M_{3 \times 4}$. Then we have $A = \sum_{i=1}^3\sum_{j=1}^4a_{ij}E_{ij}$. This shows that $\{E_{ij} \mid 1 \leq i \leq 3,\, 1 \leq j \leq 4\}$ spans $M_{3 \times 4}$, so $\dim (M_{3 \times 4}) \leq 12$. Showing that $\{E_{ij} \mid 1 \leq i \leq 3,\,1 \leq j\leq 4\}$ is linearly independent ensures that $\dim(M_{3 \times 4}) \geq 12$.

So we conclude that $\dim(M_{3 \times 4}) = 12$.

Another way is noticing that $$M_{3 \times 4} \ni (a_{ij}) \mapsto (a_{11},a_{12},\cdots, a_{21},\cdots, a_{31},\cdots, a_{34})\in \Bbb R^{12}$$ is a bijection which preserves vector sum and multiplication by scalar, hence $\dim(M_{3 \times 4}) = \dim(\Bbb R^{12})=12.$

• How would you approach in showing that they are linearly independent? – nbro Mar 21 '15 at 3:35
• Using the definition: check that $$\sum_{i=1}^3\sum_{j=1}^4a_{ij}E_{ij} = 0_{3\times 4}\implies a_{ij} = 0,\,\forall\,i,\,\forall\,j.$$ – Ivo Terek Mar 21 '15 at 3:38
• Ok, but how would you check? I have done it with vectors, by using $RREF$ (if I am not wrong), but for matrices, how would you proceed? – nbro Mar 21 '15 at 3:42
• You don't need RREF to do anything of this. Bear in mind the definition of the $E_{ij}$ and that the $a_{ij}$ in my last comment are only real numbers, suck it up and expand that double summation. It might be useful recalling the definition of linear independence too. – Ivo Terek Mar 21 '15 at 3:44
• Expanding the double summation is good for seeing what happens. If you convince yourself that the $ij$-entry of $\sum_{i=1}^3\sum_{j=1}^4a_{ij}E_{ij}$ (remember, this crap is a matrix!) is only $a_{ij}$, then there's nothing else to do. And while we're at it, notice how irrelevant the numbers $3$ and $4$ are, here. Our reasoning goes for $M_{n \times m}$, $\sum_{i=1}^n\sum_{j=1}^ma_{ij}E_{ij} = 0_{n\times m}$, etc. – Ivo Terek Mar 21 '15 at 3:48