# Dimension of the vector space? $M_n(K)$?

How do you calculate the dimensions of a vector space more generally.

For any field $K$ and $n \in \mathbb{N}$, $M_{n}(K)$ is an algebra over K. The notes says that the vector space dimension is $n^2$.

I generally know the answer, but don't see the motivation behind the word dimension.

To me a dimension is something like allowing a Euclidean vector some sort of movement i.e. $v=(t,0,0,0)^T$, however in $M_n(K)$ you have no way of getting a vector like that. So what does it mean?

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Can you do the case where $n=2$? –  Mariano Suárez-Alvarez Nov 30 '11 at 0:20
I understand you have $e_{(1,2)}$,$e_{(1,1)$, $e_{(2,1)}$,$e_{(2,2)}$ –  simplicity Nov 30 '11 at 0:23
Can you do $n=3$? –  Mariano Suárez-Alvarez Nov 30 '11 at 0:24

Canonical basis of matrices $\{E_{ij}\}$ where i,j range from 1 to n and $E_{ij}$ is the matrix with i-j entry 0 if i is not equal to j and 1 otherwise.

edit; For finite dimensional vector spaces (that is, spaces that are spanned by a finite set of vectors), a basis is a spanning set that is linearly independent. Using the Steinitz exchange lemma it is trivial to prove all basis have the same cardinilty, viz. a viz. a dimension of n (where n is the size of the set).

Consider R^2 and R. {(0,1),(1,0)} and {(1,1),(2,1)} arw both basis of R^2 while 2 and pi are basis of R.

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It is usually best if answers are composed of complete sentences—yours is missing a verb. :) –  Mariano Suárez-Alvarez Nov 30 '11 at 0:23
See I don't see how that is a basis. It's like an individual piece of a column and a row, well it the i-j entry. Surely the dimension should be a vector. So you would need to make that entry into some vector like this. –  simplicity Nov 30 '11 at 0:26
@simplicity, the dimension is a number. –  Mariano Suárez-Alvarez Nov 30 '11 at 0:27
@MarianoSuárez-Alvarez yeah but it makes no sense. It just meaningless. –  simplicity Nov 30 '11 at 0:32
@simplicity, you are asking about a vector space whose elements are matrices. The basis for such a vector space will be a set of matrices. Adam has given you such a set. The dimension of the vector space is the number of elements in the basis; in this case, that's the number of matrices in Adam's list, which number is $n^2$ because $i$ and $j$ each range from $1$ to $n$. OK? –  Gerry Myerson Nov 30 '11 at 0:32