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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 $n^2$.

This is the part of linear algebra I can't get. Please help. Extremely stuck.

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 $n=3$?2011-11-30

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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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    Dimension has uses because it is invariant under isomorphism (bijective linear map). Generally, if a quantity is invariant under isomorphism it is a deep property of the structure being studied.2011-11-30