I'm a bit confused on the notion of the dimension of a matrix, say $\mathbb{M}_{mn}$. I know how this applies to vector spaces but can't quite relate it to matrices.

For example take this matrix: $$ \left[ \begin{array}{ccc} a_{11}&\cdots&a_{1n}\\ \vdots & \vdots & \vdots \\ a_{m1}&\cdots&a_{mn} \end{array} \right] $$

Isn't the set of matrices $\mathbb{M}_{mn}$ with exactly one entry $a_{ij}$ set to $1$ on each matrix and $m\times n$ total matrices a basis for $\mathbb{M}_{mn}$? In the sense that we can take some linear combination of them and add them up to create:

$$ a_{11} \cdot \left[ \begin{array}{cccc} 1&\cdots&\cdots&0\\ \vdots & \vdots & \vdots & \vdots \\ 0&\cdots&\cdots&0 \end{array} \right] + a_{12} \cdot \left[ \begin{array}{cccc} 0 &1 &\cdots&0\\ \vdots & \vdots &\vdots & \vdots \\ 0&0&\cdots&0 \end{array} \right] + \cdots = \left[ \begin{array}{ccc} a_{11}&\cdots&a_{1n}\\ \vdots & \vdots & \vdots \\ a_{m1}&\cdots&a_{mn} \end{array} \right] $$

So the dimension of all $\mathbb{M}_{mn}$ is $m\times n$?


Yes, that is a basis for matrices, and yes that shows the dimension is $mn$.

When you say you know how it applies to vector spaces but not matrices: remember that a vector space is a collection of objects (called vectors) which you can add together to get another vector as well as multiply by a scalar to get another vector, and also there is a vector which does nothing when you add it to another vector (called the "zero vector"). Matrices are vectors in this sense because you can add matrices together and multiply them by scalars. The all-zero matrix is the zero vector.

If you still want to think in terms of column vectors, imagine taking each column of a matrix and stacking them all on top of each other to get a column vector with $mn$ entries.


The term ''dimension'' can be used for a matrix to indicate the number of rows and columns, and in this case we say that a $m\times n$ matrix has ''dimension'' $m\times n$.

But, if we think to the set of $m\times n$ matrices with entries in a field $K$ as a vector space over $K$, than the matrices with exacly one $1$ entry in different positions and all other entries null, form a basis as find in OP, and the vector space has dimension $m n$.


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