Finding the basis of a vector space out of matrices I am to find a basis for the vector space $M$ formed by all $(n \times n)$-matrices. Now, I am finding this to be quite different from previous exercises with bases, where I only have had to construct a set of linearly independent vectors. How am I to do this with matrices? Should I allow each matrix to act as $n$ vectors?
 A: If there is no further restriction on the matrix than it is simply an $n$ x $n$ matrix, then consider that it is composed of $n^2$ entries, each of which has absolutely no impact on any other. To construct a basis for this then, all you need is a set of matrices to go in your basis so that each matrix contributes to only a single element of the set. 
I'm not so good with formatting here, but to illustrate a little, the first matrix would have to contribute only to the first slot, so it would have a 1 in the "11" position, and 0's everywhere else. The second matrix would then have a 1 in the "12" position, and else 0's, and so on, so that the nth matrix has 1 in the "1n" position, and so on, with the $n^2$th matrix have a 1 in the "nn" slot.
Does this help?
A: The space of $\Re^{m\times n}$ matrices behaves, in a lot of ways, exactly like a vector space of dimension $\Re^{mn}$. To see this, chose a bijection between the two spaces. For instance, you might considering the act of "stacking columns" as a bijection. You take a matrix, you pull out its first column, those are the first $m$ elements of your vector, the next $m$ elements will be the second column and so on. 
So, now,
The standard basis in $\Re^{mn}$ would be a set of $mn$ vectors $e_i$. They have a $1$ at the $i^{th}$ location and $0$ everywhere else. If you take this back into the matrix space, you have a $mn$ matrices with $1$ at a particular element and $0$ everywhere else. So, can you form any matrix in the space using this hypothesized basis? Certainly! Because you are simply taking a weighted sum of the basis where the weights are the values of the matrix at the $(i,j)$ location.
Now, as an exercise, think about the space of Symmetric matrices. Do you need a basis of $mn$ elements or can you get away with less? 
A: You can see a square matrix as a vector and easily build a basis by using matrices that have in one entry 1 and 0 in all the others for each position in an $n\times n$ matrix. In fact every matrix can be written as a l.c. of these matrices.
