Solution

Any subspace of has dimension at most and all the numbers from to are possible. Let be the standard basis consisting of the columns of identity. To make a two dimesional subspace, you can start with and throw in any three combinations of them to make a list of 5 vectors. Similar work is done for other dimensions. For dimension zero, you must take all vectors to be zero. Thus, if you have 5 distinct vectors, then you must have dimension at least 1.

The dimension can never be more than 4, since otherwise, you can find 5 independent vectors in . This is known to be impossible. What is the proof?

Here, we can argue thus:

Take your 5 vectors and write them as columns forming a 4 by 5 matrix. Such a matrix has rank less than or equal to 4 since it has only 4 rows. But for a set of columns to be independent, the rank must match the number of columns. Hence we have a contradiction.

You should also think how the argument goes in an abstract 4 dimensional space. Take your 5 vectors and write them as combinations of the four basis vectors , thus each has a form . Make a column of the coefficients and then make a 4 by 5 matrix from the columns of coefficients. The original vectors are independent iff the rank of the resulting matrix is exactly 5. Thus we get the same reasoning.

This idea of reducing any abstract problem to a matrix problem is crucial!