Say there is a matrix A with 77 columns and the rank of AA is 55. Columns 1,3,4,51, 3,4, 5, and 77 are linearly independent (not necessarily pivot columns). Is it true or false that these linearly independent columns form a basis for the column space of A?

I think that this statement is true. My reasoning is that if these 5 columns are linearly independent, then columns 2 and 6 can be written as linear combinations of these linearly independent columns. Therefore, since those 5 columns can generate columns 22 and 66, they span the column space of A. So, since they are in and span the column space of AA, and they are linearly independent, that fulfills the definition of a basis for a subspace, and that must mean those 55 columns do form a basis for the column space of AA.

Is this statement true or false? And is the reasoning alright? Any insight would be awesome.

(Note: I am currently taking linear algebra. I’m not a math major or anything so bear with me if the explanation is somewhat mediocre.)

Thank you

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2 Answers

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My definition of (column) rank is dimension of the column space.

So if rank is 55 and you’ve found 55 linearly independent columns then it is automatic that these form a basis for column space, because any linearly independent set with 55 elements must be a basis.

But note that just the fact that 55 columns are linearly independent does not prove that they are the basis – in R2\mathbb{R}^2 (1,0)(1,0) is linearly independent but it is not a basis

This depends on the number of rows in the matrix. Consider each column as a m-dimensional vector, where m is the number of rows. So if the matrix has 5 rows (as its rank, or the number of linearly independent rows/columns) then these 5 columns indeed form a basis for the column space. But if the number of rows is greater than 5 (it can’t be less, of course), then 5 vectors are not enough in order to form a basis.

In other words, for a group of linearly-independent columns to form a basis for the column space, the size of the group should be the number of rows.