# Basis for a column space of a matrix

Say there is a matrix A with $7$ columns and the rank of $A$ is $5$. Columns $1, 3,4, 5$, and $7$ 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 $2$ and $6$, they span the column space of A. So, since they are in and span the column space of $A$, and they are linearly independent, that fulfills the definition of a basis for a subspace, and that must mean those $5$ columns do form a basis for the column space of $A$.

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

So if rank is $5$ and you've found $5$ linearly independent columns then it is automatic that these form a basis for column space, because any linearly independent set with $5$ elements must be a basis.
But note that just the fact that $5$ columns are linearly independent does not prove that they are the basis - in $\mathbb{R}^2$ $(1,0)$ is linearly independent but it is not a basis