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