Does a tall matrix with linearly independent rows have linearly independent columns?

Question

I recently came across the following problem. Let $A$ be an $n\times m$ matrix with $n\ge m$, so that the matrix $A$ is "tall". I'm interested in computing a left inverse of $A$ but, for this, I need first to verify that $A$ is indeed left-invertible.

By rank-nullity, for this, I only need to verify that $A$ has full column rank. So far, this is what I can do: Denote the rows of $A$ by $a_1,\dots,a_n$. Suppose that I can show that, amongst these $n$ vectors $\{a_i\}_{i=1}^n$ there is a set of $\{a_{i_j}\}_{j=1}^m$ which is linearly independent (I can do this by checking that the Gram sub-matrix $(a_{i_j}^{\top}a_{i_k})_{j,k=1}^m$ is invertible).

Does this necessarily imply that the original matrix $A$ had an entire column rank?

Answer 1

The column rank and row rank are equal (https://en.wikipedia.org/wiki/Rank_(linear_algebra)#Proofs_that_column_rank_=_row_rank).

Answer 2

If you are saying that for a Matrix $A^{m \times n}$ with m>n. Then by the property of Rank of Matrix

$R(A)\le n$ $\ni$ Row-Rank[$R(A_R)$]= Col-Rank[$R(A_C)$]= $R(A)$

Check the following thread for more info . Looking for an intuitive explanation why the row rank is equal to the column rank for a matrix