# Linearly independent rows and matrices

Let M be an $m \times n$ matrix whose rows are linearly independent. Suppose that $k$ columns $c_{i_1}, ... , c_{i_k}$of M span the column space of M. Let C be the matrix obtained from M by deleting all columns except $c_{i_1}, ... , c_{i_k}$. Show that the rows of C are also linearly independent.

Since the rows are linearly independent in M, we know that every row has a pivot position.

We know that the row rank is equal to the column rank. Since the row rank is at least m and the column space is at most k, we know that k is either equal to m or greater than it.

Case 1

Let k=m. Then $c_{i_1}, ... , c_{i_k}$ is both a linearly independent and a spanning set, so the rows must also be linearly independent.

Case 2

Let k>m. Since $c_{i_1}, ... , c_{i_k}$ is a spanning set, it must contain the linearly independent set. In other words, it must contain all columns in M that have a pivot position. So the rows of matrix C also have a pivot in every row, which implies that the rows are linearly independent.

Do you think my answer is correct?

I think your "proof by cases" works just fine, though I personally would do three cases, with first case "if $k < m \implies$ Contradiction", basically, then writing the very reasons you concluded that $k \geq m$.
• Nice my dear friend. You really $\textbf{did}$ it. Jun 26 '13 at 4:48