# Rows linearly independent implies columns linearly independent

If you have an $$m \times n$$ matrix and the rows are linearly independent, are the columns linearly independent too?

I know that it's the case for $$n \times n$$ matrices.

I've yet tried to think about it, and I think the answer is yes, for example the matrix: $$\begin{bmatrix} 1 & 2 & 3 \\ a & 2a & 3c \end{bmatrix}$$ has linearly dependent rows and columns, but I'm not sure that it works for every $$m \times n$$ matrix.

• Say you have $m$ rows and $n$ columns with $n>m.$ You can never have $n$ linearly independent $m$-dimensional vectors. – saulspatz Jun 8 '19 at 13:34
• – Jyrki Lahtonen Nov 1 '20 at 15:50

Take

$$\begin{pmatrix}1 &1 &0 \\2 &0 &1\end{pmatrix}$$

then the rows are linear independent, but the columns aren't, since for example for the first column you have$$\begin{pmatrix} 1 \\ 2\end{pmatrix}=\begin{pmatrix}1 \\0\end{pmatrix}+2\begin{pmatrix}0 \\ 1\end{pmatrix}$$

• thanks a lot ;) – Mari3 Jun 8 '19 at 13:29
• Your set of vectors that form the matrix are linearly independent, iff your matrix is invertible.

• only square matrices are invertible.

• Therefore, only square matrices are linearly Independent.

If you have an m x n matrix and the rows are linearly independent, are the columns linearly independent too?

This is only necessarily true when the matrix is square.

In $$m\times n$$ matrix, the maximum number of independent rows or columns possible is the order of the largest square you can get from it. If $$m >n$$ then order of the largest square is n, so you can get at most n linearly independent rows or columns (and vice versa). If you get x linearly independent rows then you will also get same number of linearly independent columns also (and vice versa), even for a rectangular matrix.