Linear Algebra – Linear Independence of Vectors in Non-Square Matrix

linear algebra

We can easily find out if the vectors are dependent or not by finding determinant of matrices formed from the vectors. But if the matrix is not a square matrix, we cannot find the determinant. Then how do we know if the vectors are linearly independent?

Best Answer

There are a few ways to test if $m$ vectors in $\mathbb{R}^n$, where $m \le n$, are linearly independent. Here are the ones I can think of off the top of my head:

  1. Form an $n \times m$ matrix by placing the vectors as columns into a matrix, and row-reducing. The vectors are linearly independent if and only if the there is a pivot in each column of the row-echelon matrix formed. (This method is easy to verify, as it follows basically from the definition of linear independence.)

  2. Form an $m \times n$ matrix by placing the vectors as rows into the matrix, and row-reducing. The vectors are linearly independent if and only if the resulting row echelon form has no zero rows. (Each row operation doesn't change the rowspace, so a row of zeros corresponds to the original row space being of dimension smaller than $m$.)

  3. Form both the matrix $A$ from point 1, and $B$ from point 2. Notice that $A = B^T$. Then $BA$ is an $m \times m$ matrix. The vectors are linearly independent if and only if $\det(BA) \neq 0$. (The vectors are linearly independent if and only if $\operatorname{rank}(A) = m$, and $\operatorname{rank}(A) = \operatorname{rank}(AB)$.)

Related Question