Prove that a set of vectors are linearly independent

Question

I read the following explanation that if I take a set of vectors and transfer it into a matrix, I can use gaussian elimination to check if I can reach a row of zeroes. If it's non reachable then the vectors are linearly independent.

I guess I have some sort of hole in my knowledge of using gaussian elimination method.

suppose I have the following set of vectors : $(1,6,2),(2,3,-3),(1,5,4)$

writing in matrix form: $\begin{bmatrix} 1& 6 &2 \\ 2& 3 &-3 \\ 1& 5 & 4 \end{bmatrix}\rightarrow \begin{bmatrix} 1& 6 &2 \\ 0& -9 &-7 \\ 0& -1 & 2 \end{bmatrix}\rightarrow\begin{bmatrix} 1& 6 &2 \\ 0& -9 &-7 \\ 0& 0 & -25 \end{bmatrix}$

This example is given, and it says that there is no way to reach a row of zeroes.

My question is what is the mark that any furter operations will not get me to the desired outcome of a row of zeros. I know that using gaussian elimination I can add rows and multiply each row by scalar. But when do I stop doing furter manipulations on the matrix?

Answer 1

The determinant of a matrix is invariant under any elementary row or column operations$^{(*)}$.

Once you have reached a triangular matrix, the determinant of all these matrices is the same and is equal to the product of the diagonal entries in this triangular matrix. In particular it is non-zero. If you could reach a row of zeroes, the determinant would be zero.

$^{(*)}$[By elementary row/column operations I mean adding a multiple of one rown/column to another row/column. Advanced row/column operations consist in multiplying a row/column by a non-zero constant, or switching two rows/columns; advanced operations do change the determinant, but you don't need them to reach the row echelon form].

Answer 2

Elementary row operations don't change the span of the row vectors: indeed, the rows after the operations are linear combinations of the original rows. Since row operations are reversible also the converse is true.

By invariance of dimension, you get that the number of nonzero rows in the echelon form is exactly the dimension of the span. Why is this true? Because the nonzero rows in the echelon form are certainly linearly independent and therefore they are a basis of their span.

In particular, if the echelon form has no zero row, the original vectors form a basis of their span and are linearly independent.

Answer 3

Hey don't get much confused. If matrix is to get a row of zeros while performing your steps of gaussian elimination, it will be automatically popped up at the end of your process or somewhere in the middle of your process. After you are done with entire process, just check if you have got row full of zeros, if not, then no way you can get it by doing any sort of further row operations. In your case, as there is no row full of zeros, given three guys are independent.

You can try these examples: $1. (1,1,1), (1,2,3), (2,3,4)$ $2. (1,2,3,4), (2,3,4,5), (1,3,5,7)$.

The logic behind doing this is explained in answer by arnaud

Answer 4

Put the vectors as rows then row reduce the matrix to its unique reduced echelon matrix, observe that the row vectors in the reduced echelon matrix are linearly independent and so the original vectors must be linearly independent.