Column vectors being linearly dependent IFF there exists a column having no pivot

Question

Proof: "$\implies$:

Let the column vectors of an $m \times n$ matrix form a linearly dependent set. Then, by hypothesis, there is a vector $\vec{0} \neq \vec{x} = (x_1, \dots, x_n)^t$ such that $\vec{0}=\sum_{i=1}^{n}x_{i}\vec{v}_{i}$.

A column having no pivot is equivalent to a zero column vector.

Choose an $i \in \{1, \dots, n\}$ with $x_{i} \neq 0$. Without loss of generality, let be $i = n$.

Then,

$\vec{0} = \sum_{i=1}^{n}x_{i}\vec{v}_{i} \Rightarrow \vec{v}_{n} = -\left(\frac{1}{x_{n}}\right)\sum_{i=1}^{n - 1}\vec{v}_{i} \Rightarrow \vec{v}_{n} + \left(\frac{1}{x_{n}}\right)\sum_{i=1}^{n - 1}\vec{v}_{i} = \vec{0}$

The last implication implies that a linear combination of column vector can be added to another column vector to yield the zero column vector.

How should I take this further?

Answer 1

You should specify how you choose the pivot element.

For example, the matrix

$\begin{pmatrix} 5 & 5 \\ 5 & 5 \\ \end{pmatrix}$

can be manipulated using the Gauss elimination algorithm with pivot choice 5. Thus, there is no column without pivot element. But the set of column vectors is clearly linearly dependent.

Hence, your conjecture is false.