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 $i = n$.


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

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?

  • $\begingroup$ I edited your question: MathJax is now used and the proof direction is stated explicitely. Nice trick: Your life will get easier if you subdivide the proof of "iff" conjectures into a "$\implies$" section and a "$\impliedby$" section. (Exception: Proof through a chain of biimplications) $\endgroup$ – user7427029 Sep 16 '18 at 23:32
  • $\begingroup$ @user7427029 I know - I've already graduated. I just thought the veteran audience would be aware of the direction of my proof without it being stated explicitly. I am addressing the 'Only if' implication. $\endgroup$ – Mathematicing Sep 17 '18 at 12:32
  • $\begingroup$ O. k., my fault. Thanks for the info. $\endgroup$ – user7427029 Sep 17 '18 at 14:20

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.


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