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

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?

• 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) – user7427029 Sep 16 '18 at 23:32
• @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. – Mathematicing Sep 17 '18 at 12:32
• O. k., my fault. Thanks for the info. – user7427029 Sep 17 '18 at 14:20

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