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True or false: The non-pivot columns of a matrix are always linearly dependent.

This is false, I just don't really understand why. Thanks for any help!!

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  • $\begingroup$ Non-pivot columns need not to be linearly dependent as a subset of the matrix columns but non-pivot columns are linear combinations of the pivot columns. $\endgroup$ – Algebraic Pavel May 13 '15 at 10:05
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What can you say about the following matrix? $$\begin{pmatrix}1&0&1&2\\0&1&1&0\end{pmatrix}$$

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  • $\begingroup$ is it because each of the non-pivot columns only have trivial solutions? $\endgroup$ – mataxu May 13 '15 at 5:48
  • $\begingroup$ @mataxu: Yes, $$c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 \begin{pmatrix} 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ has only trivial solution. $\endgroup$ – Ángel Mario Gallegos May 13 '15 at 5:52
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Hint In the matrix $$\begin{pmatrix}1 & 1\end{pmatrix}$$ the second column is the only nonpivot column and it is nonzero.

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