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Say we suspect the columns of a matrix are independent and want to verify that fact quickly by hand. What is the best way to do it?

I'm currently studying MITx 18.033 where they recommend checking if the nullspace is $\{\mathbf 0\}$ by reducing into row echelon form, but how can I check quickly (ideally in my head for a small matrix such as below)?

$$ \begin{pmatrix} 2 & 1 & 9 \\ 3 & 2 & 11 \\ -1 & -3 & 8 \\ 4 & 6 & -4 \end{pmatrix} $$

I suspect columns are independent iff any 3 $2 \times 2$ submatrices with distinct rows and distinct columns are independent. This allows us to prove a matrix is independent by finding those submatrices:

$$ \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}, \ \begin{pmatrix} 2 & 11 \\ -3 & 8 \end{pmatrix}, \ \begin{pmatrix} 2 & 9 \\ -1 & 8 \end{pmatrix} $$

Is this correct? If so, why? If not, how do I prove a matrix's columns are independent quickly?

Zaz
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  • The question is confused. The rows of that matrix are linearly dependent, and you can ascertain that by simply counting rows and columns. Whether the columns are linearly independent is another (and less obvious) question. – Marc van Leeuwen Jan 25 '23 at 10:17
  • @MarcvanLeeuwen: I corrected the question to ask about columns. Thank you. – Zaz Jan 25 '23 at 18:34

2 Answers2

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The magic words are "row reduction".

Igor Rivin
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  • Thank you. I clarified my question: I am asking if there is a faster/easier way than than, a way I could reliably do in my head for small matrices. – Zaz Jan 25 '23 at 02:39
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The '$3$ $2\times2$' condition is necessary (consider the independence of $(c_1,c_2), (c_2, c_3), (c_1, c_3)$ but not sufficient. Consider the matrix: $$\left( \begin{matrix} 1& 1& 2\\ 1& 2& 3\\ 1& 2& 3\\ 3& 1& 4\\ \end{matrix} \right) $$, which has submatrices with distinct columns & row sets $\left( \begin{matrix} 1& 1\\ 1& 2\\ \end{matrix} \right) \left( r_1,r_3,c_1,c_2 \right) ,\left( \begin{matrix} 1& 2\\ 1& 3\\ \end{matrix} \right) \left( r_1,r_3,c_1,c_3 \right) ,\left( \begin{matrix} 2& 3\\ 1& 4\\ \end{matrix} \right) \left( r_3,r_4,c_2,c_3 \right) $ but $c_1+c_2=c_3$.

I'm not sure if there exists any simpler methods, but personally I think row reduction should should work pretty fast with this sort of problems.

erpxyr2001
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  • To make sure I'm understanding this right: $c_i$ represents column $i$, so the top right $1$ should be a $2$? And pairwise independence doesn't imply independence so any technique that just looks at two columns at a time will not work? – Zaz Jan 25 '23 at 06:32
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    Sorry for the error which I've now fixed. And your interpretation is correct - that's why the construction of the counterexample is relatively easy. – erpxyr2001 Jan 25 '23 at 06:55