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The following is a question from the sample GRE Mathematics Subject Test found on the ETS website:

Let $M$ be a $5\times 5$ real matrix. Exactly four of the following five conditions on $M$ are equivalent to each other. Which of the five conditions is equivalent to NONE of the other four?

(A) For any two distinct column vectors $u$ and $v$ of $M$, the set {$u,v$} is linearly independent.

(B) The homogeneous system $Mx=0$ has only the trivial solution.

(C) The system of equations $Mx=b$ has a unique solution for each real $5\times 1$ column vector $b$.

(D) The determinant of $M$ is nonzero.

(E) There exists a $5\times 5$ real matrix $N$ such that $NM$ is the $5\times 5$ identity matrix.

Apparently, the correct answer is (A), but I can't figure out why this is true. If $M$ is nonsingular, as is implied by statements (B)-(E), then isn't that equivalent to its column vectors being linearly independent? And if the 5 column vectors are independent, then I can easily show that each pair of vectors are independent. What am I missing?

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3 Answers 3

up vote 5 down vote accepted

Linear independence of $n$ vectors, for $n>2$, is not equivalent to "pairwise" independence. Take three different vectors in the plane, for an example.

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Thanks. I realized right after I posted the question that although linear independence of the $n$ vectors implies pairwise linear independence, the other way around is false. I appreciate your quick answer though. –  jake Oct 14 '10 at 3:35
    
@jake: there are in fact somewhat (at first) unexpected instances in mathematics of things being pairwise something but not collectively, for example independent events in probability, pairwise independence is much weaker, it doesn't imply independence of n events. –  Weltschmerz Oct 14 '10 at 3:38
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HINT $\rm\ \ u,v\ $ linear independent $\rm\ \Rightarrow\:\ u + v,\:u+2v,\: u+3v,\:\ldots\: $ pairwise independent

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Suppose you have a vector $v$ for which is nonzero. Then $v \not= 2v$ but $M(2v) = 2M(v)$ and $M(v)$ are not linearly independent. Conditions B-E are all equivalent.

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