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My book presents the following matrix:

\begin{bmatrix}1&0&3&0&-4\\0&1&-1&0&2\\0&0&0&1&-2\\0&0&0&0&0\end{bmatrix}

The book denotes the columns as $a_1, \ldots, a_5$ and asks us to say whether the following columns are linearly independent:

$$\{a_1, a_2, a_4\}, \{a_1, a_2, a_3\}, \{a_1, a_3, a_5\}$$

The book's answer is as follows:

Because $a_3 = 3a_1 - a_2, \{a_1, a_2, a_3\}$ is linearly dependent. So, $\{a_1, a_2, a_4\}$ and $\{a_1, a_3, a_5\}$ are linearly independent.

I already understand how to get the correct answer for this problem; you just look at the matrix and see whether one of them can be represented as a linear combination of the others or not. I'm just confused on the book's way of doing it. I'm confused on where they say "So, $\{a_1, a_2, a_4\}$ and $\{a_1, a_3, a_5\}$ are linearly independent" -- it seems like they're coming to that conclusion from seeing that $a_3 = 3a_1 - a_2$ , and I don't understand how those other sets' linear independence can be concluded from that?

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  • $\begingroup$ The only wrong thing is the usage of the word "So" in the 2nd sentence; it shall be "Also" (see the answer by egreg or see en.wikipedia.org/wiki/Gaussian_elimination . Please, is the book originally in English or is it a translation? $\endgroup$ – yo' Jan 9 '17 at 5:46
  • $\begingroup$ @yo' I agree with you about the misuse of “so”. Linear independence of the other two sets does not follow from the linear dependence of the first one, but instead from considering row echelon forms. $\endgroup$ – egreg Jan 9 '17 at 16:36
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You are right that the book should not conclude it that way unless you have additional information such as only one of those sets are not linearly independent.

Otherwise, without additional information about $a_4$ and $a_5$ we cannot make such conclusion, for example, I can replace $a_4$ by $a_1+a_2$ and the conclusion would have been false.

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  • $\begingroup$ Okay, thanks, I'm just going to ignore that part of the answer then $\endgroup$ – dagny Jan 8 '17 at 7:59
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    $\begingroup$ You do have additional information: the matrix is in reduced row echelon form $\endgroup$ – egreg Jan 8 '17 at 9:53
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    $\begingroup$ I don't want to sound harsh, but with this knowledge, you wouldn't pass the math exam at my university. Please look at the Gaussian elemination and you'll see that you know a lot about the matrix! en.wikipedia.org/wiki/Gaussian_elimination $\endgroup$ – yo' Jan 9 '17 at 5:41
  • $\begingroup$ @yo' thanks for the feedback. my maths knowledge is really limited and I agree that I won't pass your math exam. egreg's answer is correct (note that i did not downvote him) provided those justification are provided in the solution itself. with the current justification given in the original solution in the book, he only justifies $\frac13$ of the answer. I am glad that you agree there is a problem with the word "so", which is what the question is asking. there is indeed additional information, I have the whole matrix, but it is not used in the original solution in the book. $\endgroup$ – Siong Thye Goh Jan 9 '17 at 11:39
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Note that the matrix is in reduced row echelon form.

Since the set $\{a_1,a_2,a_4\}$ consists of the pivot (or dominant) columns, it is linearly independent.

Moreover, the nonpivot columns can be easily written as linear combination of the pivot ones, namely $$ a_3=3a_1-a_2,\quad a_5=-4a_1+2a_2-2a_4 $$ This implies that $\{a_1,a_2,a_3\}$ is linearly dependent.

For the last set, we can reason as follows: remove columns $2$ and $4$ to get a matrix in (nonreduced) row echelon form, where every column is dominant.

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    $\begingroup$ The downvoter is apparently not aware of very simple methods connected with Gaussian elimination. $\endgroup$ – egreg Jan 8 '17 at 9:52

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