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Maybe it's because you can't ask those questions to wolfram & I should use a matrix instead but when imputting

linear independence {$t$, $t^2+1$, $t^2+1-t$}

It says the three functions are linearly independent when the third one is clearly a linear combination of the other two. How should I input this to get a valid answer? I wanna check whether or not my results are correct.

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  • $\begingroup$ Entering linear independence {t, t, t} also gives (t,t,t) is linearly independent. $\endgroup$
    – naslundx
    Aug 11, 2016 at 7:50
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    $\begingroup$ linear independence {(0, 1, 0), (1, 0, 1), (1, -1, 1)} $\endgroup$
    – msm
    Aug 11, 2016 at 7:55
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    $\begingroup$ This kind of problem is the price to pay if you use some more or less guessed pseudo-syntax, in which brackets have no proper fixed meaning. Wolfram|Alpha is a very useful tool, but don't forget it's more a google than a CAS. $\endgroup$
    – leftaroundabout
    Aug 11, 2016 at 9:48
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    $\begingroup$ @leftaroundabout what a nice, polite way of saying "rtfm" :-) $\endgroup$
    – Carl Witthoft
    Aug 11, 2016 at 11:21
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    $\begingroup$ @Carl: There's an m? $\endgroup$
    – user14972
    Aug 11, 2016 at 15:34

2 Answers 2

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It's because WolframAlpha interprets your input as one vector, i.e. the space of the single vector $(t, t^2+1, t^2+1-t)$.

An appropriate input would be (treating $1$, $t$ and $t^2$ as basis vectors):

linear independence (0,1,0), (1,0,1), (1,-1,1)

which outputs linearly dependent.

You can find other input examples for linear algebra here.

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    $\begingroup$ Ah, I see now thank you for your answer. $\endgroup$
    – Heat Ice
    Aug 11, 2016 at 8:07
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    $\begingroup$ Can wolfram only do polynimials? What if I want to check the linear dependence of $\{ \sin x, x \}$ for example? $\endgroup$
    – Ovi
    Aug 11, 2016 at 15:45
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Another way to check for linear independence in W|A is to compute the Wronskian, say with the input "wronskian(($t$, $t^2+1$, $t^2+1-t$), $t$)", which results in $0$ so the set of functions is indeed linearly dependent.

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