$A = $ {$1+x,1+x+x^2,1$} is a basis for $P2$ ($P2$ signifies the set of all polynomials of degree 2 or less). It is linearly independent and spans $P2$ (since any arbitrary vector in $P2$ can be written as a linear combination of $1+x,1+x+x^2,$ and $1$).

So the dimension of $P2$ is 3. The set $B = $ {$1+x,1+x+x^2,x$} is linearly independent and has 3 vectors. Furthermore, it even spans all of $P2$ since any arbitrary vector in $P2$ can be written as a linear combination of $1+x,1+x+x^2,$ and $x$ .

As it turns out, the matrix I made in an attempt to prove that set $A$ spans $P2$ is just one row switch away from the matrix corresponding to set B. I feel that has something to do with my confusion but I'm not sure what.

My textbook asked me to extend {$1+x,1+x+x^2$} into a basis for $P2$. In order to do that I added $1,x,$ and $x^2$ to the set and tried to find which of those three were linearly dependent on the existing two vectors and thus could be removed. I found that none of them are linearly dependent, but my textbook only gave set $A$ as an answer...what is going on here? Where is the mistake in my thinking? Could any combination of three vectors here work? Any help is greatly appreciated.

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    $\begingroup$ They are both bases, both extending $\{1 + x, 1 + x + x^2\}$. In most cases, there will be multiple bases that extend a given linearly independent set. In this case, there are infinitely many, so you'll have to excuse your textbook for not listing them all! :-) $\endgroup$ – Theo Bendit Feb 1 at 1:54
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    $\begingroup$ @TheoBendit I see, it's truly an odd question then seeing as there are three right answers and only one was specified...thank you for the response, you've put my mind at ease! $\endgroup$ – James Ronald Feb 1 at 1:55
  • $\begingroup$ Well actually, like I said, there are infinitely many. To properly extend a linearly independent set $S$ by one element, all you need to do is add in a vector that doesn't lie in the span of $S$. Since $S$ spans a plane in the three-dimensional space $P_2$, almost all vectors would work! $\endgroup$ – Theo Bendit Feb 1 at 2:01
  • $\begingroup$ @TheoBendit Ahh right of course, I should've said infinitely many answers as opposed to just three. Thanks again! $\endgroup$ – James Ronald Feb 1 at 2:43

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