let us consider following problem

  1. A subsepace $S$ of a vector space $V$ is given.

    Determine a basis for $S$ and extend your basis for $S$ to obtain a basis for $V$.

    $V=P_2$, $S$ is the subspace consisting of all polynomials of the form $$(2a_1+a_2)x^2+(a_1+a_2)x+3(a_1-a_2).$$

we know that for polynomials of order $2$ basis can be represented by following set of vectors

$(1,x,x^2)$ now we want to extend this basis for vector space $V$,does it means that for vector space $V$ linear combination of basis of $S$ will be another basis or?clearly if $V$ is vector space of order more then $2$,we can add more element like $x^3$,but in this case what we should do?thanks in advance


1 Answer 1


If you rewrite the defining equation for $S$ as all polynomials of the form $a_1(2x^2+x+3)+a_2(x^2+x-1)$, then it becomes clear that $S$ is defined as the span of $2x^2+x+3$ and $x^2+x-1$, which you can verify are linearly independent, hence form a basis for $S$. All that remains to do, then, is to find a third linearly independent vector to extend this to a basis for $V$.

  • $\begingroup$ should i search it using brute force method or just by guessing or is there any other method?thanks for reply $\endgroup$ Dec 23, 2013 at 18:48
  • $\begingroup$ There are many methods; it shouldn't be difficult to guess and verify (in which case it will be easier to verify spanning rather than linear independence). $\endgroup$ Dec 23, 2013 at 18:51
  • $\begingroup$ does there exist any matrix form?because mostly my friend was interested in this problem for his exam $\endgroup$ Dec 23, 2013 at 18:52
  • $\begingroup$ Well, this is of limited use because it only works in three dimensions, but in the circumstance of a three dimensional vector space and two vectors written in terms of an obvious basis (here $1$, $x$, and $x^2$), you can use a method you might remember from calculus of finding a third perpendicular vector. $\endgroup$ Dec 23, 2013 at 18:54
  • $\begingroup$ but how to apply cross product there?by the way thanks very much $\endgroup$ Dec 23, 2013 at 18:56

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