2
$\begingroup$

Here's the question: find a basis for the subspace of $\mathbb{P_{3}}$ consisting of all vectors of the form $at^3-bt^2+ct+d$, where $c=a-2d$ and $b=5a+3d$. And here's my solution: First of all, I put the values that mentioned in the question and I got: $at^3-(5a+3d)t^2+(a-2d)t+d=a(t^3-5t^2+t)+d(-3t^2-2t+1)$ So, it can be seen that it spans, say $\mathbb{W}$. More rigorously $\{t^3-5t^2+t,-3t^2-2t+1\}$ spans $\mathbb{W}$. Now, we must check whether it is linearly independent. In order to do that we must attain some scalars: $t^3(a_{1})+t^2(-5a_{1}-3a_{2})+t(a_{1}-a_{2})+a_{2}=0$. One can easily see that $a_{1}=a_{2}=0$. Therefore $\{t^3-5t^2+t,-3t^2-2t+1\}$ is a basis for $\mathbb{W}$ Do I make any make? Are there any gaps? Thanks for any help.

$\endgroup$
  • 2
    $\begingroup$ Looks fine to me. $\endgroup$ – Andrei Jan 13 at 20:27
  • $\begingroup$ Okay thank you so much, it helped! $\endgroup$ – beingmathematician Jan 13 at 20:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.