I have a subspace $U = \langle x^2-x+4,x-1,x^2+x \rangle $ of $P_2$ over $\mathbb R$. I need to find a basis of $U$.

We know already that these $3$ vectors span $U$ so we need to check for linear independence

I wrote these vectors as columns in a matrix which has $3$ pivots. So this means those $3$ vectors already form a basis of $U$ as they are also linearly independent.

Does this mean that $U$ has dimension $3$ and hence these $3$ vectors infact form a basis of $P_2$?

Working for matrices:

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

  • $\begingroup$ Yes, of course. It would have been interesting to see your work on matrices, wouldn’t it? via begin{bmatrix} & & \ & & \ & & end{bmatrix} $\endgroup$ Apr 14 at 13:01
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    $\begingroup$ @StéphaneJaouen Please check my edit, thank you. $\endgroup$
    – adisnjo
    Apr 14 at 13:06
  • $\begingroup$ You may check out a similar post, regarding vector space for polynomials. Link: math.stackexchange.com/questions/546152/… Hope it helps! $\endgroup$
    – Eddy Y
    Apr 14 at 13:08
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    $\begingroup$ Thanks. The sign $=$ can't be used here. These two matrices ARE NOT equal. You can use $\equiv$ maybe. $\endgroup$ Apr 14 at 13:19

2 Answers 2


HINT.- Do as usually with vectors $u,v,w$ and formed $$\lambda(x^2-x+4)+\mu(x-1)+\rho(x^2+x)=0$$ You know that if the only possibility for $\lambda, \mu,\rho$ is zero then the vectors are linearly independent.You have $$(\lambda+\rho)x^2+(-\lambda+\mu+\rho)x+(4\lambda-\mu)=0$$ This is valid for all value of $x$ because this polynomial is zero in the ring of polynomials. then you have a system in $\lambda, \mu,\rho$. Solving this system you verify that your initial vectors are independent and because there are $3$ vectors and the degree is $2$ you can choose the simplest base which is $\{1,x,x^2\}$.


Indeed! You have just shown that your set of vectors are linearly independent (just augment those matrices with $0$'s) and that work says that the solution of $a(x^2-x+4)+b(x-1)+c(x^2+x)=0$ has to be $a=b=c=0$. Which means they are linearly independent. Thus, you have a linearly independent spanning set, which is of course a basis. Since the set has 3 elements, it must be a basis of $P_2$.


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