# Finding a basis of a Subspace

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:

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

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

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 $$P_2$$.