Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am given two sub-spaces, v1 and v2. They are in the vector space $\mathbb{R}[x]_{\deg < 4}$.

$v_1=\text{span} \left( {x}^{3}+4\,{x}^{2}-x-3,{x}^{3}+5\,{x}^{2}+5 ,3\,{x}^{3}+10\,{x}^{2}-5\,x+3\right) $ $v_2=\text{span} \left( {x}^{3}-5\,x,{x}^{2}+x,{x}^{3}+2\,{x}^{2}-3 \,x\right)$

I am told that one sub-space is included in the other, and I need to

a. determine which subspace is included in the other

b. find the base of the smaller subspace

c. complete the base from the part b of the question so that it is the base of the larger subspace.

So far I've understood that $v_2$ is part of $v_1$ because $v_1$ has scalars without x-dependence, and $v_2$ does not have any. So $v_1$ includes $v_2$. Next for b I rref-ed the matrix of $v_2$ and found that the 3 vectors are linearly independent, and since I am told they are span therefore I know it is the base. Found. For c I need to add something so that it is the base of $v_1$. This is where I'm not certain how to proceed.

share|cite|improve this question
$\mathbb{R}^4[x]$ is not a field. – lhf Dec 27 '11 at 20:42
My guess is that $\mathbb R^4[x]$ is supposed to be the degree $\leq 4$ (or $< 4$? it doesn't seem to matter) polynomials with coefficients in $\mathbb R$. – Dylan Moreland Dec 27 '11 at 20:45
Yeah, OP most probably meant "vector space $\mathbb R^4[x]$. – Patrick Da Silva Dec 27 '11 at 20:46
Are you sure that the three vectors given for $v_2$ are linearly independent? It seems like \[ (x^3 - 5x) + 2(x^2 + x) - (x^3 + 2x^2 - 3x) = 0 \] is a nontrivial relation. – Dylan Moreland Dec 27 '11 at 21:23
I should remark that I knew that something was wrong just because $\dim v_2 = 3$, $\dim v_1 \leq 3$ (since $v_1$ can be generated by three elements) and $v_2 \subset v_1$ would imply that $v_1 = v_2$, which is impossible. – Dylan Moreland Dec 27 '11 at 21:30
up vote 2 down vote accepted

wrt the basis $x^3,x^2,x,1$, the spaces are $$V_1=\langle(1,4,-1,-3),(1,5,0,5),(3,10,-5,3)\rangle$$ $$ V_2=\langle(1,0,-5,0),(0,1,1,0),(1,2,-3,0)\rangle $$ with a little manipulation (rref on the appropriate matrices) you get $$V_1=\langle(1,0,-5,0),(0,1,1,0),(0,0,0,1)\rangle$$ $$ V_2=\langle(1,0,-5,0),(0,1,1,0)\rangle $$ so you can see that $V_2\subset V_1$ and that $V_1=V_2+\langle(0,0,0,1)\rangle$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.