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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.

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$\mathbb{R}^4[x]$ 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
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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
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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
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1 Answer

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 $V_1=V_2+\langle(0,0,0,1)\rangle$

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