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vector space => $V=\operatorname{Span}(1+x,1+x+x^2,x^2,2+2x+x^2) \subset R[x]$ -> $R[x]$ stands for the vector space of polynomials in x.

$V_1=1+x$, $V_2=1+x+x^2$, $V_3=x^2$, $V_4=2+2x+x^2$

$\operatorname{span}(V_1,V_2,V_3,V_4)= \{aV_1+bV_2+cV_3+dV_4\}$ => $a(1+x)+b(1+x+x^2)+c(x^2)+d(2+2x+x^2)$

I figured out to the part where it is linearly dependent.

  1. $a+b+2d =0$ $\Rightarrow$ $a+b+2(-b-c)=0$ $\Rightarrow$ $a=b+2c$
  2. $b+c+d=0$ $\Rightarrow$ $d=-b-c$

Let $b=c=1$, $a=3$, $d=-2$ => $3V_1+V_2+V_3-2V_4=0$

But I don't know how to go from here to find the basis with a linearly dependent span vector space.

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  • $\begingroup$ The equality you got means that $v_1,v_2,v_3,v_4$ are not linearly independent. Note that $v_3=v_2-v_1$ and $v_4=v_1+v_2.$ $\endgroup$ – mfl Nov 29 '16 at 20:16
  • $\begingroup$ Therefore (mfl's comment), it is sufficient to use $v_1$ and $v_2$ as basis vectors and the dimension of the vector space is 2. $\endgroup$ – user247327 Nov 30 '16 at 15:11
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Solution:$$1+x+x^2=(1+x)+x^2$$ $$2+2x+x^2=2(1+x)+x^2$$ $$\mbox{$1+x+x^2$ and $2+2x+x^2$ can be expressed as a \\linear combination of $\{1+x,x^2\}$}$$ $$V=span(1+x,x^2,1+x+x^2,2+2x+x^2)$$ $$\Rightarrow V=span(1+x,x^2)$$ $$ \{1+x,x^2\} \mbox{ is linearly dependent and it spans V} $$ $$\mbox{Therefore basis of V is $\{1+x,x^2\}$}$$ $$\mbox{Dimension of V=2}$$

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