1
$\begingroup$

Given that we have:

\begin{align} V&=\{p\in\Bbb{R}_4(x) : p'(0)=0 \wedge p(1)=p(0)=p(-1)\} \\ W&=\{p\in\Bbb{R}_4(x) : p(1)=0\}. \end{align}

I am asked to find the dimensions of $V,W,V+W,V \cap W $ and their respective bases.

Rewrite $V$ and $W$: \begin{align} V&=\{p\in\Bbb{R}_4(x): x^2 (x-1)(x+1) \text{ divides } p\} \\ W&=\{p\in\Bbb{R}_4(x): (x-1) \text{ divides } p \} \end{align}

In this case I get that $V \cap W=V$

and the base of $V$ is $\{x^4-x^2\}$ the base for $W$ is $\{x-1,x^2-x,x^3-x^2,x^4-x^3\}$ thus the dimensions are $1$ and $4$... the only thing I'm not sure about is the sum of those two spaces any help with that?

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    $V \cap W = V$ means $V \subseteq W$, right? – 2017-01-09
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    Well in that case would the sum be $W$? – 2017-01-09
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    i would say so yes :) – 2017-01-09
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    Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. I have [edit]ed your question to reflect this principle. – 2017-12-21

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