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Given the following scenarios:

(1) What number x satisfies $10x=3$

(2) What 3-vector u satisfies (1,1,0) x u = (0,1,1)

(3) What polynomial p satisfies $\int_{-1}^1p(y)dy=0$ and $\int_{-1}^1yp(y)dy=1$

I know that they can, in some way, be written as $Lv=w$ where $L:V\to W$, where $L$ maps the set of vectors V to the set of vectors W. I'm being asked to write the sets V and W where the vectors v and w originate from.

I know that for (1), L is 10, v is $(\frac 3{10})$ and w is $(3)$ but I'm lost on how to write the correct set notation and whether the set contains multiple vectors.

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The problem seems to be to translate these into linear algebra problems. So for the first one, we could treat it as about the linear map $\mathbb{R} \to \mathbb{R}$ that takes an element $x\in \mathbb{R}$ and sends it to $10x$.

So an option for (1) is $V=W=\mathbb{R}$ (we could also take $\mathbb{Q}$ or $\mathbb{C}$ here).

For (2), we could take $V=W= \mathbb{R}^3$, because it takes a vector $u\in \mathbb{R}^3$ and sends it to the vector $(1,1,0)\times u\in \mathbb{R}^3$ (it's important to check this is indeed a linear map).

The third one could be take $V=\mathbb{R}[x]$ the vector space of polynomials in $x$, and $W=\mathbb{R}^2$. The linear map takes a polynomial $p(x)\in \mathbb{R}[x]$ and sends it to the pair of real numbers $(\int_{-1}^1 p(y)dy, \int_{-1}^1 yp(y)dy)$.

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  • $\begingroup$ Thanks for the reply! Just for confirmation's sake, for (2), is there an actual vector u to make the equation a true statement? $\endgroup$ – William Jan 23 '17 at 16:29
  • $\begingroup$ No, there is no such vector. If $a \times b = c$ for vectors $a,b,c\in \mathbb{R}^3$, then $c$ is orthogonal to both $a$ and $b$. But in the question, $(1,1,0)$ is not orthogonal to $(0,1,1)$, so there is no such $u$. $\endgroup$ – Jonathan Grant Jan 23 '17 at 16:39

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