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.


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 $(\int_{-1}^1 p(y)dy, \int_{-1}^1 yp(y)dy)$.

  • $\begingroup$ 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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