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Let $\mathbb{R}_3[x]$ be a vector space of polynomials p with degree $\leq3$ and show that $\phi: \mathbb{R}_3[x]\rightarrow\mathbb{R}^3, \phi(p):=[p(-1), p(0), p(1)] $ is a linear transformation.

Now I know that transformation is linear if these two conditions are true:

A linear transformation between two vector spaces $V$ and $W$ is a map $T:V\rightarrow W$ such that the following hold:

  1. $T(v_1+v_2)=T(v_1)+T(v_2)$ for any vectors $v_1$ and $v_2$ in $V$, and

  2. $T(\alpha v)=\alpha T(v)$ for any scalar $\alpha$.

Let $p(x)=\alpha p_1(x)+\beta p_2(x)$, now we have $\phi(p(x)) = \phi(\alpha p_1(x)+\beta p_2(x))=\left[\begin{array}{c}\alpha p_1(-1)+\beta p_2(-1)\\\alpha p_1(0)+\beta p_2(0)\\\alpha p_1(1)+\beta p_2(1)\end{array}\right]=\alpha\left[\begin{array}{c} p_1(-1)\\p_1(0)\\p_1(1)\end{array}\right]+\beta\left[\begin{array}{c} p_2(-1)\\p_2(0)\\p_2(1)\end{array}\right]$

Does this prove that the transformation is linear?

  • 2
    Yep. That should do it.2012-11-23
  • 0
    +1 for showing your work, including the criteria for any transformation to be linear, and taking the time to format!2012-11-23
  • 0
    Thanks. I know that people will answer question if it is properly formatted. Now about the prove, they always give me headache because I am never really sure if that is all I have to do:)2012-11-23
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    @Mathfan The proof is already done! More explicitly, you get condition $1$ by setting $\alpha = \beta = 1$ and condition $2$ by setting $p_2(x) = 0$. I don't think you'll need to explain that to your tutor, though.2012-11-23

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