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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?

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2  
Yep. That should do it. –  Isaac Solomon Nov 23 '12 at 19:36
    
+1 for showing your work, including the criteria for any transformation to be linear, and taking the time to format! –  amWhy Nov 23 '12 at 19:45
    
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:) –  Mathfan Nov 23 '12 at 20:12
    
@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. –  Cocopuffs Nov 23 '12 at 20:45

1 Answer 1

Apparently, the problem is essentially solved within the formulation and comments, but because there are a lot of upvotes and no answer, I propose to write one up.

As OP rightly observes, the map in question obeys the rule: $$\phi( \alpha p + \beta q ) = \alpha \phi(p) + \beta \phi(q)$$ (note that the expression $\alpha \begin{bmatrix} p(-1) \\ p(0) \\ p(1) \end{bmatrix} + \beta \begin{bmatrix} q(-1) \\ q(0) \\ q(1) \end{bmatrix} $ is $\alpha \phi(p) + \beta \phi(q)$ ).

It remains to see that this condition implies linearity of $\phi$, which is defined by the following contidions:

  1. For any $p,q \in \mathbb R_3[x]$, it holds that $\phi(p + q) = \phi(p) + \phi(q)$.
  2. For any $p \in \mathbb R_3[x]$, $\alpha \in \mathbb R$, it holds that $\phi(\alpha p) = \alpha \phi(p)$.

We know that $\phi$ satisfies the condition:

  1. For any $p,q \in \mathbb R_3[x]$, $\alpha,\beta \in \mathbb R$, it holds that $\phi(\alpha p + \beta q) = \alpha \phi(p) + \beta \phi(q)$.

Now, putting $\alpha = \beta = 1$ in 3., we recover condition 1. Putting $\beta = 0$ and $q$ arbitrary, we recover condition 2. So, both conditions hold, and $\phi$ is indeed linear.

As an aside, note that if $\phi$ is linear, i.e. 1. and 2. hold, then: $$ \phi(\alpha p + \beta q) = \phi( \alpha p) + \phi(\beta q)= \alpha \phi(p) + \beta \phi(q)$$ so condition 3. follows. This means that 3. is in fact a different formulation of linearity (in fact, it might be taken as an alternative definition).

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