# Show that T is a linear map

Let $f(x) = x^{4} - 1$ and $g(x)= x^{4}- 4$.

Let $V$ be the vector space consisting of real polynomials of degree less than or equal to $3$.

Let $T:V \longrightarrow V$ be the map such that for $h \in V$, $T(h)$ is the remainder of the Euclidean division of $fh$ by $g$.

Show that $T$ is a linear map.

I thought I could start by proving that $T(0) = 0$? How would I proceed after that?

-
After that, prove $T(a+b)=T(a)+T(b)$ and $T(ra)=rT(a), r\in \mathbb{R}$. That's assuming $V$ is a vector space of real polynomials-you didn't specify. – Kevin Carlson Oct 17 '12 at 22:11

Think the point in this exercise is to remember the definition of the Euclidian division.

Namely, you should remember that, given polynomials $p, q \neq 0$ there always exist polynomials $c, r$ such that

$$p = qc + r \qquad \text{with} \quad r = 0 \quad \text{or}\quad \mathrm{deg}(r) < \mathrm{deg}(q) \ .$$

Moreover, the quotient $c$ and the remainder $r$ are unique verifying this equation with this condition.

Right?

So, let's try to prove that $T(h + h') = T(h) + T(h')$, for instance. By definition of $T$, we have:

Now, let's add up both equations:

$$f(h+h') = g (c_h + c_{h'}) + T(h) + T(h') \ .$$

This looks familiar, doesn't it?

Indeed, with we think for a moment, we conclude that we also have

$$T(h) + T(h') = 0 \quad \text{or} \quad \mathrm{deg}(T(h) + T(h')) < \mathrm{deg}(g) \ ,$$

haven't we? But these are exactly the conditions that, by definition, the reminder $T(h+h')$ verifies:

$$f(h+h') = g c_{h+h'} + T(h + h') \qquad \text{with} \quad T(h +h') = 0 \quad \text{or} \quad \mathrm{deg}(T(h) + T(h')) < \mathrm{deg}(g) \ .$$

But, remember that this equation and condition defined both the quotient and the remainder uniquely. Hence we must have

$$T(h+h') = T(h) + T(h') \ .$$

(And also $c_{h+h'} = c_h + c_{h'}$, but they didn't ask us to prove that.)

Ok, so now try to prove

$$T(\lambda h) = \lambda T(h)$$

and you're done.

-