What exactly does the following linear transformation represent

Question

Let n>2 be even

$$T:P_n \rightarrow P_{\frac n2} $$

$$T(a_0 + a_1x + \ldots + a_nx^n) = a_0 + a_2x + \ldots + a_{n-2}x^{\frac n2 -1} +a_n x^{\frac n2} $$

I am very deeply confused on what the following linear transformation does to the set of polynomials of degree $n$ and why it's considered a linear transformation. From my understanding this breaks the second axiom of linear transformation of

$$T(rv) = rT(v)$$

it changes the coefficients of the variables so I feel as if though this axiom wouldn't hold, but then again I'm also not 100% sure what the transformation is doing to the polynomials.

Answer 1

If $n=2$ and $v(x) = 1+2x+x^2 \in P_2$ then under your above notation $a_0 = 1, a_1 = 2, a_2 = 1$ and $$ T(v) = T(1+2x+x^2) = a_0 + a_2x = 1+x $$ Now,for any scalar $r$, $$ T(rv) = T(r+2rx+rx^2) = r + rx = r(1+x) = rT(v) $$ We will now generalize for polynomials of degree $n$ where $n$ is an even number. If $$ v(x) = a_0 + a_1x + \dots + a_n x^n $$ and $r$ is a scalar then \begin{align*} T(rv) = T(ra_0 + ra_1x + \dots + ra_n x^n) &= ra_0 + ra_2 x + \dots + ra_n x^{n/2}\\ &=r\left(a_0 + a_2 x + \dots + a_n x^{n/2}\right)\\ &=rT(v) \end{align*}

I hope this clarifies the problem.

Answer 2

\begin{align}T(r(a_0 + a_1x + \ldots + a_nx^n))&= T((ra_0) + (ra_1)x + \ldots + (ra_n)x^n))\\&=(ra_0)+(ra_2)x+\ldots +(ra_n)x^\frac{n}2\\&= r(a_0 + a_2x + ... + a_{n-2}x^{\frac n2 -1} +a_n x^{\frac n2} )\end{align}

The second axiom of linear transformation is not broken.

The transformation maps polynomial of degree $n$ to a polynomial of degree $\frac{n}2$.

For example $1+2x+3x^2$ is being mapped to $1+3x$. It is a mapping that drop all the odd terms for the polynomial in the original space and then map the $i$-th coefficient to the $\frac{i}{2}$-th coefficient.

Answer 3

I suspect you may be making a very common mistake in linear transformations involving polynomials. To show that $T$ preserves scalar multiplication you have to show that $T(rp)=rT(p)$, and you interpret this as $$T(a_0+a_1(rx)+\cdots+a_n(rx)^n)=r(a_0+a_2x+\cdots+a_nx^{n/2})\ .$$ This is wrong because you have an incorrect left hand side. Multiplying a polynomial by a scalar means you multiply the coefficients by the scalar, not the variable. For example, if $$r=4\quad\hbox{and}\quad p=1+2x+3x^2\ ,$$ then $$rp=4+8x+12x^2\qquad\hbox{not}\qquad 1+2(4x)+3(4x)^2\ .$$ If you take $p=a_0+a_1x+\cdots+a_nx^n$ (better assume $n$ is even) and multiply carefully, you get $$\eqalign{ T(p)&=a_0+a_2x+\cdots+a_nx^{n/2}\cr rp&=(ra_0)+(ra_1)x+\cdots+(ra_n)x^n\cr T(rp)&=(ra_0)+(ra_2)x+\cdots+(ra_n)x^{n/2}\cr &=r(a_0+a_2x+\cdots+a_nx^{n/2})\cr &=rT(p)\ .\cr}$$ As to "what it does to the polynomial", you could interpret it as $$T(p(x))=p(\sqrt x)+p(-\sqrt x)$$ - do the algebra on the right hand side to check this. A slight caveat: if you treat this as a numerical equality of real numbers it only works for $x\ge0$. But as a formal expression it works for any (indeterminate) $x$.

Finally, as to "it changes the coefficients", this is exactly what it should do - the coefficients define the polynomial, so to create a new polynomial you have to change the coefficients. A simpler example: differentiating $1+2x+3x^2$ "changes the coefficients" and gives the result $2+6x$.