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I was trying to solve this problem but I got a little stuck in the second point:

  1. Let T be the linear transformation T: $R[x] \rightarrow R[x]$,the polynomials of x with real coefficients, such that: $T(p(x))=p(3x)$.

    a) Show that T is bijective.

    b) Find the eigenvalues of T.

    c) Prove that does not exit a polynomial $p(x) \in R$ such that: $T^{-1}=p(T)$

It is easy to prove a), but when I tried to prove b) I got some problems, because $R[x]$ is not a finite dimensional space so I can not proceed as I always do finding the matrix of the transformation. So I don't really know how to find the eigenvalues of T when the space does not have a finite basis. I would really appreciate any hint or advice you could give me.

Thanks.

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  • $\begingroup$ Recall the meaning of an eigenvector (a vector which is mapped to a constant multiple of itself, under the transformation) and the corresponding eigenvalue; does that help? $\endgroup$ – StackTD Mar 14 '16 at 16:44
  • $\begingroup$ Is $R[x]$ the polynomials of $x$ with real coefficients? $\endgroup$ – Rory Daulton Mar 14 '16 at 16:44
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    $\begingroup$ Start from the definition of an eigenvalue $T(p(x))=\lambda p(x)$ which gives $p(3x)=\lambda p(x)$. For what $p(x)$ can this be satisfied? $\endgroup$ – eyedropper Mar 14 '16 at 16:45
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Nobody has said anything about part c), so here are my two cents:

Hint: Note that the eigenvalues of $T^{-1}$ are $\{3^{-i}\}$ where $i\geq 0$ is an integer. The eigenvalues of $p(T)$, on the other hand, are $\{p(3^i)\}$ where $i \geq 0$ is an integer.

Note that for any polynomial $p$, we have $$ \lim_{i \to \infty} |p\left( 3^i\right)| = \infty $$ but the eigenvalues of $T^{-1}$ form a bounded set.

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Suppose $\;\lambda\;$ is an eigenvalue with eigenvector $\;h(x)=\sum\limits_{k=0}^n a_kx^k\in\Bbb R[x]\;$ , then

$$h(3x)=T(h(x))=\lambda h(x)\iff \sum_{k=0}^n a_k3^kx^k=\sum_{k=0}^n a_k\lambda x^k\iff$$

$$\iff\,\forall\,k=0,1,...,n\;,\;\;a_k3^k=a_k\lambda\iff\lambda=3^k\;\;\text{for at least one}\;\;k$$

and the only possibility the above is a constant is if $\;k=0\implies \lambda=1\;$. So all the constant polynomials are eigenvectors of $\;T\;$ belonging to the eigenvalue $\;1\;$ .

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  • $\begingroup$ Thank you, I really forgot to use the definition of eigenvector, it was not as hard as I was thinking. $\endgroup$ – Felipe Mar 14 '16 at 16:51
  • $\begingroup$ @David My pleasure. Indeed, sometimes the basic definitions are the easiest way to go. $\endgroup$ – DonAntonio Mar 14 '16 at 16:53
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    $\begingroup$ I don't think that this is the correct solution. $a_k$ can be non-negative just for one $k$. $\endgroup$ – eyedropper Mar 14 '16 at 16:59
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    $\begingroup$ To add on to what @eyedropper said, the condition $a_k 3^k = a_k\lambda \iff \lambda = 3^k$ only holds under the assumption that $a_k\neq 0$. $\endgroup$ – Jason DeVito Mar 14 '16 at 17:13
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You should immediately note from $p(3x)=\lambda p(x)$ that $p(x)$ cannot be a polynomial having any $x$ of different degrees. Let $p(x)=ax^k $ with $a\in \mathbb{R}$. Then

$$T(p(x))=ap(3x)=a(3x)^k=a3^kx^k=3^k(ax^k)=3^kp(x)$$

Thus $\lambda = 3^k$. We deduce that $T$ has an infinite number of eigenvectors that come from the set $\{x^k|k\in \mathbb{N}_0\}$ for the eigenvalues $\{3^k|k\in \mathbb{N}_0\}$.

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  • $\begingroup$ Thanks, I was checking an as you say for a polynomial $p(x)=a_n x^n + ... + a_0$ I would have that $3^i = \lambda$ and $a_0=\lambda a_0$ so $\lambda=1$. The thing is get confused with the fact that $\lambda = 3^i$ and $\lambda = 1$, but as you say then T will have an infinite number of eigenvalues. $\endgroup$ – Felipe Mar 14 '16 at 17:35

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