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$T: \mathbb{C}[x] → \mathbb{C}[x]$ be the $\mathbb{C}$-linear transformation defined on the complex vector space $\mathbb{C}[x]$ of one variable complex polynomials by $T (f(x)) = f(x + 1)$. How many eigenvalues does T have?

The basis of $\mathbb{C}[x]$ over $\mathbb{C}$ are $\{1,x,x^2,\cdots , x^{n} ,\cdots\}$.

Then, $$T(1) = 1$$ $$T(x)= x+1 $$ $$T(x^2) = (x+1)^2 = x^2 + 2x + 1$$ $$T(x^n) = (x+1)^n = x^n + (n_{C_1})x^{n-1}+ \cdots 1$$

Now my intuition is that all the diagonal entries of the matrix will be $1$ and it will be an upper triangularized matrix then the eigen-value of this linear transformation is $1$.We need to find the number of times it occurs

Let $f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots a_0$ be a polynomial such that $T(f(x)) = f(x) \implies f(x+1) = f(x)$ from this equation we can conclude that the only possible polynomial that will satisfy this equation is $c$ where $c \in \mathbb{C}$.

Hence , $T(1) = 1$ so $1$ is the only possible eigen-vector and $1$ is the only eigen-value.

Is my way of approaching the problem correct?

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  • $\begingroup$ You have to be a bit careful when referring to "the matrix" with respect to an infinite basis. $\endgroup$
    – Christoph
    Commented Jul 20, 2021 at 11:59
  • $\begingroup$ @Christoph I get your point. Can you suggest a way in which I can correct that?@Samuel has a nice approach but I want to retain the idea of matrix.Is there a way out? $\endgroup$
    – Antimony
    Commented Jul 20, 2021 at 12:02
  • $\begingroup$ I've completed my answer to give a full matrix-based proof using restrictions to finite-dimensional $T$-invariant subspaces. $\endgroup$
    – Christoph
    Commented Jul 20, 2021 at 12:26
  • $\begingroup$ @Christoph Yes, I got it. $\endgroup$
    – Antimony
    Commented Jul 20, 2021 at 12:32

3 Answers 3

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One has to be careful when dealing with "infinite matrices". For example, the infinite upper triangular matrix $$ \begin{pmatrix} 0 & 1 & 0 & 0 & \cdots \\ 0 & 0 & 1 & 0 & \cdots \\ 0 & 0 & 0 & 1 & \ddots \\ \vdots & \vdots & \vdots & \ddots & \ddots \end{pmatrix} $$ has the "eigenvector" $(1,-1,1,-1,1,-1,\dots)^{\mathrm t}$ with respect to the "eigenvalue" $-1$. (I wrote those in quotes because there is a problem of the vector not having finite support here.)

However, there is an easy way around this in your case:

The linear map $T\colon\mathbb C[x]\to\mathbb C[x]$ given by $(Tf)(x) = f(x+1)$ preserves the degree: $$ \deg(Tf) = \deg(f). $$ Hence, for each $k\ge 0$ the subspace $\mathbb C_k[x]$ consisting of polynomials $f$ with $\deg(f)\le k$ is $T$-invariant. That is, $T$ restricts to a linear map $$T|_{\mathbb C_k[x]}\colon\mathbb C_k[x]\to\mathbb C_k[x].$$ Note that we really only need $\deg(Tf)\le\deg(f)$ for this, so it also works for other transformations like the derivative.

Now let $f$ be an eigenvector of $T$. Since $f\in \mathbb C_k[x]$ for $k=\deg(f)$, we can consider $T_{\mathbb C_k[x]}$ instead of $T$, which has the finite basis $1,x,x^2,\dots,x^k$ and whose (honest to god) matrix is indeed upper triangular with ones on the diagonal. Indeed, the matrix is given by $$ A_k = \begin{pmatrix} 1 & 1 & 1 & 1 & \cdots & 1 \\ & 1 & 2 & 3 & \cdots & \binom{k}{1}\\ & & 1 & 3 & \ddots & \binom{k}{2}\\ & & & 1 & \ddots & \vdots \\ & & & & \ddots & \binom{k}{k-1}\\ & & & & & 1 \end{pmatrix} \in \mathbb C^{(k+1)\times (k+1)}. $$ Hence, $1$ is the only eigenvalue of $T|_{\mathbb C_k[x]}$ and also of $T$ itself. Inspecting the matrix of the restriction we see that the only eigenvectors are multiples of $1$, since $A_k-I$ is of rank $k$.

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Alternative approach to prove that $f(x)=f(x+1) \implies f$ is a constant polynomial.

Let’s suppose that $f$ has degree $n\geq 1$. Because of the Fundamental Theorem of Algebra, $f$ has at least a complex root, let’s call it $\alpha$. Since $f(x)=f(x+1)$, $\alpha\in\mathbb{C}$ is a root of $f$ iff $\alpha + 1$ is a root of $f$. Finally, by recursion, this would imply that $\alpha+m$ is a root of $f$ for all $m\in\mathbb{Z}$, which would result in $f$ having an infinite number of roots, which is absurd (a polynomial of degree $n\geq 1$ has at most $n$ roots).

P.S.: a similar proof using extension fields is valid for any infinite field, so it is not necessary to invoke the FTA, but in this case I decided to keep it as simple as possible.

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You have to justify your claim that $f(x+1)=f(x)$ implies that $f$ is a constant.

By iteration you get $f(x)=f(x+n)$ for all $n \in \mathbb N$ . Any non-constant polynomial has the property that $|f(x) | \to \infty$ as $ x\to \infty$. Letting $n \to \infty$ in $f(x)=f(x+n)$ we get a contradiction.

[$\sum\limits_{k=0}^{n}a_kx^{k}=x^{n}[\sum\limits_{k=0}^{n}a_kx^{k-n}] \to \pm \infty$ according as $a_n >0$ or $a_n <0$].

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  • $\begingroup$ Is the rest of my answer ok? $\endgroup$
    – Antimony
    Commented Jul 20, 2021 at 11:48
  • $\begingroup$ @Antimony Yes, your answer is correct : $1$ is the only eigen value. $\endgroup$ Commented Jul 20, 2021 at 11:48
  • $\begingroup$ Since this works over any infinite field, I wouldn't use the topology of $\mathbb C$ here. Any polynomial will infinitely many roots must be the zero polynomial. $\endgroup$
    – Christoph
    Commented Jul 20, 2021 at 11:52

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