# FInding the number of eigenvalues of the given $\mathbb{C}$ linear transformation.

$$T: \mathbb{C}[x] → \mathbb{C}[x]$$ 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?

• You have to be a bit careful when referring to "the matrix" with respect to an infinite basis. Commented Jul 20, 2021 at 11:59
• @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? Commented Jul 20, 2021 at 12:02
• I've completed my answer to give a full matrix-based proof using restrictions to finite-dimensional $T$-invariant subspaces. Commented Jul 20, 2021 at 12:26
• @Christoph Yes, I got it. Commented Jul 20, 2021 at 12:32

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 $$k$$.

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.

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 $$a_n <0$$].

• Is the rest of my answer ok? Commented Jul 20, 2021 at 11:48
• @Antimony Yes, your answer is correct : $1$ is the only eigen value. Commented Jul 20, 2021 at 11:48
• 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. Commented Jul 20, 2021 at 11:52