# Chebyshev polynomials via linear algebra

## Branko Ćurgus

The Chebyshev differential equation

The Chebyshev differential equation is $(1-t^2)\, y^{\prime\prime}(t) - t\, y^{\prime}(t) + \alpha^2 y(t) = 0.$ In this equation $\alpha$ is a parameter and we are seeking to find all functions $y(t)$ which satisfy this differential equation. Problems like this are studied in some differential equations classes. (However, at present, neither Math 331 nor Math 432 at WWU covers this topic.) I would like to point out here that using linear algebra we can find all coefficients $\alpha$ for which the Chebyshev differential equation has polynomial solutions.
A specific vector space of polynomials

To be specific consider the vector space of all polynomials of degree less than or equal to $5$. This space is denoted by ${\mathbb P}_5$ in the textbook. Set $\phi_0(t) = 1, \quad \phi_n(t) = t^n, \quad n=1,2, \ldots, \qquad \mathcal B = \bigl\{\phi_0, \phi_1, \phi_2, \phi_3, \phi_4, \phi_5 \bigr\}.$ Notice that $\mathcal B$ is the standard basis for ${\mathbb P}_5$.
Linear transformation - differential operator

Inspired by the Chebyshev differential equation we define a linear transformation $T : {\mathbb P}_5 \to {\mathbb P}_5$ by $(Tp)(t) = (1-t^2)\, p^{\prime\prime}(t) - t\, p^{\prime}(t), \quad p \in {\mathbb P}_5.$ It should be clear that $T$ maps ${\mathbb P}_5$ to ${\mathbb P}_5$ and that it is a linear transformation.
Corresponding matrix representation

Based on what we learned in Section 5.4 we find the $\mathcal B$-matrix, call it $M$, of $T$. To find $M$ we need to calculate $(T\phi_0)(t) = 0, \quad (T\phi_1)(t) = -t, \quad (T\phi_n)(t) = n(n-1)\,t^{n-2} - n^2 t^n, \ n=2,3,4,\ldots.$ That is $T\phi_0 = 0, \quad T\phi_1 = \phi_1, \quad T\phi_n = n(n-1)\, \phi_{n-2} - n^2 \phi_n, \ n=2,3,4,\ldots.$ Based on these calculations $\mathcal B$-matrix of $T$ is $M = \left[ \begin{array}{rrrrr} 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & -1 & 0 & 6 & 0 & 0 \\ 0 & 0 & -4 & 0 & 12 & 0 \\ 0 & 0 & 0 & -9 & 0 & 20 \\ 0 & 0 & 0 & 0 & -16 & 0 \\ 0 & 0 & 0 & 0 & 0 & -25 \end{array} \right].$
Eigenvalues and eigenvectors

The eigenvalues of $M$ are $0, -1, -4, -9, -16, -25$. Using row reduction for each of these eigenvalues we can find the corresponding eigenvectors. For example, for $\lambda = -16$ to find the corresponding eigenspace we need to solve $\left[ \begin{array}{rrrrr} 16 & 0 & 2 & 0 & 0 & 0 \\ 0 & 15 & 0 & 6 & 0 & 0 \\ 0 & 0 & 12 & 0 & 12 & 0 \\ 0 & 0 & 0 & 7 & 0 & 20 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -9 \end{array} \right] \, \left[ \begin{array}{r} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{array} \right] = \left[ \begin{array}{r} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array} \right].$ Even without row reduction, just by back-substitution we find that $x_6 = 0, \quad x_5, \quad x_4 = 0, \quad x_3 = - x_5, \quad x_2 = 0, \quad 8 x_1 = x_5.$ Thus, an eigenvector corresponding to $\lambda = -16$ is for example $\bigl[ 1 \ \ 0 \ \ -8 \ \ 0 \ \ 8 \ \ 0 \bigr]^T$. Similarly we calculate other eigenvectors:
 $0$ $-1$ $-4$ $-9$ $-16$ $-25$ $\left[\begin{array}{r} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right]$ $\left[\begin{array}{r} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right]$ $\left[\begin{array}{r} -1 \\ 0 \\ 2 \\ 0 \\ 0 \\ 0 \end{array}\right]$ $\left[\begin{array}{r} 0 \\ 0 \\ -3 \\ 0 \\ 4 \\ 0 \end{array}\right]$ $\left[\begin{array}{r} 1 \\ 0 \\ -8 \\ 0 \\ 8 \\ 0 \end{array}\right]$ $\left[\begin{array}{r} 0 \\ 5 \\ 0 \\ \!\!-20 \\ 0 \\ 16 \end{array}\right]$

Back to the differential operator

What does this tell us about $T$? The eigenvalues of $T$ are same as the eigenvalues of $M$. The polynomials whose coefficients are the coordinates of the eigenvectors are eigen-polynomials of $T$. For example with $q_4(t) = 1 \, \phi_0(t) - 8 \, \phi_2(t) + 8 \, \phi_4(t) = 1 - 8 \, t^2 + 8 \, t^4$ we have $T \, q_4 = -16\, q_4$ The complete list of six eigen-polynomials is \begin{alignat*}{5} q_0 & = \phi_0 & \phantom{xxx} & \text{that is} & \phantom{xxx} & & q_0(t) & = 1 \\ q_1 & = \phi_1 & & \text{that is} & & & q_1(t) & = t \\ q_2 & = -\phi_0 + 2 \phi_2 & & \text{that is} & & & q_2(t) & = -1 + 2t^2 \\ q_3 & = -3\phi_1 + 4\phi_3 & & \text{that is} & & & q_3(t) & = -3t + 4t^3 \\ q_4 & = \phi_0 - 8 \phi_2 + 8 \phi_4 & & \text{that is} & & & q_4(t) & = 1 - 8 t^2 + 8 t^4 \\ q_5 & = 5 \phi_1 - 20 \phi_3 + 16 \phi_5 & & \text{that is} & & & q_5(t) & = 5 t - 20 t^3 + 16 t^5 \end{alignat*}
Back to the Chebyshev differential equation

What does this tell us about the Chebyshev differential equation. We just found out that the Chebyshev differential equation has polynomial solutions for $\alpha = 0, 1, 2, 3, 4, 5, \ldots$. We found specific formulas for the first six solutions. To find the polynomial solutions for larger positive integers $\alpha$ we would have to work with larger matrices.
An amazing connection

One amazing aspect of classical orthogonal polynomials is that often they are connected to formulas that were encountered in other contexts. The Chebyshev polynomials are related to the formulas in Exercise 17 in Section 4.7.
• Recall Exercise 17 in Section 4.7 in which we were asked to explore connections between trigonometric functions ${\mathbf x}_k(t) = (\cos t)^k, \quad k = 0, 1, 2, 3, \ldots$ and the functions ${\mathbf y}_k(t) = \cos (k\,t), \quad k = 0, 1, 2, 3, \ldots.$ We found the following formulas: \begin{alignat*}{5} {\mathbf y}_0 & = {\mathbf x}_0 &\phantom{xxx} &\text{that is}& \phantom{xxx} & & 1 & = 1 \\ {\mathbf y}_1 & = {\mathbf x}_1 & & \text{that is} & & & \cos t & = \cos t \\ {\mathbf y}_2 & = -{\mathbf x}_0 + 2 {\mathbf x}_2 & & \text{that is} & & & \cos(2\,t) & = -1 + 2(\cos t)^2 \\ {\mathbf y}_3 & = -3{\mathbf x}_1 + 4{\mathbf x}_3 & & \text{that is} & & & \cos(3\,t) & = -3 \cos t + 4(\cos t)^3 \\ {\mathbf y}_4 & = {\mathbf x}_0 - 8 {\mathbf x}_2 + 8 {\mathbf x}_4 & & \text{that is} & & & \cos(4\,t) & = 1 - 8 (\cos t)^2 + 8 (\cos t)^4 \\ {\mathbf y}_5 & = 5 {\mathbf x}_1 - 20 {\mathbf x}_3 + 16 {\mathbf x}_5 & & \text{that is} & & & \cos(5\,t) & = 5 \cos t - 20 (\cos t)^3 + 16 (\cos t)^5 \\ {\mathbf y}_6 & = -{\mathbf x}_0 + 18 {\mathbf x}_2 - 48 {\mathbf x}_4 + 32 {\mathbf x}_6 & & \text{that is} & & & \cos(6\,t) & = -1 + 18 (\cos t)^2 - 48 (\cos t)^4 + 32 (\cos t)^6 \end{alignat*}
• Now substituting the variable $t$ in the last column of formulas by $\arccos t$ and comparing to the formulas for the Chebyshev polynomials we get \begin{alignat*}{2} 1 & = 1 & & = q_0(t) \\ \cos(\arccos t) & = t & & = q_1(t) \\ \cos\bigl(2\,\arccos t \bigr) & = -1 + 2t^2 & & = q_2(t) \\ \cos\bigl(3\,\arccos t \bigr) & = -3 t + 4 t^3 & & = q_3(t) \\ \cos\bigl(4\,\arccos t \bigr) & = 1 - 8 t^2 + 8 t^4 & & = q_4(t) \\ \cos\bigl(5\,\arccos t \bigr) & = 5 t - 20 t^3 + 16 t^5 & & = q_5(t) \\ \cos\bigl(6\,\arccos t \bigr) & = -1 + 18 t^2 - 48 t^4 + 32 t^6 & & = q_6(t) \end{alignat*}
• In fact, it is universally true that $q_n(t) = \cos\bigl(n\,\arccos t \bigr), n = 0,1,2,\ldots$.