# Characteristic polynomial of a simple matrix: Chebyshev?

In my recent MO question, Darij Grinberg mentioned a closely related (structure-wise) determinant, that is, $$\det\left(x_{\min\{i,j\}}\right)_{i,j}^{1,m}=x_1(x_2-x_1)(x_3-x_2)\cdots(x_m-x_{m-1}).$$ In particular, $$\det(\min\{i,j\})=1$$. This prompted me to ask:

QUESTION. Is this true? The characteristic polynomial of the matrix $$\mathbf{N}_m=(\min\{i,j\})_1^m$$ equals $$P_m(\lambda)=\sum_{k=0}^m(-1)^{m-k}\binom{2m-k}k\,\lambda^k.$$

Remark 1. It's worth pointing out that $$P_m(\lambda^2)=(-1)^m\lambda^{2m}\,U_{2m}\left(\frac1{2\lambda}\right),$$ where $$U$$ is Chebyshev polynomial of the second kind.

Remark 2. Needless to say, $$P_m(\lambda)$$ implies a formula for all elementary polynomials $$\mathbf{e}_j$$ of the roots of characteristic polynomial for the given matrix $$\mathbf{N}_m$$.

Remark 3. As a "fun" aside, the entries of the matrix $$\mathbf{N}_m$$ offer a systematic way of calculating the total number of squares (of all sizes) in an $$m\times m$$ grid; i.e. $$\sum_{i,j=1}^m\min\{i,j\}=1^2+2^2+\cdots+m^2.$$

Yes, the characteristic polynomial is given by $$(-1)^m U_{2m}(1/2\lambda ) \lambda^{2m}$$.

The inverse matrix is given by $$\begin{pmatrix} 2 & -1 & 0 & 0 & \dots \\ -1 & 2 & -1 & 0 & \dots \\ 0 & -1 & 2 & -1 & \dots \\ 0 & 0 & -1 & 2 & \dots % \\ 0 & 0 & 0 & -1 & \dots \\ \dots & \dots & \dots & \dots & 1\end{pmatrix}$$ where the $$1$$ in the bottom -right corner denotes that the bottom-right entry (and no other diagonal entry) is a $$1$$. (This can be seen by writing your matrix as $$A^T A$$, where $$A$$ has $$1$$s on the diagonal and upper triangle and $$0$$s in the lower triangle, and taking the inverse of $$A$$.)

The characteristic polynomial is $$Q_m(\lambda) + Q_{m-1}(\lambda)$$, where $$Q_m$$ is the characteristic polynomial of the $$m \times m$$ matrix with $$2$$s on the diagonal, $$-1$$s adjacent to the diagonal, and $$0$$s elsewere.

Laplace expansion gives $$Q_m(\lambda)= (\lambda-2) Q_{m-1}(\lambda) - Q_{m-2} (\lambda)$$ which gives the generating function $$\sum_{m=0}^{\infty} Q_m(\lambda) t^m = \frac{1}{ 1 - (\lambda -2) t + t^2}$$ so the characteristic polynomial of this matrix has the generating function $$\frac{1+t }{ 1 - (\lambda -2) t + t^2}.$$

Since the determinant is $$1$$, we can obtain the characteristic polynomial of the inverse matrix by substituting $$\lambda^{-1}$$ for $$\lambda$$ and multiplying by $$(-\lambda)^m$$, i.e. substituting $$(-\lambda t)$$ for $$t$$, getting $$\frac{1 - \lambda t }{ 1+ (1 -2\lambda ) t + \lambda^2 t^2}$$

as the generating function for the characteristic polynomial of your matrix.

The Chebyshev polynomial has the generating function $$\sum_{n=0}^{\infty} U_n(x) t^n = \frac{1}{ 1- 2x t+ t^2}$$ so

$$\sum_{m=0}^{\infty} U_{2m}(x) t^{2m}=\frac{1}{2} \left( \frac{1}{ 1- 2x t+ t^2} + \frac{1}{ 1+ 2xt + t^2}\right) = \frac{ 1 + t^2 } { 1 + 2t^2 + t^4 - 4 x^2 t^2 }$$

and thus $$\sum_{m=0}^{\infty} (-1)^m U_{2m}(1/2\lambda ) \lambda^{2m} t^{2m} = \frac{ 1 - \lambda^2 t^2 } { 1 - 2\lambda^2 t^2 + \lambda^4 t^4 + t^2 }$$

which is the same after substituting $$\lambda^2$$ for $$\lambda$$ and $$t^2$$ for $$t$$.

• I've edited $AA^T$ to $A^TA$. Feb 13, 2022 at 19:05
• @T.Amdeberhan Thanks! I had about a 50/50 chance of getting that one right... Feb 13, 2022 at 21:20