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.
$$\sum_{i,j=1}^m\min\{i,j\}=1^2+2^2+\cdots+m^2.$$
Best Answer
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 $t$.