Extending a Chebyshev-polynomial determinant identity

The following $n\times n$ determinant identity appears as eq. 19 on Mathworld’s entry for the Chebyshev polynomials of the second kind:

$$U_n(x)=\det{A_n(x)}\equiv \begin{vmatrix}2 x& 1 & 0 &\cdots &0\\ 1 & 2x &1 &\cdots &0 \\
0 & 1 & 2x &\cdots &0\\0 & 0 & 1 & \ddots & \vdots \\
\vdots & \ddots & \ddots &\ddots & 1\\ 0 & 0 & \cdots & 1 & 2x\end{vmatrix}$$

as can be proven (for example) by expanding by minors to get the recurrence relation for $U_n(x)$.

While working on a spectral problem for my research, I noticed that this result can be extended. Suppose we consider the determinant of $A_n(x)+t\,\mathbf{e}_k \mathbf{e}^T_k$ where $t$ is some free parameter and $k$ is some index. Then the $k$-th column vector may be expressed as $$\mathbf{e}_{k-1}+(2x+t)\mathbf{e}_k+\mathbf{e}_{k+1}=
(\mathbf{e}_{k-1}+2x\,\mathbf{e}_k+\mathbf{e}_{k+1})+t\,\mathbf{e}_k.$$ Since the determinant is an linear function of its $k$-th column vector, we can expand in two terms: The first is just $\det{A_n(x)}=U_n(x)$, and for the second we can expand by minors to get a block diagonal matrix $\text{diag}(A_{k-1}(x),A_{n-k}(x))$ with determinant simplying to $U_{k-1}(x)U_{n-k}$. Putting these together gives the result $$\det{A_n(x)+t\,\mathbf{e}_k\mathbf{e}^T_k}=U_n(x)+t \, U_{k-1}(x)U_{n-k}(x).$$

One can similarly introduce a second parameter at a different row vector and compute the resulting determinant. Hence there should be a well-defined answer to the following question:

Given a set of $n$ parameters $\{t_k\}$, express $\det{(A_n(x)+\text{diag}(\{t_k\}))}$ in terms of $\{U_n(x)\}.$

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For tridiagonal matrices we can derive these by induction:

$$U_n(x,\{t_1, \dots, t_n\})=\det{A_n(x,\{t_1, \dots, t_n\})}\equiv \begin{vmatrix}2 x+t_1& 1 & 0 &\cdots &0\\ 1 & 2x + t_2 &1 &\cdots &0 \\
0 & 1 & 2x + t_3 &\cdots &0\\0 & 0 & 1 & \ddots & \vdots \\
\vdots & \ddots & \ddots &\ddots & 1\\ 0 & 0 & \cdots & 1 & 2x+t_n\end{vmatrix}$$

Using row operations from linear algebra we can expand along the first row to get a linear expansion in $t_1$.

$$ U_n(x,\{t_1, \dots, t_n\}) =
U_n(x,\{0,t_2, \dots, t_n\}) + t_1 U_{n-1}(x,\{t_2, \dots, t_n\})$$

The term of the form $t_1 f(x)$ can be found by setting the remaining variables to $0$. It is:

$$ t_1 U_{n-1}(x)$$

Since the determinant is multilinear, it is possible to take two partial derivatives and find the $t_1 t_2$ term.

$$ t_1 t_2\left|\begin{array}{cc|ccc}
1& 0 & 0 &\cdots &0\\
0& 1 & 0 &\cdots &0 \\ \hline
0& 0 & 2x + t_3 &\cdots &0\\
0 & 0 & 1 & \ddots & \vdots \\
\vdots & \ddots & \ddots &\ddots & 1\\ 0 & 0 & \cdots & 1 & 2x+t_n\end{array}\right| = t_1 t_2 U_{n-1}(x,\{t_2, \dots, t_n\})$$

I realized after a long time, these new determinants are not symmetric in $t_1, \dots, t_n$. These methods still hold and you can recover the individual terms as products of Chebyshev polynomials. For example , for $a < b < c$ the $t_a t_b t_c$ term is:

$$ t_a t_b t_c U_{a-1}(x) U_{b-a-1}(x)U_{c-b-1}(x)U_{n-c}(x) $$

Notice the total degree (including the $t$ factors) are $n$. My mnemonic is simply to draw a row of $n$ letters and count them

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U4 U6 U4 U6