I would like to know if this is a correct proof to show why no two successive leading principal minors of a tridiagonal matrix have common eigenvalues.
Let's assume that there exist two successive leading principal minors of a tridiagonal matrix that have common eigenvalues. Let $A$ be our tridiagonal matrix, and denote its determinants of the first $k$ and $k+1$ leading principal minors by $D_k$ and $D_{k+1}$, respectively. Then, we have (for $k=1$):
$$ D_k = a_1\\ D_{k+1} = a_1b_2 - c_2^2 $$ where $a_i, b_i,$ and $c_i$ are the diagonal, superdiagonal, and subdiagonal elements of $A$, respectively.
If $D_k$ and $D_{k+1}$ have a common eigenvalue, then there must exist a non-zero vector $x$ such that
$$A^{k+1}x = \lambda A^k x$$
where $\lambda$ is the common eigenvalue. However, we can rewrite this equation as
$$a_1A^k x + b_2A^{k-1}x - c_2^2A^{k-2}x = \lambda a_1A^k x$$
Since $a_1$ is non-zero, we can divide both sides by $a_1A^k x$ to get
$$1 + \frac{b_2}{a_1} \frac{1}{A} - \frac{c_2^2}{a_1} \frac{1}{A^2} = \lambda$$
where $A$ is the matrix associated with the linear transformation induced by $A$.
Now, notice that the left-hand side of this equation defines a polynomial in the matrix inverse of $A$. Specifically, define
$$p(x) = 1 + \frac{b_2}{a_1}x - \frac{c_2^2}{a_1}x^2$$
Then, we have shown that $p(A^{-1})$ has $\lambda$ as an eigenvalue. However, since $A$ is a tridiagonal matrix, its inverse is also tridiagonal. Therefore, $p(A^{-1})$ is also a tridiagonal matrix. However, we know that a tridiagonal matrix can have at most $k$ eigenvalues. Therefore, $p(A^{-1})$ cannot have a distinct eigenvalue for each $k$, which contradicts our assumption that $D_k$ and $D_{k+1}$ share an eigenvalue.
Hence, we must conclude that no two successive leading principal minors of a tridiagonal matrix can share an eigenvalue.