A real tridiagonal matrix A satisfying $ a_{i,i+1}a_{i+1,i} \geq 0 $ has real eigenvalues? I have shown that a real tridiagonal matrix $A$ for which $ a_{i,i+1}a_{i+1,i} > 0 $ holds, is similar to a real symmetric (and therefore Hermitian) matrix, and hence has real eigenvalues. I have done this by explicitly computing the product $ D^{-1}AD $, for a diagonal matrix $D$ and shown that the matrix $D$ may be chosen in a way that will make the product symmetric. However, I don't know how to extend this to the limiting case where only $ a_{i,i+1}a_{i+1,i} \geq 0 $ holds. On Wikipedia it says that in the limiting case, it still holds that $A$ has real eigenvalues, although it is not necessarily similar to a Hermitian matrix. I have obtained the coupled equations $$ \frac{d_{i}}{d_{i+1}} a_{i,i+1} = \frac{d_{i+1}}{d_{i}} a_{i+1,i} ,$$ from which one may obtain a recursive scheme for calculating $D$ using an arbitrary nonzero number for $d_{11} = d_1$. From this, I don't know how to proceed.
Does anyone know how to carry out the limiting procedure?
 A: If $a_{i,i+1}a_{i+1,i}=0$, then one of these off-diagonal entries is zero (or both). For example, if $a_{i+1,i}=0$ the matrix $A$ has the block form
$$
A=\pmatrix{A_{11}&A_{12}\\0&A_{22}}, \quad 
$$
where $a_{i+1,i}$ is the top-right element of the zero block and $A_{12}$ has at most one nonzero entry (the bottom-left entry is $a_{i,i+1}$). The eigenvalues of $A$ are then given by the eigenvalues of the diagonal blocks $A_{11}$ and $A_{22}$. On the other hand, if $a_{i,i+1}$ is zero then $A$ can be partitioned in a similar manner to the lower block triangular form and if the entries are both zero, the matrix is block diagonal.
Now if $A_{11}$ and $A_{22}$ are such that no entry in the first super or subdiagonal are zero, you can apply what you already know (the blocks are similar to a real symmetric matrix and the eigenvalues are real). Otherwise, you can devise a similar partitioning and apply the argument recursively.
Note that if only one of the entries $a_{i,i+1}$ and $a_{i+1,i}$ is zero, you cannot make this matrix similar to a symmetric matrix (at least not by a diagonal scaling).
