非对称三对角矩阵转换为对称三对角矩阵

Hey there! Let's break down your questions one by one—great to see you diving into these matrix transformation details.

关于复Hermitian非对称三对角矩阵的情况 #

First, let's clear up a key definition here: a Hermitian matrix is conjugate-symmetric by definition, meaning for all indices $i,j$, $A_{ij} = \overline{A_{ji}}$. If you're dealing with a tridiagonal Hermitian matrix, that translates to:

• All main diagonal elements are real (since $A_{ii} = \overline{A_{ii}}$ requires them to equal their own conjugate)
• Each subdiagonal element $A_{i+1,i}$ is the complex conjugate of the corresponding superdiagonal element $A_{i,i+1}$

In the complex linear algebra context, this is already the direct analog of a real symmetric matrix—Hermitian matrices have all the nice "symmetric-like" properties: real eigenvalues, orthonormal (unitary) eigenvectors, and are diagonalizable via unitary transforms.

If you're thinking of a tridiagonal matrix that looks unsymmetric (e.g., super/subdiagonal elements aren't conjugate pairs), that matrix can't be Hermitian—those two properties contradict each other. So for a truly Hermitian tridiagonal matrix, you don't need a similarity transform to make it "symmetric" (in the appropriate complex-space sense) because it already fits that structure. If your end goal is to map it to a real symmetric matrix, you could use a unitary transform to diagonalize it, but that's a different task than converting to symmetric tridiagonal form.

关于带周期边界的非对称三对角矩阵的情况 #

These matrices are often called cyclic tridiagonal matrices (or periodic tridiagonal), where the top-right ($A_{1,n}$) and bottom-left ($A_{n,1}$) elements are non-zero, in addition to the standard tridiagonal entries.

For real non-periodic unsymmetric tridiagonal matrices, the similarity transform to symmetric form relies on diagonal scaling: you find a diagonal matrix $D$ such that $D^{-1}AD$ is symmetric. This works because the constraints are local (only adjacent row/column pairs matter).

But for cyclic tridiagonal matrices, this local scaling approach falls apart—the off-corner elements introduce a global constraint that can't be satisfied with a simple diagonal transform. In general, there is no similarity transform (using diagonal, unitary, or arbitrary invertible matrices) that can convert an arbitrary cyclic unsymmetric tridiagonal matrix to symmetric form.

Why? Let's look at similarity invariants: symmetric matrices have only real eigenvalues and are always diagonalizable. If your cyclic unsymmetric tridiagonal matrix has complex eigenvalues, or is not diagonalizable (e.g., has Jordan blocks), it can't be similar to a symmetric matrix—those properties are preserved under similarity transforms.

That said, there are special cases where it might work:

• If the cyclic matrix is Hermitian (i.e., $A_{1,n} = \overline{A_{n,1}}$ and sub/superdiagonals are conjugate pairs), then it's already in a conjugate-symmetric form, so no transform is needed.
• If it's normal ($AA^* = A^*A$), it's unitarily similar to a diagonal matrix, but that's not the same as being similar to a symmetric tridiagonal matrix.

总结 #

• For complex Hermitian tridiagonal matrices: They are already conjugate-symmetric (the complex equivalent of real symmetric matrices), so no transform is required to achieve a "symmetric" structure in the appropriate context.
• For cyclic (periodic) unsymmetric tridiagonal matrices: A general similarity transform to symmetric form does not exist, unless the matrix has special properties (like being Hermitian or normal) that already impose symmetric-like structure.

备注：内容来源于stack exchange，提问作者CW279