[proofplan]
We show that any two TST (tridiagonal symmetric Toeplitz) matrices $A$ and $B$ share a common eigenbasis. The argument proceeds in two stages. First, we observe that every TST matrix is an affine function of a single tridiagonal matrix $T$ (the one with zeros on the diagonal and ones on the sub- and super-diagonals), so $A$ and $B$ commute. Second, since both $A$ and $B$ are real symmetric and hence diagonalisable by the spectral theorem, the commutativity guarantees simultaneous diagonalisability — that is, the existence of an orthonormal basis of common eigenvectors.
[/proofplan]
[step:Express every TST matrix as an affine function of a single tridiagonal matrix $T$]
Let $T \in \mathbb{R}^{m \times m}$ denote the tridiagonal matrix with zero diagonal and unit sub- and super-diagonal entries:
\begin{align*}
T_{ij} = \begin{cases} 1 & \text{if } |i - j| = 1, \\ 0 & \text{otherwise}. \end{cases}
\end{align*}
A TST matrix with diagonal parameter $\alpha \in \mathbb{R}$ and off-diagonal parameter $\beta \in \mathbb{R}$ has entries $A_{ij} = \alpha \delta_{ij} + \beta T_{ij}$, where $\delta_{ij}$ is the Kronecker delta. In matrix form:
\begin{align*}
A = \alpha I + \beta T.
\end{align*}
Similarly, a second TST matrix $B$ with parameters $\alpha', \beta' \in \mathbb{R}$ satisfies $B = \alpha' I + \beta' T$.
[guided]
A TST matrix is defined by exactly two parameters: a constant $\alpha$ on the main diagonal and a constant $\beta$ on both the sub- and super-diagonals. Writing this in terms of the identity matrix $I$ and the "shift" matrix $T$ (which has ones on the first sub- and super-diagonals and zeros elsewhere) gives
\begin{align*}
A = \alpha I + \beta T, \qquad B = \alpha' I + \beta' T.
\end{align*}
This decomposition is the key structural observation: all TST matrices live in the two-dimensional subspace $\operatorname{span}\{I, T\}$ of $\mathbb{R}^{m \times m}$. Every algebraic property of $A$ and $B$ is therefore controlled by $I$ and $T$.
[/guided]
[/step]
[step:Verify that $A$ and $B$ commute]
Since $I$ commutes with every matrix, we compute:
\begin{align*}
AB &= (\alpha I + \beta T)(\alpha' I + \beta' T) = \alpha\alpha' I + \alpha\beta' T + \beta\alpha' T + \beta\beta' T^2, \\
BA &= (\alpha' I + \beta' T)(\alpha I + \beta T) = \alpha'\alpha I + \alpha'\beta T + \beta'\alpha T + \beta'\beta T^2.
\end{align*}
Both expressions equal $\alpha\alpha' I + (\alpha\beta' + \beta\alpha') T + \beta\beta' T^2$, so $AB = BA$.
[/step]
[step:Apply simultaneous diagonalisability of commuting symmetric matrices]
Both $A$ and $B$ are real symmetric matrices: $A^\top = (\alpha I + \beta T)^\top = \alpha I + \beta T^\top = \alpha I + \beta T = A$, since $T$ is symmetric (its sub-diagonal and super-diagonal entries are both $1$, so $T_{ij} = T_{ji}$). The same argument gives $B^\top = B$.
By the [Spectral Theorem](/theorems/???), every real symmetric matrix is orthogonally diagonalisable, so both $A$ and $B$ are diagonalisable with real eigenvalues.
We now invoke the following standard result: if two diagonalisable matrices commute, they are simultaneously diagonalisable — that is, there exists an invertible matrix $P$ such that both $P^{-1}AP$ and $P^{-1}BP$ are diagonal.
To verify the hypotheses: $A$ and $B$ are diagonalisable (as real symmetric matrices) and $AB = BA$ (as shown in the previous step). The conclusion follows: there exists an orthonormal basis $\{v_1, \ldots, v_m\}$ of $\mathbb{R}^m$ consisting of simultaneous eigenvectors of $A$ and $B$. In particular, $A$ and $B$ share the same set of eigenvectors.
[guided]
We have shown that $A$ and $B$ are both real symmetric and that they commute. Why does this force a shared eigenbasis?
The key fact is: **commuting diagonalisable matrices are simultaneously diagonalisable.** Here is the argument in outline. Let $\lambda$ be an eigenvalue of $A$, and let $E_\lambda = \ker(A - \lambda I)$ be the corresponding eigenspace. For any $v \in E_\lambda$:
\begin{align*}
A(Bv) = B(Av) = B(\lambda v) = \lambda(Bv),
\end{align*}
so $Bv \in E_\lambda$. This means $B$ maps each eigenspace of $A$ into itself: $B(E_\lambda) \subseteq E_\lambda$. Since $B$ is real symmetric, the restriction $B|_{E_\lambda} : E_\lambda \to E_\lambda$ is also symmetric (with respect to the standard inner product restricted to $E_\lambda$) and hence diagonalisable within $E_\lambda$ by the spectral theorem. Choosing an orthonormal eigenbasis of $B|_{E_\lambda}$ within each eigenspace $E_\lambda$ of $A$ and taking the union over all eigenvalues $\lambda$ of $A$ produces an orthonormal basis of $\mathbb{R}^m$ consisting of simultaneous eigenvectors of both $A$ and $B$.
The symmetry of $A$ and $B$ is essential: two commuting matrices that are not diagonalisable need not share an eigenbasis. For instance, $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $B = A$ commute, but $A$ has only a one-dimensional eigenspace and no basis of eigenvectors.
[/guided]
[/step]
