linear algebra - Eigenvalues of tridiagonal symmetric matrix with diagonal entries 2 and subdiagonal entries 1

Problem:

Let A be a square matrix with all diagonal entries equal to 2, all entries directly above or below the main diagonal equal to 1, and all other entries equal to 0. Show that every eigenvalue of A is a real number strictly between 0 and 4.

Attempt at solution:

• Since A is real and symmetric, we already know that its eigenvalues are real numbers.




• Since the entries in the diagonal of A are all positive (all 2), A is positive definite iff the determinants of all the upper left-hand corners of A are positive. I think this can be proven via induction (showing that each time the dimension goes up, the determinant goes up too)




• Since A is symmetric and positive definite, eigenvalues are positive, i.e. greater than 0.

But I can't get the upper bound of 4. Any help would be appreciated. Thank you.

Answer

The characteristic polynomial of $A-2I$ is the $n\times n$ determinant $D_n(X)$ of the matrix with entries $-1$ directly above or below the main diagonal, entries $X$ on the main diagonal, and entries $0$ everywhere else, hence

$$D_{n+2}(X)=XD_{n+1}(X)-D_n(X),$$
for every $n\geqslant0$ with $D_1(X)=X$ and the convention $D_0(X)=1$.
This recursion is obviously related to Chebyshev polynomials and one can prove:

For every $u$ in $(0,\pi)$ and $n\geqslant0$, $D_{n}(2\cos(u))=\dfrac{\sin((n+1)u)}{\sin(u)}$.

Assume that this holds for $n$ and $n-1$ for some $n\geqslant1$, then

$$D_{n+1}(2\cos(u))\sin(u)=2\cos(u)\sin(nu)-\sin((n-1)u)=\sin((n+1)u).$$
Since $D_1(2\cos(u))=2\cos(u)=\sin(2u)/\sin(u)$ and $D_0(2\cos(u))=1=\sin(u)/\sin(u)$, this proves the claim. Hence $x=2\cos(k\pi/(n+1))$ solves $D_n(x)=0$ for every $1\leqslant k\leqslant n$. These $n$ different values are the eigenvalues of $A-2I$.

Since the eigenvalues of $A-2I$ are all in the interval $[-2\cos(\pi/(n+1)),+2\cos(\pi/(n+1))]$, the eigenvalues of $A$ are all in the interval $]0,4[$.