Eigenvalues of tridiagonal symmetric matrix with diagonal entries $2$ and subdiagonal entries $1$

Question

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 $4$.

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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 1

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 $]0,4[$.

Answer 2

Gershgorin circle theorem help us to conclude that the eigenvalues of $A$ are in the interval $(0,4)$.

Actually, we can compute them. First, we notice that $A$ is similar to the matrix which has $2$ on the diagonal, $-1$ directly above and below it, and the other entries are $0$ (multiply on the left and on the right by $\operatorname{diag}((-1)^j,1\leq j\leq n)$. If $x=(x_1,\dots,x_n)$ is a eigenvector for the eigenvalue $\lambda$, we should have $$-x_{j-1}+(2-\lambda)x_j-x_{j+1}=0, 2\leq j\leq n-1.$$ We can assume that $\lambda$ has the form $2(1-\cos \theta)$. Then we should have $$-x_{j-1}+2\cos \theta \cdot x_j-x_{j+1}=0.$$ We want the initial condition to be $x_0=0$ and $x_1\neq 0$. Such a recurrence relation can be solved by $a \alpha^k+b\beta^k$, where $\alpha$ and $\beta$ are the roots of $X^2-2\cos \theta X+1$. These one are $e^{i\theta}$ and $e^{-i\theta}$, and we have $a=-b$, hence we can take $x_j=\sin(j\theta)$. The last equation should be $$2\cos\theta \sin(n\theta)-\sin((n-1)\theta)=0,$$ hence $$2\cos\theta\sin(n\theta)-\sin(n\theta)\cos\theta)+\cos(n\theta)\sin\theta=0,$$ which gives $$\cos\theta\sin(n\theta)+\cos(n\theta)\sin\theta=0,$$ so $\sin((n+1)\theta)=0$. Hence the eigenvalues of $A$ have the form $$\lambda_k:=2\left(1-\cos\left(\frac{k\pi}{n+1}\right)\right)=4\sin^2\left(\frac{k\pi}{2(n+1)}\right),1\leq k\leq n.$$