Solving Linear Algebra Problem from Berkeley Qual Spring 1990 Using Calculus

Question

I am looking for a calculus solution of the following problem:

Problem. (Berkeley Qual Spring 1990):

Let $n$ br a positive integer, and let $A_n:= (a_{ij}) \in M_n(\mathbb{R})$ be a matrix given by $a_{ii} =2,\ a_{i,i+1} = -1,\ a_{ij} = 0$ otherwise. For instance,

$A_4 = \begin{pmatrix} 2 & -1 & 0 & 0\\ -1 & 2 & -1 &0 \\ 0 & -1 & 2 & -1\\ 0 & 0 & -1 & 2 \end{pmatrix} $

Prove that every eigenvalue of $A$ is a positive real number.

Here is the solution I am trying to complete.

We proceed by induction on $n$. Let $p_n(t) := \det(A_n-t Id)$, i.e. the characteristic polynomial of $A_n$. This satisfies the recursive rule $p_{n+2}(t) =(2-t)p_{n+1}(t) - p_n(t)$. It suffices to show that $p_n>0$ for all $t<0$.

The base case is easy. For the induction step, it is sufficient to prove the bound $|p_n(t)| \le (2-t)p_{n+1}(t)$ for $t<0$. How do can we prove this bound?

It's clear from $\deg p_n(t) = n$ that for $t \to - \infty$, the bound is satisfied, but this is not enough.

Note: my question is not a solution to the Problem above (see Note 2), but rather a solution using my particular approach (or a similar approach doing calculus on characteristic polynomials).

Note 2: I already know two solutions; one is to prove $A_n$ is positive definite, and another is to prove some bounds for entries of eigenvectors. See Problem 7.5.27 in Berkeley Problems in Mathematics. The point of my question is to solve the Problem in as many was as possible.

Answer 1

$$\left( \begin{array}{rr} 1 & 0 \\ - \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rr} 2 & 0 \\ 0 & \frac{ 3 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rr} 1 & - \frac{ 1 }{ 2 } \\ 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rr} 2 & - 1 \\ - 1 & 2 \\ \end{array} \right) $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 1 & 0 \\ 0 & - \frac{ 2 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 3 }{ 2 } & 0 \\ 0 & 0 & \frac{ 4 }{ 3 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & - \frac{ 2 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 2 \\ \end{array} \right) $$

$$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ 0 & - \frac{ 2 }{ 3 } & 1 & 0 \\ 0 & 0 & - \frac{ 3 }{ 4 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & \frac{ 3 }{ 2 } & 0 & 0 \\ 0 & 0 & \frac{ 4 }{ 3 } & 0 \\ 0 & 0 & 0 & \frac{ 5 }{ 4 } \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 1 & - \frac{ 2 }{ 3 } & 0 \\ 0 & 0 & 1 & - \frac{ 3 }{ 4 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 2 & - 1 & 0 & 0 \\ - 1 & 2 & - 1 & 0 \\ 0 & - 1 & 2 & - 1 \\ 0 & 0 & - 1 & 2 \\ \end{array} \right) $$ $$\left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 1 & 0 & 0 & 0 \\ 0 & - \frac{ 2 }{ 3 } & 1 & 0 & 0 \\ 0 & 0 & - \frac{ 3 }{ 4 } & 1 & 0 \\ 0 & 0 & 0 & - \frac{ 4 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrrr} 2 & 0 & 0 & 0 & 0 \\ 0 & \frac{ 3 }{ 2 } & 0 & 0 & 0 \\ 0 & 0 & \frac{ 4 }{ 3 } & 0 & 0 \\ 0 & 0 & 0 & \frac{ 5 }{ 4 } & 0 \\ 0 & 0 & 0 & 0 & \frac{ 6 }{ 5 } \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & - \frac{ 1 }{ 2 } & 0 & 0 & 0 \\ 0 & 1 & - \frac{ 2 }{ 3 } & 0 & 0 \\ 0 & 0 & 1 & - \frac{ 3 }{ 4 } & 0 \\ 0 & 0 & 0 & 1 & - \frac{ 4 }{ 5 } \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrr} 2 & - 1 & 0 & 0 & 0 \\ - 1 & 2 & - 1 & 0 & 0 \\ 0 & - 1 & 2 & - 1 & 0 \\ 0 & 0 & - 1 & 2 & - 1 \\ 0 & 0 & 0 & - 1 & 2 \\ \end{array} \right) $$

https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia#Statement_in_terms_of_eigenvalues

Answer 2

We will find a formula for the eigenvalues of $n \times n$ matrix

$A = \begin{pmatrix}2 & -1 & 0 & \cdots & 0 \\-1 & 2 & -1 & \ddots & \vdots\\ 0 & -1 & \ddots & \ddots & 0\\ \vdots & \cdots & \ddots & 2 & -1 \\ 0 & \cdots & 0 & -1 & 2 \end{pmatrix}$

Solution:

Let $x = \begin{pmatrix}x_{1} \\ x_{2} \\ \vdots \\ x_{n-1} \\ x_{n} \end{pmatrix} $

The components in $(A - \lambda I)x = 0$ are $-x_{k-1} + (2-\lambda)x_{k} - x_{k+1} =0, k = 1, 2, \cdots, n$ with $x_{0} = x_{n+1} = 0$.

Treating that recurrence relation like a differential equation, we seek a solution of the form $x_{k} = cr^{k}$ and on plugging in we get:

$r^{2} - \gamma r + 1 = 0$, where $\gamma = 2 - \lambda$.

Note that we must have two distinct solutions for this quadratic equation since otherwise we will get the solutions $x_{k} = \alpha \rho^{k} + \beta k \rho^{k}$ which becomes zero for all $k$ since $x_{0} = 0$ and $x_{n + 1} = 0$ and since we are seeking an eigenvector this would not be allowed. Hence, we must have two distinct roots $r_{1}, r_{2}$ and the solutions are given by $x_{k} = \alpha r_{1}^{k} + \beta r_{2}^{k}$. Note that $x_{0} = 0 \implies \alpha = -\beta$ and this together with $x_{n + 1} = 0 \implies r_{2}^{n+1} - r_{1}^{n + 1} = 0 \implies$ that $\frac{r_{1}}{r_{2}}$ is a root of unity. Hence $r_{1} = r_{2} e^{2\frac{i \pi j} {n + 1}}$

Factoring and comparing the coefficients:

$r^{2} - \gamma r + 1 = (r - r_{1}) (r- r_{2}) = r^{2} - (r_{1} + r_{2}) r + r_{1}r_{2} \implies r_{1}r_{2} = 1, r_{1} + r_{2} = \gamma$.

We have $r_{1} = e^{\frac{i \pi j} {n + 1}}$, $r_{2} = e^{-\frac{i \pi j} {n + 1}}$

Hence, the eigenvalues of $A$ are $\lambda_{j} = 2- \gamma_{j} = 2- 2\cos \frac{j \pi}{n + 1} $ for $j = 1, 2, \cdots, n$

It follows that $\lambda_{j} > 0$.

Answer 3

I'm not sure I follow why working with a recurrence and applying triangle inequality is 'calculus'. The below is a proof that comes down to (1.) using the recurrence and (2.) applying Cauchy Interlacing.

The recurrence
$p_{n+2}(t) =(2-t)p_{n+1}(t) - p_n(t)$
where $p_n(t) := (-1)^n\cdot \det\big(tI_n - A_n\big)= \det\big(A_n-tI_n \big)$
i.e. it is $(-1)^n$ times the characteristic polynomial.

By induction hypothesis we have $A_n$ having all positive eigenvalues, that is
$p_n(t)= \big(\lambda_1^{(n)}-t\big)\big(\lambda_2^{(n)}-t\big)...\big(\lambda_n^{(n)}-t\big)$
with each $\lambda_k^{(n)}\gt 0$

We want to prove this implies $\lambda_k^{(n+1)}\gt 0$ for $k\in \{1,2,...,n,n+1\}$

Now suppose $p_{n+1}(t^*) =0$ for some $t^*\in(-\infty, 0]$ and apply the recurrence twice.
$p_{n+2}(t^*) =(2-t^*)p_{n+1}(t^*) - p_n(t^*) = 0 - p_n(t^*) \lt 0$
$p_{n+3}(t^*) =(2-t^*)p_{n+2}(t^*) - p_{n+1}(t^*) = (2-t^*)p_{n+2}(t^*)-0 \lt 0$

The eigenvalues in each of these polynomials interlace
$0=p_{n+1}(t^*)= \Big(\big(\lambda_1^{(n+1)}-t^*\big)...\big(\lambda_n^{(n+1)}-t^*\big)\Big)\cdot \big(\lambda_{n+1}^{(n+1)}-t^*\big)$
Interlacing of eigenvalues of $A_{n}$ and $A_{n+1}$ tells us the first $n$ terms in the above are necessarily positive and the last term is zero.

$p_{n+3}(t^*)= \Big(\big(\lambda_1^{(n+3)}-t^*\big)...\big(\lambda_n^{(n+3)}-t^*\big)\big(\lambda_{n+1}^{(n+3)}-t^*\big)\Big)\cdot\Big(\big(\lambda_{n+2}^{(n+3)}-t^*\big)\big(\lambda_{n+3}^{(n+3)}-t^*\big)\Big)$
Since $t^*$ is not a root for $p_{n+3}$ but the eigenvalues of $A_{n+3}$ interlace with those of $A_{n+1}$ we see the first $n+1$ terms are all positive and the last 2 terms are negative giving us
$0\gt p_{n+3}(t^*)= \Big(\gt 0 \Big)\cdot \Big(\gt 0\Big)\gt 0$
which is a contradiction.

Thus $p_{n+1}(t^*) \neq 0$ for any $t^*\in(-\infty, 0]$ and the positivity of all eigenvalues follows.

addendum
to see the interlacing of e.g. eigenvalues of e.g. the $n$ and $n+1$ dim case, consider with standard basis vectors $\mathbf e_k\in \mathbb R^{n+1}$
$B := \bigg[\begin{array}{c|c|c|c|c} \mathbf e_1 & \mathbf e_2 &\cdots &\mathbf e_n \end{array}\bigg]$
$B^TB = I_n$ and
$B^T\big(A_{n+1}\big)B = A_n$