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Consider the matrix

$$A = \begin{pmatrix} f_1 & & \cdots & & f_n\\ p_1 & 0 & \cdots & \cdots & 0\\ 0&\ddots &&&\vdots\\ \vdots& &\ddots&&\vdots\\ 0 & & &p_{n-1} & 0 \end{pmatrix}$$

where $f_1,\ldots,f_n,p_1,\ldots,p_{n-1}>0$. My lecturer states that is is easy to show this matrix has one positive eigenvalue however i'm not so sure.


Firstly, how do we find the characteristic polynomial for this matrix? Secondly how would one find a solution to this characteristic polynomial? It seems impossible since the matrix entries are not known.. Is there a theorem we can use to prove this instead? I have been taught the Perron-Frobenius theorem however that only tells us about the maximum eigenvalue, it doesn't tell us about whether the other eigenvalues are going to be negative.

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  • $\begingroup$ The matrix in question has an Hessenberg form, for which many properties are known. $\endgroup$ – Jean Marie Apr 14 '16 at 20:38
  • $\begingroup$ @JeanMarie "It is known is not an answer". I asked "how" do we show this. $\endgroup$ – user2850514 Apr 14 '16 at 21:00
  • $\begingroup$ @user2850514: a comment is not an answer. You'd better have a look a the hint. $\endgroup$ – Yves Daoust Apr 14 '16 at 21:21
  • $\begingroup$ I don't pretend it is an answer ! It was only a comment. $\endgroup$ – Jean Marie Apr 14 '16 at 21:22
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You can use the basic definition of eigenvalue/eigenvector.

you want to find $\lambda$ and some $v \in \mathbb{R}^n$ such that $A v = \lambda v$.

This has an easy form, notice that $Av = \left( \begin{array}{c} < f, v > \\ p_1 v_1 \\ p_2 v_2 \\ \vdots \\ p_{n-1} v_{n-1} \end{array} \right) $

So by looking for $\lambda$, pose $Av = \lambda v$ and you get $n$ equations. Notice that $p_i v_i = \lambda v_{i+1} \ \forall i < n$, thus we can express all the $v_i$ as a function of $v_1$, defining $\tilde p_k $ as the product of the first $k$ elements of vector $p$ (and $\tilde p_0 = 1$ ), we have $$ \prod_{i=1}^k p_i \lambda^{-k+1} v_1 = v_k \Leftrightarrow \tilde p_k \lambda^{-k+1} v_1 = v_k \quad \forall k > 1 $$

With this the first equation is reduced to: \begin{eqnarray} <f,v> = & \lambda v_1 \Leftrightarrow \sum_{i=1}^n v_i f_i = \lambda v_1 \\ & v_1 \sum_{i=1}^n \tilde p_i \lambda^{-i+1} f_i = \lambda v_1 \\ & \Rightarrow \sum_{i=1}^n \frac{\tilde p_i f_i}{\lambda^{i}} = 1 \end{eqnarray}

If lambda is small the sum goes to infinity, if lambda is big the sum goes to 0. Hence by continuity there exists one positive lambda that solves this equation. Then notice that the function is estrictly monotonous with respect to lambda (each term is), thus you have unicity.

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  • $\begingroup$ Hi, what does the $< f,v >$ notation mean? $\endgroup$ – user2850514 Apr 14 '16 at 21:06
  • $\begingroup$ it's the scalar product, $\sum f_i v_i $. $\endgroup$ – pancho Apr 14 '16 at 21:07
  • $\begingroup$ @ Francisco Jose Romero Hinrichs This is strikingly close to the structure of the eigenvectors of a companion matrix (it is a particular case of the problem above) to a polynomial $P(x)$ (a companion matrix is diagonalized by the Vandermonde matrix $V(r_1,r_2,\cdots,r_n)$ where the $r_k$ are the roots of (P(x)$)en.wikipedia.org/wiki/Companion_matrix $\endgroup$ – Jean Marie Apr 14 '16 at 21:37
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$$|A-\lambda I| = \left| \begin{pmatrix} f_1 - \lambda & & \cdots & & f_n\\ p_1 & -\lambda & \cdots & \cdots & 0\\ 0&\ddots &&&\vdots\\ \vdots& &\ddots&&\vdots\\ 0 & & &p_{n-1} & -\lambda \end{pmatrix} \right|$$ now add $\frac{p_{n-1}}{\lambda}$ times of last column to $(n-1)$'th column then add $\frac{p_{n-2}}{\lambda}$ times of $(n-1)$'th column to$(n-2)$'th column and so on... till you get this: $$|A-\lambda I| = \left| \begin{pmatrix} D_{11} & & \cdots & & f_n\\ 0 & -\lambda & \cdots & \cdots & 0\\ 0&\ddots &&&\vdots\\ \vdots& &\ddots&&\vdots\\ 0 & & &0 & -\lambda \end{pmatrix} \right|$$ in which $D_{11} = f_1-\lambda + \frac{p_1 f_2}{\lambda} + \frac{p_1 p_2 f_3}{\lambda ^2} + ... + \frac{p_1 p_2 ... p_{n-1} f_n}{\lambda^{n-1}}$ so the characteristic polynomial is $$P(\lambda) = (- \lambda)^{n-1}\times D_{11} $$ for showing that there exist at least one positive solution to $P(\lambda) =0$ notice that $P(0)= (-1)^{n-1} p_1p_2...f_{n-1}$ and as $\lambda \to \infty$ we get $P(\lambda) \to (-1)^{n}\times \infty$ so we conclude that always there exist a positive solution.

to show there is exactly one positive root we proceed like below: $$P(\lambda)=\begin{cases} \lambda^n - (f_1 \lambda^{n-1}+...+p_1 p_2 ... p_{n-1} f_n), P(0)<0 \; ,P(\infty)>0 & \text{if $n$ is even} \\ -\lambda^n + (f_1 \lambda^{n-1}+...+p_1 p_2 ... p_{n-1} f_n),P(0)>0 \; ,P(\infty)<0 & \text{if $n$ is odd} \end{cases}$$

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  • $\begingroup$ I want to show there is exactly one positive solution, not atleast one. $\endgroup$ – user2850514 Apr 14 '16 at 21:06
  • $\begingroup$ Looking at the form of $P(\lambda)$, how to be assure that there is exactly one positive root. $\endgroup$ – Sry Apr 15 '16 at 5:04
  • $\begingroup$ I should prove after $P(\lambda)$ crosses zero it's derivative stays positive in case $n$ is even and similarly derivative of $P(\lambda)$ in $n$ odd case stays negative after crossing zero but it's not a trivial task and still searching for a way. $\endgroup$ – K.K.McDonald Apr 15 '16 at 5:07

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