2

Let $A$ be a $n\times n$ matrix over $\mathbb{R}$ with positive entries. Assume $v$ is a generalized eigenvector of $A$ with $(A-\lambda I)^r v=0$, where $r\geq 2$ and $|\lambda|<1$. The unique dominant eigenvalue of $A$ is $\hat{\lambda}=1$. As part of intermediate steps of a homework solution, I need to show $\lim_{k\to \infty} A^k v=0$.

I know that if $v'$ is an ordinary eigenvector with $v'=(A-\lambda I)^{r-1} v$, then $A^k v'=\lambda^k v'$. Also, since $v$ is an generalized eigenvector, $(A^k-\lambda^k I)^r v=0$. However, the way to solve this problem is still unclear to me. Is there any hint or comment?

Also, if this statement is indeed not true, please let me know so that I can try another approach to my original homework question.

Update:

Can I do something like this (probably not):

$(A^k-\lambda^k I)^{r-1} (A^k-\lambda^k I) v=0$ implies

$(A^k-\lambda^k I)^{r-1} A^k v=(A^k-\lambda^k I)^{r-1} \lambda^k v$.

So, $A^k v=\lambda^k v$. Since $|\lambda|<1$, we get $\lim_{k\to \infty} A^k v=0$.

2 Answers2

0

Suppose for simplicity that $A$ is a single Jordan block. Then $e_r$ (the $r$th standard basis vector) is an $r$th-order generalized eigenvector.

By induction you can explicitly write down $A^k$, and its $r$th column is $A^k e_r$. The nonzero entries of this column are of the form $\binom{k}{l} \lambda^{k-l}$ for $l = 0, \ldots, r-1$. As $k \to \infty$, we have $|\binom{k}{l} \lambda^{k-1}| \le k^l |\lambda|^{k-l} \to 0$ because exponential growth is faster than polynomial growth. Thus $A^k e_r \to 0$.

For a general $A$, you can use its Jordan form to reduce the problem to the situation described above.

I am not implying that the Jordan form is necessary to solve this problem; I am using it just for convenience. I think you can solve the problem directly by using the definition of generalized eigenvalue to write $A^k v$ as a linear combination of $v, Av, \ldots, A^{r-1} v$, for any $k \ge r$, and show that the coefficients of this linear combination tend to $0$ as $k \to \infty$. I suspect the computation will be equivalent to the "computing the $k$th power of a Jordan block" result I was using above.

angryavian
  • 93,534
  • Thanks! Your answer is correct only if all eigenvalues of $A$ have magnitudes less than $1$. But indeed, $A$ has a unique dominant eigenvalue equals to $1$ in my homework question. Sorry, I forgot to mention it in my original post. –  Apr 15 '18 at 01:29
  • @ChunjingGU I think the only eigenvalue that matters is the eigenvalue of your particular generalized eigenvector $v$. I think that as long as that eigenvalue $\lambda$ is less than $1$ in magnitude, the above will hold, even if other eigenvalues are larger than $1$. – angryavian Apr 15 '18 at 01:52
  • No. If the dominant eigenvalue $1$ is at the entry $J_{11}$, where $J$ is the jordan form of $A$, then $J^k$ will not vanish as $k\to\infty$, since $J_{11}^k=1$ as $k\to\infty$. –  Apr 15 '18 at 02:10
  • @ChunjingGU We are not talking about $A^k$ vanishing, we are only talking about $A^k v$ vanishing for a particular $v$. In terms of the Jordan form $J$ of $A$, you are asking for a particular column of $J$ vanishing. For this, the only eigenvalue that matters is the eigenvalue of the Jordan block that the column intersects. – angryavian Apr 15 '18 at 02:36
  • Now I understand a little bit. Can we first orthonormalize the basis of generalized eigenvectors, then argue that $A^k v''$ is a vanishing column? Here, $v''$ is the corresponding orthonormal eigenvector corresponding to $v$. The problem is that, if $v\neq e_r$, then the first entry of $J^k v$ would be the first entry of $v$, which can well be nonzero. I still assume $J_{11}=1$. –  Apr 15 '18 at 03:16
0

Well, if $v\in Ker(A-\lambda I)^r$, then there are several invariant subspaces $W_i$ such that $\oplus W_i$ contains $v$. And the restriction of $A$ on $W_i$ can be represented by a Jordan block, say $J_s(\lambda)$, once choosing basis properly. This is basically the theory of Jordan normal form. Now every entry of $J_s(\lambda)^k$ tends to $0$ as you can easily verify. So $A^kv$ tends to $0$.

Y. Hu
  • 417
  • I am still uncomfortable with the unique dominant eigenvalue $1$. Can it make $J^k$ not tend to $0$? –  Apr 15 '18 at 06:07
  • @ChunjingGU Generally, $A^k$ won't tend to $0$. The point is that $A^kv$ lies in an invariant subspace, on which the restriction of $A^k$ tends to $0$. – Y. Hu Apr 15 '18 at 06:21
  • Sorry, but what is the invariant subspace that $A^k v$ lies in? And why the restriction of $A^k$ tends to $0$? –  Apr 15 '18 at 06:28
  • Now I understand what you are saying. Thanks a lot! –  Apr 15 '18 at 06:54