2
$\begingroup$

Consider $A=$$\left (\begin{array}{} 3 & 1\\ 2 & 4 \end{array} \right)$

Consider the set $E=\{a\in \Bbb R:\lim a^n A^n\text{exists and is different from zero}\}$

What will be $E$?

The eigen values of $A$ are $5,2$. If $v$ is an eigen vector of $A$ corresponding to eigen value $\lambda$ then $A^nv=\lambda^nv\implies \lambda^n$ is an eigen value of $A^n$ corresponding to eigen vector $v$.

Now $(a^nA^n) v=a^n(A^nv)=(a^n \lambda^n)v$

Also $\lim (a^n \lambda^n)$ exists and equals non-zero only when $|a\lambda|= 1\implies |a|=\dfrac{1}{\lambda}\implies |a|=\dfrac{1}{2},\dfrac{1}{5}$

So $E=\{\pm \dfrac{1}{5},\pm\dfrac{1}{2}\}$

Will yo please check my solution?Is it okay or what are the edits required?

  • 1
    $a=\pm1/2$ doesn't work, since if $v$ is an eigenvector of eigenvalue $5$, then $a^nA^nv=\pm2.5^nv$, so the entries of $a^nA^n$ must be unbounded as $n$ increases. – 2017-01-20
  • 0
    Will you please suggest how to do it then@Arthur – 2017-01-20
  • 0
    I am. You have shown that $E\subseteq \{\pm1/2,\pm1/5\}$, and it's well done. I'm just telling you how you could narrow it down even further. – 2017-01-20
  • 0
    @Arthur; in your first comment wont $a^nA^nv=\pm \dfrac{5^n}{2^n}v$? – 2017-01-20
  • 0
    Moreover I feel none of entries in the set $\{\pm \dfrac{1}{2},\pm \dfrac{1}{5} \}$ will work since if $a=\pm \dfrac{1}{5}$ then also if $v$ is an eigen vector corresponding to $2$ then $a^nA^nv=({\dfrac{2}{5}})^n\to 0$@Arthur – 2017-01-20
  • 0
    But that doesn't prove that all entries of $\frac1{\pm5^n}A^n$ go to zero, only that if the limit exists, then that limit is singular. As to your other comment, wouldn't $\frac{5^n}{2^n}$ be equal to $2.5^n$? – 2017-01-20
  • 0
    Yes ,you are right – 2017-01-20

2 Answers 2

1

Here is an outline of how I would solve the problem: $A$ has eigenvalues $5$ and $2$. That means that there is a basis of $\Bbb R^2$ using corresponding eigenvectors of $A$, say $v_5$ and $v_2$.

Take any vector $v\in \Bbb R^2$. It can be written as $c_5v_5+c_2v_2$ for some real numbers $c_5,c_2$. Now study $a^nA^nv$, and find for what values of $a$ both the following are satisfied:

  • $\lim_{n\to\infty}\|a^nA^nv\|$ is not infinite for any $v$
  • There is some $v$ for which $\lim_{n\to\infty}a^nA^nv$ exists and is non-zero
0

As Arthur pointed out my mistake, let me try to answer it again.

EDIT:Since eigenvalues are distinct so $A$ is diagonalizable and suppose $P$ be the corresponding non-singular modal matrix with $D=diag(2,5)$. Then

$A=P^{-1}DP\implies (aA)^n=P^{-1}(aD)^nP$

which exists and non-zero IFF $(aD)^n\neq O$ for which you need $a=+ \frac{1}{5}$.

$lim_{n\rightarrow\infty}(\frac{1}{5}D)^n=diag(0,1)$

$lim_{n\rightarrow\infty}(\frac{-1}{5}D)^n=diag(0,\lim_{n\rightarrow\infty}(-1)^n)$

doesn't exist uniquely.

$lim_{n\rightarrow\infty}(\pm\frac{1}{2}D)^n=$ doesn't exist.

  • 0
    @Arthur...thanks for pointing out my mistake. See my edit. – 2017-01-20
  • 0
    Much better. Still, you haven't disproven $a = -\frac15$. – 2017-01-20
  • 0
    @Arthur...thanks a lot again. – 2017-01-21
  • 0
    Sorry ,I did not get why $\lim (\pm \dfrac{1}{2}D)^n=\text{diag}(0,0)$ – 2017-01-22
  • 0
    $\lim (\pm \dfrac{1}{2}D)^n=\text{diag}(\pm 1,\pm \dfrac{5}{2})^n\neq \text{diag}(0,0)$ – 2017-01-22