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Say $A$ is a complex $n\times n$ matrix. $A$ is invertible, and moreover, supposer that the set $\{||A^n||:n=1,2,...\}$ is bounded. Prove that $A$ is diagonalizable.

My attempt was to do the following. Say $J$ is the Jordan Canonical form of $A$. Since norm is invariant under similar matrices, we have that $J^n$ is bounded as well. Let $B_1,..,B_k$ be the Jordan blocks appearing in $J$. Assume for sake of contradiction that there is a $j$ such that $B_j$ has size bigger than $1$. Then

$B_j=\lambda I +S$, where $S$ is the matrix with $1$'s in the entries above the diagonal and $0$ elsewhere.

I also know that $\lambda\neq 0$, but I dont know how to finish the problem. Any thoughts?

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what do you mean, precisely, by $\parallel A^n \parallel ?$ Also, you should not use $n$ for both the number of rows and columns of $A$ and as the exponent. –  Will Jagy Sep 21 '12 at 4:35
    
You should be able to work out what powers of $\lambda I+S$ look like (you could just calculate a few, and look for patterns), and see if you can deduce that the entries increase without bound. –  Gerry Myerson Sep 21 '12 at 5:12
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You can compute $B_j^n$ explicitly; you can take the size to be $2$ for starters. You will either find that its norm is always unbounded, or a get counterexample to what you have to prove. It seems to me the latter is actually the case, when $|\lambda|<1$. –  Marc van Leeuwen Sep 21 '12 at 5:15
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2 Answers

up vote 3 down vote accepted

This is false as stated.

Note that $$ \left( \begin{array}{rr} \frac{1}{b} & 0 \\ 0 & 1 \end{array} \right) \cdot \left( \begin{array}{rr} a & b \\ 0 & a \end{array} \right) \cdot \left( \begin{array}{rr} b & 0 \\ 0 &1 \end{array} \right) = \left( \begin{array}{rr} a & 1 \\ 0 & a \end{array} \right). $$ So that shows you how to find the Jordan form of such a 2 by 2 matrix, as soon as $b \neq 0.$

Next, take your $$ A \; = \; \left( \begin{array}{rr} \lambda & 1 \\ 0 & \lambda \end{array} \right). $$

We calculate $$ A^n \; = \; \left( \begin{array}{rr} \lambda^n & n \, \lambda^{n-1} \\ 0 & \lambda^n \end{array} \right). $$

In turn, the Jordan form of $A^n$ is $$ J_n \; = \; \left( \begin{array}{rr} \lambda^n & 1 \\ 0 & \lambda^n \end{array} \right). $$

So, if $| \lambda | \leq 1,$ the norm of $A^n$ stays bounded. Depending on the norm you mean, it may require $| \lambda | < 1$ for the norm to stay bounded.

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Sorry. I read the assigment again, it says $\{|A^k|: k \in \mathbb{Z}\}$, so the bound is also on the negative powers of $A$, and I know how to do it in this case. Thanks. –  Daniel Montealegre Sep 21 '12 at 5:42
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Take $J = \begin{bmatrix} \frac{1}{2} && 1 \\ 0 && \frac{1}{2} \end{bmatrix}$. Then a slightly tedious computation gives $J^n = \begin{bmatrix} \frac{1}{2^n} && n\frac{1}{2^{n-1}} \\ 0 && \frac{1}{2^n} \end{bmatrix}$.

Since $J^n \to 0$, it is clear that the sequence is bounded, but $J$ is not diagonalizable.

Answer to modified question (purely for my own interest):

Since $A^k v = \lambda^k v$ for an eigenvector $v$, it is clear that if $A^k$ is bounded for all integers $k$, then $|\lambda| = 1$, ie, all eigenvalues lie on the unit circle. Suppose $A$ is not diagonalizable, then $A$ has a Jordon normal form with a Jordan chain at least 2 long, ie, there exists unit vectors $v_1,v_2$ such that $Av_1 = \lambda v_1$ and $A v_2 = v_1 + \lambda v_2$. Another tedious computation shows that $A^k v_2 = k \lambda^{k-1} v_1 + \lambda^k v_2$, which contradicts $A^k$ being bounded since $\|A^k\| \geq \| A^k v_2 \| \geq k - 1$. Hence all Jordan chains are of length 1, which implies that $A$ is diagonalizable.

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For a less tedious computation, look at $\lambda^{-1} A$ which has the same norms and diagonalizability as $A$. Then the relation is just $Av_2 = v_1 + v_2$, which is easy to iterate. –  Erick Wong Sep 21 '12 at 6:37
    
@ErickWong: Neat idea! –  copper.hat Sep 21 '12 at 6:44
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