Suppose $A$ is an $n\times n$ matrix with complex entries such that there exists strictly positive constants $c_1<1<c_2$ so that $$c_1<\frac{\|A^Nx\|}{\|x\|}<c_2$$ for any integer $N\geq 1$ and $x\neq 0$

(a) Show that if $\lambda $ is an eigenvalue of $A$, then $|\lambda|=1$

(b) Show that $A$ is similar to a unitary matrix.

How would utilize the inequality? Let $\{v_1,...,v_n\}$ be the eigenvectors of $A$ corresponding to eigenvalues $\lambda_1,...,\lambda_n$ then $A^Nv=c_1\lambda_1^Nv_1+...+c_n\lambda_n^Nv_n$. However, in here I already assumed $A$ is diagonalizable. Any hints/ideas? thanks

  • $\begingroup$ How did you get $A^N = \lambda_1^Nv_1+ \cdots + \lambda_n^Nv_n$? The left side is an $n \times n$ matrix while the right side is an $n \times 1$ vector. $\endgroup$ – JimmyK4542 Jul 22 '15 at 4:23
  • $\begingroup$ @JimmyK4542 my bad it should be $A^N v=c_1\lambda_1^Nv_1+...$ $\endgroup$ – nerd Jul 22 '15 at 4:37
  • $\begingroup$ For part (b), have you been taught about the Jordan canonical form of a matrix? $\endgroup$ – JimmyK4542 Jul 22 '15 at 4:41
  • $\begingroup$ Yeas but i m rusty at using it $\endgroup$ – nerd Jul 22 '15 at 5:46

For part (a): Suppose $\lambda$ is an eigenvalue of $A$ and $v$ is a corresponding eigenvector.

Then, $A^Nv = \lambda^Nv$ for every integer $N \ge 1$. Hence, $\dfrac{\|A^Nv\|}{\|v\|} = \dfrac{\|\lambda^Nv\|}{\|v\|}= |\lambda|^N$.

Now, suppose $|\lambda| < 1$ or $|\lambda| > 1$. Can you show that $c_1 < \dfrac{\|A^Nx\|}{\|x\|} < c_2$ is violated for $x = v$ and some large integer $N$?

For part (b): Let $A = VJV^{-1}$ be the Jordan canonical form of $A$. From part (a), you already know that the diagonal entries of $J$ (the eigenvalues of $A$) have magnitude $1$. If $J$ is strictly diagonal, then $J^*J = JJ^* = I$ is trivial. So suppose $J$ has a Jordan block of size $2$ or larger and try to derive a contradiction. You can do this by constructing a vector $x \neq \vec{0}$ such that $\|A^Nx\|$ gets arbitrarily large as $N \to \infty$. Then, this vector $x$ will violate $c_1 < \dfrac{\|A^Nx\|}{\|x\|} < c_2$ for some large integer $N$.

For instance, in the specific case where $J = \begin{bmatrix}\lambda&1\\0&\lambda\end{bmatrix}$, set $x = Vy$, where $y = \begin{bmatrix}0 \\ 1\end{bmatrix}$. Then, we get $A^Nx = VJ^NV^{-1}Vy = VJ^Ny = V\begin{bmatrix}\lambda^N & N\lambda^{N-1} \\ 0& \lambda^N\end{bmatrix}\begin{bmatrix}0 \\ 1\end{bmatrix} = V\begin{bmatrix}N\lambda^{N-1}\\\lambda^N\end{bmatrix}$. Hence,

$\|A^Nx\|$ $= \left\|V\begin{bmatrix}N\lambda^{N-1}\\\lambda^N\end{bmatrix}\right\|$ $= \left\|\lambda^NV\begin{bmatrix}0\\1\end{bmatrix}+N\lambda^{N-1}V\begin{bmatrix}1\\0\end{bmatrix} \right\|$ $\ge N\left\|V\begin{bmatrix}1\\0\end{bmatrix}\right\| - \left\|V\begin{bmatrix}0\\1\end{bmatrix}\right\|$ $\to \infty$ as $N \to \infty$. (Note that $V\begin{bmatrix}1\\0\end{bmatrix} \neq 0$ since $V$ is invertible). Thus, $\|A^Nx\| \to \infty$ as $N \to \infty$.

I'll let you work out the general case where $J$ has several Jordan blocks with at least one that is of size $n_i \ge 2$.

  • $\begingroup$ I just figured out a), sorry I asked an easy question (having a bad day :( ). To show $A$ is similar to unitary matrix, do we use the Jordan form? $\endgroup$ – nerd Jul 22 '15 at 4:43
  • $\begingroup$ Yes, the Jordan form is exactly what is needed. $\endgroup$ – JimmyK4542 Jul 22 '15 at 5:40
  • $\begingroup$ @JimmyK4542 Thanks so much for that dedicated answer. Appreciated $\endgroup$ – nerd Jul 22 '15 at 5:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.