# matrix with eigenvalues on the unit circle

Suppose $$A$$ is an $$n\times n$$ matrix with complex entries such that there exists strictly positive constants $$c_1<1 so that $$c_1<\frac{\|A^Nx\|}{\|x\|} 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 $$\{v_1,\ldots,v_n\}$$ be the eigenvectors of $$A$$ corresponding to eigenvalues $$\lambda_1,\ldots,\lambda_n$$ then $$A^Nv=c_1\lambda_1^Nv_1+\ldots+c_n\lambda_n^Nv_n$$. However, in here I already assumed $$A$$ is diagonalizable. Any hints/ideas? thanks

• 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. Jul 22, 2015 at 4:23
• @JimmyK4542 my bad it should be $A^N v=c_1\lambda_1^Nv_1+...$
– nerd
Jul 22, 2015 at 4:37
• For part (b), have you been taught about the Jordan canonical form of a matrix? Jul 22, 2015 at 4:41
• Yeas but i m rusty at using it
– nerd
Jul 22, 2015 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 $n_i \ge 2$.

• 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?
– nerd
Jul 22, 2015 at 4:43
• Yes, the Jordan form is exactly what is needed. Jul 22, 2015 at 5:40
• @JimmyK4542 Thanks so much for that dedicated answer. Appreciated
– nerd
Jul 22, 2015 at 5:47