# Bounded distance of powers of a matrix from identity (Frobenius norm) implies that the matrix is identity

Let $A=\left( { a_{i,j} }\right)$ be an invertible $n \times n$ matrix over $\mathbb{C}$. Denote $\|A\|^2 := \sum_{i,j} |a_{i,j}|^2$ the Frobenius norm.

Suppose that for all integer $k$ we have $\|A^k - I\| < \frac{1}{2}$.

Does this imply that $A$ is the identity matrix?

I tried to use the Jordan form of $A$ and sub-multiplicity. I think I know how to show that all eigenvalues of $A$ have absolute value $1$.

Yes.

First notice that $A^k$ is bounded.

If you write $A$ in the Jordan form, you can see that all eigenvalues must satisfy $|\lambda^k-1| \le {1 \over 2}$. In particular, we must have $|\lambda| =1$, and hence $\lambda =1$.

Now consider a Jordan block $J$ of $A$, I claim that it must have size one.

If not, then the $(1,2)$ element of $J^k$ is $k$, which is a contradiction.

Hence $A$ consists of Jordan blocks of size one with eigenvalue one and so $A=I$.

• We get that $| \lambda^k - 1 | < \frac{1}{2}$ since that norm is sermi-multiplicative and therefore is not less than the absolute value of the largest eigenvalue, right? – darkl Mar 12 '16 at 22:21
• Since you are using the Frobenius norm, we have $|A_{ij}| \le \|A\|_F$ for all $i,j$. – copper.hat Mar 12 '16 at 22:21
• But why can one assume that the matrix's norm equals to its Jordan form's norm? – darkl Mar 12 '16 at 22:23