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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 $1$.

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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 $A=I$.

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  • $\begingroup$ 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? $\endgroup$ – darkl Mar 12 '16 at 22:21
  • $\begingroup$ Since you are using the Frobenius norm, we have $|A_{ij}| \le \|A\|_F$ for all $i,j$. $\endgroup$ – copper.hat Mar 12 '16 at 22:21
  • $\begingroup$ But why can one assume that the matrix's norm equals to its Jordan form's norm? $\endgroup$ – darkl Mar 12 '16 at 22:23

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