Jordan Block of a complex matrix, with $A^4=I$

The following statement is false or true:

If $$A \in M(n, \mathbb{C})$$ is a matrix with complex entries of order $$n$$ such that $$A^4=I$$ then $$\begin{pmatrix} i & 1\\ 0 & i \end{pmatrix}$$ Can be a Jordan Block of $$A$$.

I believe this statement is false, but I could not formalize the demonstration.

Observe that

$$A^4=I\implies (A-I)(A+I)(A^2+I)=(A-I)(A+I)(A-iI)(A+iI)$$

Thus, over $$\;\Bbb C\;$$ , the matrix's minimal polynomial decomposes as a product of different linear factors and is thus diagonalizable, which means it cannot have a Jordan Block as the one you wrote.

• Yes, except you're not saying what you mean. In fact $(A-I)(A+I)(A^2+I)=(A-I)(A+I)(A-iI)(A+iI)$ regardless of whether $A^4=I$; also a polynomial cannot be diagonalizable. – David C. Ullrich Jun 23 at 23:08
• I had thought of that, but I thought more arguments were needed. Thank you @DonAntonio – Mrcrg Jun 24 at 0:33
• @DavidC.Ullrich 1) That decomposition you mention is due precisely to the fact that $\;A^4=I\;$ . It is not independent of it but resulting from it, (2) Yes, my grammar made it sound like the polynomial is the one which is (or is not) diagonalizable, but making a little effort, like the OP did, I think it can be understood correctly...in particular because immediately before that I wrote " the matrix's polynomial..." . The OP had to round up and finish the argument: a square matrix is diagonalizable iff its minimal polynomial can be written as a product of different linear factors. – DonAntonio Jun 24 at 7:17
• "That decomposition you mention is due precisely to the fact that $A^4=I$. It is not independent of it but resulting from it," What??? Can you give an example of $A$ where $(A-I)(A+I)(A^2+I)\ne(A-I)(A+I)(A-iI)(A+iI)$? No. What's due to $A^4=I$ is the fact that $(A-I)(A+I)(A-iI)(A+iI)=0$. Which of course is what matters, but which you didn't mention explicitly. – David C. Ullrich Jun 24 at 13:17
• " I think it can be understood correctly": Yes of course. I didn't say it couldn't be understood correctly, I said that you didn't say what you meant. – David C. Ullrich Jun 24 at 13:22

An alternative approach: we have $$B = iI + N$$, where $$I$$ denotes the identity matrix and we have $$N = \pmatrix{0&1\\0&0}.$$ Notably, $$iI$$ and $$N$$ commute (so the binomial theorem applies) and $$N^2 = 0$$. Thus, we have $$B^4 = (iI + N)^4 = (iI)^4 + 4(iI)^3N + 6(0) + 4(0) + 0 = I -4i N\\ = \pmatrix{1&-4i\\0&1}.$$ Thus, if $$B$$ is a Jordan block of the Jordan form $$J$$ of $$A$$, then $$J^4$$ will have $$B^4$$ as a diagonal submatrix. So, we cannot have $$J^4 = I$$. It follows that $$A^4 \neq I$$.