# If $\det(A^2)= \det(A)$, then $A^2 \in \{A, A^{-1}, -A, -A^{-1}\}$

Is the following statement always true?

Let $A \in GL_n( \mathbb Z)$. If $\det(A^2)= \det(A)$, then $A^2 \in \{A, A^{-1}, -A, -A^{-1}\}$ .

Note: $GL_n(\mathbb Z)$ is defined as the set of all invertible $n \times n$ matrices over $\mathbb Z$.

I think that the statement is true because $det(A^2)=det(A)$ is true if $A = E_n$ and so $\det(A^2)=\det(A) \equiv \det(A) \cdot \det(A)=\det(A) \equiv 1 \cdot 1 = 1$ and thus $A^2 \in \{A, A^{-1}, -A, -A^{-1}\}$

Question: Is that guess correct?

The question is related to that post.

• The correct equation would be $det(A)\left[det(A)-1\right]=0$. The matrix $A$ can be any with zero determinant. – HBR Jul 4 '17 at 9:05
• No it cannot be with zero determinant because $A \in GL_n(\mathbb{Z})$, and therefore invertible – AsafHaas Jul 4 '17 at 9:12
• Plenty of counterexamples. For example $$A=\pmatrix{2&3\cr3&5\cr},\quad A^2=\pmatrix{13&21\cr21&34\cr}$$ both have determinant $=1$, but the conclusion is false because $$A^{-1}=\pmatrix{5&-3\cr-3&2\cr}.$$ You get something similar with consecutive Fibonacci numbers of your choice (other than that if you have a starting index of wrong parity, then $\det A=-1$). – Jyrki Lahtonen Jul 4 '17 at 10:08

The condition $\det(A)=\det(A^2)$ is equivalent to $\det(A)=1$, but that doesn't mean that $A^2$ should be equal to $\pm A$ or $\pm A^{-1}$.
For example, take $$A=\begin{pmatrix}1&1\\0&1\end{pmatrix}.$$ Then $$A^2=\begin{pmatrix}1&2\\0&1\end{pmatrix}\quad \text{and}\quad A^{-1}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}$$ so that $\det (A)=1=\det(A^2)$, but $A^2\notin \left\{A,A^{-1},-A,-A^{-1}\right\}$.
$|AB| = |A||B|$ for square matrices $A$ and $B$ of the same order. So $|A^2| = |A|^2$. Hence $|A^2| = |A|$ is equivalent to $|A| = 0$ or $|A| = 1$. This we can conclude.