0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The correct equation would be $det(A)\left[det(A)-1\right]=0$. The matrix $A$ can be any with zero determinant. $\endgroup$ – HBR Jul 4 '17 at 9:05
  • 4
    $\begingroup$ No it cannot be with zero determinant because $A \in GL_n(\mathbb{Z})$, and therefore invertible $\endgroup$ – AsafHaas Jul 4 '17 at 9:12
  • 2
    $\begingroup$ 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 $\det A=-1$). $\endgroup$ – Jyrki Lahtonen Jul 4 '17 at 10:08
11
$\begingroup$

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 $A^2\notin \left\{A,A^{-1},-A,-A^{-1}\right\}$.

$\endgroup$
0
$\begingroup$

$|AB| = |A||B|$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.