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Is there a name for this type of matrix ? The matrix is always $3\times 3$.

All the $9$ elements are integers.

The determinant of the matrix is $-1$.

Two of the three eigenvalues are reciprocals of one another, and are both positive real numbers. The third eigenvalue is always $-1 $.

Since none of the eigenvalues is ever zero, the matrix is always nonsingular , and always has an inverse.

The even higher powers of this matrix have determinant $+1 $, and the odd higher powers of this matrix have determinant $-1$.

The even higher powers of this matrix have three eigenvalues, one of those eigenvalues is $+1 $, and the other two eigenvalues are reciprocals of one another.

The odd higher powers of this matrix have three eigenvalues , one of those eigenvalues is $-1 $, and the other two eigenvalues are reciprocals of one another.

Amazingly, the three eigenvectors associated with the three eigenvalues of the matrix, seem to be the same for for all the higher powers of that matrix.

Is there a name for this given type of Matrix ?

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  • $\begingroup$ Why not call it a Derek matrix? $\endgroup$ – Angina Seng Sep 7 '18 at 19:31
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    $\begingroup$ The "amazing" fact about the eigenvectors you mention is actually true for all matrices. $\endgroup$ – Eric Wofsey Sep 7 '18 at 19:31
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Most of the things you mention occur for all matrices.

If $Av=\lambda v$, then $$ A^2v=A(Av)=A(\lambda v)=\lambda AV=\lambda ^2 v. $$ Continuing, you get that $A^nv=\lambda ^n v$.

In general, if $A$ is $n\times n$ and $\lambda_1,\ldots,\lambda_n$ are the eigenvalues (counting multiplicities) with corresponding eigenvectors $v_1,\ldots,v_n$, then $A^m$ has eigenvalues $\lambda_1^m,\ldots,\lambda_n^m$ with corresponding eigenvectors $v_1,\ldots,v_n$.

Now, going back to your concrete case, you have that $A$ is $3\times 3$. If you require that $-1$ is an eigenvalue, and that $\det A=-1$, then all the other properties you mention follow right away.

Finally, I don't think there's a specific name for matrices $A\in M_3(\mathbb Z)$ with $\det A=-1$ and $-1$ as an eigenvalue.

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