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Would it be possible to have 3 square matrices (preferably 2x2 or 3x3) $A$, $B$ and $C$ such that:

  • $A\neq B \neq C$;
  • The product $A\cdot B\cdot C$ equals the Identity Matrix;
  • All 3 matrices are Involutory (They are their own inverses);
  • The products between them are commutative.

And would it be possible to generalize this for $n$ matrices instead of just 3?

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  • $\begingroup$ Take $A=B=C=I$. $\endgroup$ – Git Gud Aug 13 '14 at 18:30
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In 2D, if we need them to be distinct, then

$$ A = \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}, B = -A, C = -I $$

will work. I believe that, up to change of basis, this characterizes the situation.

For the positive direction (change of basis works to produce new examples), let $S$ be a change of basis matrix, let $A' = SAS^{-1}, B' = SBS^{-1}$, etc. Clearly these are simultaneously diagonalizable ($S^{-1}A'S = A$, which is diagonal, etc.) so they commute. The product is

$$ A'B'C' = SABCS^{-1} = SIS^{-1} = I $$

and $A'A' = SAAS^{-1} = SIS^{-1} = I$ etc.

For the negative direction, first note that it is required that $A^2 = I$ (and likewise for $B,C$). Let $\lambda$ be an eigenvalue of $A$, with eigenvector $\xi$. Then

$$ \xi = I\xi = A^2\xi = \lambda A\xi = \lambda^2 \xi $$

so $\lambda = \pm 1$.

If $A,B,C$ are a commuting family, then they are simultaneously diagonalizable, meaning $S^{-1}AS, S^{-1}BS, S^{-1}CS$ are all diagonal for some fixed $S$, and so it suffices to consider $A,B,C$ diagonal to begin with. By our observation about the eigenvalues, there are only 4 possible diagonalized matrices, the three given as an example above, and the identity. You can easily see that switching out any one of the matrices in the example for the identity will not work.

For 3D, you have more choices for the diagonalized matrices (you can put $\pm 1$ in each of the diagonal entries, so there are $8$ such diagonal $3\times 3$ matrices). You could, for instance, let

$$ A = D(-1,-1,1),\, B = D(1,1,-1),\, C = -I $$

where $D(x,y,z)$ means "the diagonal matrix with diagonal entries $x,y,z$", and then vary this a little bit for additional examples.

For $n$ matrices, you can use the same kind of analysis; you are still limited by the number of distinct diagonal matrices with only $\pm 1$ on the diagonal. For 2D, your 4th matrix can be the identity, and that's it. For 3D, the 4th matrix can be the identity. For 3D with $n = 5$, you could have $A = D(-1,1,1), B = D(1,-1,1), C = D(1,1,-1), D = -I, E = I$, etc.

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  • $\begingroup$ Forgot to mention, they all need to be different. Already made appropriate corrections to the question. $\endgroup$ – Disousa Aug 13 '14 at 18:46
  • $\begingroup$ Ok, I've addressed this. $\endgroup$ – BaronVT Aug 13 '14 at 19:37

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