Find values that k may take if $k=\det\left(A^3+B^3+C^3\right).\det\left(A+B+C\right)$

determinantlinear algebramatrices

Let $A,B,C$ be $n\times n$ matrices with real entries such that their product is pairwise commutative. Also $ABC=O_{n}$. If
$$k=\det\left(A^3+B^3+C^3\right).\det\left(A+B+C\right)$$
then find the value or the range of values that $k$ may take.

My Attempt

I tried $k=\left(\det(A+B+C)\right)^2\left(\det(A^2+B^2+C^2-AB-BC-CA)\right)$. but couldn't go further than this

Best Answer

Steps:

  1. Show that $\det (A^2 + B^2 + C^2 - AB - BC - CA ) \geq 0$. (See the linked problem.)
  2. Show that $ k \geq 0 $.
  3. For any $ k \geq 0$, find matrices $A, B, C$ that satisfy the condition.

Hints for the construction:

Diagonal matrices work.

The non-zero values are all the same, and equal to ...

The condition $ABC = 0$ means ...

If $ k > 0$, what does that imply about $A+B+C$?

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