Is this true:

Question

Is this true:

empty · Answer

$$[\det(A+A^\top)]^{2/n} + [\det(A-A^\top)]^{2/n} = 4 [\det(A)]^{2/n}$$

If A is a symmetric or skew symmetric matrix, this holds, but does it hold in general?

dan815 · Answer

so this expression where A and A^T are now general can be rewritten as the sum of sym and skesym matrices

empty · Answer

Proof for symmetric matrices that this hold (skew symmetric is not much different)
Since A is symmetric, $A=A^\top$

$$[\det(A+A^\top)]^{2/n} + [\det(A-A^\top)]^{2/n} = 4 [\det(A)]^{2/n}$$
$$[\det(A+A)]^{2/n} + [\det(A-A)]^{2/n} = 4 [\det(A)]^{2/n}$$
$$[\det(2A)]^{2/n} + [\det(0)]^{2/n} = 4 [\det(A)]^{2/n}$$
$$[\det(2I)]^{2/n} [\det(A)]^{2/n} = 4 [\det(A)]^{2/n}$$
$$[\det(2I)]^{2/n} = 4$$

empty · Answer

Well since no one seems to be too interested, the answer is simple. Despite being true for the two simple cases to check, it's not true in general, just calculate with any random 2x2 matrix that isn't symmetric or skew symmetric to provide a counter example.