Is the matrix permanent invariant under rotations?

Question

The determinant is, since it's just the product of the eigenvalues. What about the permanent?

Answer 1

When you say rotations, what do you mean.

Multiply matrix $A$ by a rotation matrix?

$R(\theta) = \pmatrix{\cos \theta &\sin\theta\\-\sin\theta&\cos\theta}$

In that case no.

e.g. let $A= \pmatrix{1&-1\\-1&1}$

$\text {perm.} A = 2$

$RA = \pmatrix{\cos\theta - \sin\theta & -\cos\theta + \sin\theta\\-\cos\theta - \sin\theta&\cos\theta + \sin\theta}$

$\text {perm.} RA = 2\cos 2\theta$

But if you mean moving the elements around the matrix like so...

$A = \pmatrix {a&b\\c&d}\\ A' = \pmatrix {b&d\\a&c}$

Then yes, as perm A is invariant under the permutation of rows or columns, and under transposition. And this transformation is invariant.

Answer 2

The permanent is a sum of products for all possible choices of $n$ elements of the matrix in different rows and columns. We get the same products if we rotate the matrix, hence the same permanent.