Matrix functional equation $ f \left( H ^ 2 \right) = \alpha f ( H ) $

Question

Could someone give me some hint about a possible method to find the function $ f $ which solve this equation: $$ f \left( H ^ 2 \right) = \alpha f ( H ) $$ where $ \alpha $ a constant with $ \alpha \in \mathbb C $ and: $$ H = \begin {bmatrix} a _ { 1 1 } & a _ { 1 2 } \\ a _ { 2 1 } & a _ { 2 2 } \end {bmatrix} $$ a matrix with complex entries? $ f $ should give a matrix as output, and the trivial solution $ f = 0 $ is to be rejected.

Answer 1

How about $$ f ( H ) = ( \log H ) ^ { \log _ 2 \alpha } \text ? $$ Well, there are problems with this definition! First of all, a logarithm of a matrix $ H $ can only be defined when $ H $ is invertible. Aside from that, it is not unique. But maybe we can make our peace with it by recalling that even simple functions like $ z \mapsto \sqrt z $ from $ \mathbb C $ to $ \mathbb C $ are multivalued. And lastly, how can we define a matrix to the power of a complex number (let's close our eyes to the fact that logarithm is a multivalued complex function, and $ \log _ 2 \alpha $ can take multiple values!)? That can be done like this: write your favorite matrix $ A $ in Jordan normal form, $ A = P J P ^ { - 1 } $, and for $ z \in \mathbb C $ let $ A ^ z = P J ^ z P ^ { - 1 } $, where: $$ J ^ z = \begin {cases} \begin {bmatrix} \lambda ^ z & 0 \\ 0 & \mu ^ z \end {bmatrix} & J = \begin {bmatrix} \lambda & 0 \\ 0 & \mu \end {bmatrix} \\ \begin {bmatrix} \lambda ^ z & z \lambda ^ { z - 1 } \\ 0 & \lambda ^ z \end {bmatrix} & J = \begin {bmatrix} \lambda & 1 \\ 0 & \lambda \end {bmatrix} \end {cases} $$ Taking $ f $ to be a multivalued function defined like above, verifying that it satisfies the desired functional equation is straightforward.