2

For two square matrices $A$ and $B$ of the same size, we have the identity $$ \det\left[ \begin{matrix} A & B \\ B & A \end{matrix} \right] \ = \ \det\left( A + B \right) \det( A - B ) \ . $$ This is supposed to hold even when the matrices are non-commuting ie. $A B \neq BA$.

In the case that $AB=BA$, I can see how the above follows from the identity $$ \det\left[ \begin{matrix} A & B \\ C & D \end{matrix} \right] \ = \ \det(D) \det\left( A - B D^{-1} C \right) $$ by taking $(C,D) \to (B,A)$ and mutiplying the argument of the determinants together, leads to $$ \det\left[ \begin{matrix} A & B \\ C & D \end{matrix} \right] \ \to \ \det(A) \det\left( A - B A^{-1} B \right) = \det\left( A^2 - B A^{-1} B A \right) = \det\left( A^2 - B^2 \right) = \det\left( A - B \right)\det\left( A + B \right) $$ but the above proof relies heavily on $AB=BA$.

How do you see this identity is true in the case that $AB\neq BA$?

Rodrigo de Azevedo
  • 1
QuantumEyedea
  • 2,805
  • 1
    Try with $$ \left( \begin{array}{cc} I&I \ 0&I \ \end{array} \right) \left( \begin{array}{cc} A&B \ B&A \ \end{array} \right) \left( \begin{array}{cc} I& -I \ 0 &I \ \end{array} \right) = \left( \begin{array}{cc} A+B& 0 \ B& A-B \ \end{array} \right) $$ As in this question: https://math.stackexchange.com/questions/3322357/proof-of-the-matrix-identity-det-beginpmatrixab-ba-endpmatrix-detab?rq=1 – rarwoan Nov 12 '20 at 18:39

1 Answers1

8

By elementary operations over the columns and the rows of the matrix, you have$$\det\left[ \begin{matrix} A & B \\ B & A \end{matrix} \right] \ = \det\left[ \begin{matrix} A+B & B \\ B+A & A \end{matrix} \right] \ = \det\left[ \begin{matrix} A+B & B \\ 0 & A-B \end{matrix} \right] $$

which gives the formula.

TheSilverDoe
  • 29,720