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Let's say that $\Phi, U_{11},U_{12},U_{21},U_{22}$ are $n\times n$ matrices with coefficients in $\mathbb{C}$. Consider the block matrix:

$$ M = \begin{bmatrix} U_{11}\Phi & U_{12}\Phi \\ U_{21}\Phi & U_{22}\Phi \end{bmatrix} $$

where $U_{ij}\Phi$ is the usual matrix product. Can anything be deduced about the determinant of this matrix in terms of the determinant of $\Phi$?

Chris
  • 678

1 Answers1

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Observe that

$M=\begin{pmatrix}U_{11}\Phi & U_{12}\Phi \\ U_{21}\Phi & U_{22}\Phi\end{pmatrix}=\begin{pmatrix}U_{11} & U_{12} \\ U_{21} & U_{22}\end{pmatrix}\begin{pmatrix}\Phi & 0_{n\times n} \\ 0_{n\times n} & \Phi\end{pmatrix}$.

Then by property $\det(AB)=\det(A)\det(B)$, the determinant of $M$ is

$\det(M)=\det\left(\begin{pmatrix}U_{11} & U_{12} \\ U_{21} & U_{22}\end{pmatrix}\right)\det\left(\begin{pmatrix}\Phi & 0_{n\times n} \\ 0_{n\times n} & \Phi\end{pmatrix}\right)=\det\left(\begin{pmatrix}U_{11} & U_{12} \\ U_{21} & U_{22}\end{pmatrix}\right)[\det(\Phi)]^2$,

where the last step comes from the fact that the determinant of a block diagonal matrix is the product of the determinants of the diagonal blocks of such matrix (see General expression for determinant of a block-diagonal matrix).