Matrix Determinant Formula

Question

Problem. Let $A$ be a non-singular $n\times n$ matrix and let $\Gamma=[\Gamma_1\quad\Gamma_2]$ be an $n\times n$ orthogonal matrix where $\Gamma_1$ is $n\times n_1$, $\Gamma_2$ is $n\times n_2$ and $n=n_1+n_2$. Show that $$\det(\Gamma_1^TA\Gamma_1)=\det(A)\det(\Gamma_2^TA^{-1}\Gamma_2).$$

My Attempts. Here we make use of the property of orthogonal matrix: \begin{align} \det(A)=\det(\Gamma^TA\Gamma)=\det\left(\begin{bmatrix} \Gamma_1^T \\ \Gamma_2^T \end{bmatrix}A\begin{bmatrix} \Gamma_1 & \Gamma_2 \end{bmatrix}\right)=\det\left(\begin{bmatrix} \Gamma_1^TA\Gamma_1 & \Gamma_1^TA\Gamma_2 \\ \Gamma_2^TA\Gamma_1 & \Gamma_2^TA\Gamma_2 \end{bmatrix}\right). \end{align} Since $A$ is non-singular, $\Gamma_1^TA\Gamma_1$ is also non-singular. Thus, \begin{align} \det(A)=\det(\Gamma_1^TA\Gamma_1)\det\left(\Gamma_2^TA\Gamma_2-\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2\right). \end{align} If the formula we want to prove is true, we would have \begin{align} 1&=\det(\Gamma_2^TA^{-1}\Gamma_2)\cdot\det\left(\Gamma_2^TA\Gamma_2-\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2\right) \\ &=\det\left(\Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_2-\Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2\right). \end{align} Nonetheless, I have no idea how to simplify the terms in the parenthesis because I only have $\Gamma_1\Gamma_1^T+\Gamma_2\Gamma_2^T=I$. Hope anyone has good suggestions.

Answer 1

Let's define $B = \Gamma^T A \Gamma$ and take the notation: $$ \begin{aligned} \Gamma_{11} = \Gamma_1^TA\Gamma_1 \\ \Gamma_{12} = \Gamma_1^TA\Gamma_2 \\ \Gamma_{21} = \Gamma_1 A \Gamma_2^T \\ \Gamma_{22} = \Gamma_2^T A\Gamma_2 \end{aligned} $$ Let's denote the Shur complement of $\Gamma_{11}$ as $B/ \Gamma_{11} = \Gamma_{2}^{T} A \Gamma_{2}-\Gamma_{2}^{T} A \Gamma_{1}\left(\Gamma_{1}^{T} A \Gamma_{1}\right)^{-1} \Gamma_{1}^{T} A \Gamma_{2}$ which is the same as $B / \Gamma_{11} = \Gamma_{22} - \Gamma_{21} \Gamma_{11}^{-1} \Gamma_{12}$. Using this notation we get: $$ \begin{aligned} &\det(A) = \det(B) = \det(\Gamma_{11})\det(B / \Gamma_{11}) \\ &\det(\Gamma_1^T A \Gamma_1) = \det(A)\det((B / \Gamma_{11})^{-1}) \end{aligned} $$ We then compute $B^{-1}$: $$ B^{-1} = \left( \begin{matrix} \Gamma_{11}^{-1} + \Gamma_{11}^{-1} \Gamma_{12}(B / \Gamma_{11})^{-1}\Gamma_{21}\Gamma_{11}^{-1} & -\Gamma_{11}^{-1} \Gamma_{12}(B / \Gamma_{11})^{-1}\\ -(B / \Gamma_{11})^{-1} \Gamma_{21} \Gamma_{11}^{-1} & (B / \Gamma_{11} )^{-1} \end{matrix} \right) $$

We also have

$$ B^{-1} = \Gamma^T A^{-1} \Gamma = \left ( \begin{matrix} \Gamma_1^T A^{-1} \Gamma_1 & \Gamma_1^T A^{-1} \Gamma_2 \\ \Gamma_1 A^{-1} \Gamma_2^T & \Gamma_2^T A^{-1} \Gamma_2 \end{matrix} \right ) $$

By checking the dimensions we get: $$ \Gamma_2^T A^{-1} \Gamma_2 = (B / \Gamma_{11} )^{-1} $$

From where we get $\det(\Gamma_1^T A \Gamma_1) = \det(A)\det(\Gamma_2^T A^{-1} \Gamma_2 )$.

Answer 2

To simplify the terms, we indeed have to use the formula $\Gamma_2\Gamma_2^T=I-\Gamma_1\Gamma_1^T$. Here \begin{align*} \Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_2&=\Gamma_2^TA^{-1}(I-\Gamma_1\Gamma_1^T)A\Gamma_2 \\ &=\Gamma_2^TA^{-1}A\Gamma_2-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1A\Gamma_2 \\ &=I-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1A\Gamma_2 \end{align*} and \begin{align*} &\quad\ \Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2 \\ &=\Gamma_2^TA^{-1}(I-\Gamma_1\Gamma_1^T)A\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2 \\ &=\Gamma_2^TA^{-1}A\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2 \\ &=\color{red}{\Gamma_2^T\Gamma_1}(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2-\Gamma_2^TA^{-1}\Gamma_1\color{red}{\Gamma_1^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}}\Gamma_1^TA\Gamma_2 \\ &=0-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1^TA\Gamma_2=-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1^TA\Gamma_2. \end{align*} Thus, \begin{align*} &\quad\ \Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_2-\Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2 \\ &=I-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1A\Gamma_2-(-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1^TA\Gamma_2)=I. \end{align*} The formula thus follows.

P.S. If anyone has simpler solutions, please share it with me!! This calculation is too complicated.