The Determinant of a Matrix of Matrices

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To find the determinant of the matrix structured as \(\begin{pmatrix}\mathbf{L}&\mathbf{A}^T\\\mathbf{A}&\mathbf{0}\end{pmatrix}\), where L is an n-by-n matrix and A is an m-by-n matrix, the discussion highlights that terms involving elements of L will contain zeroes, leading to confusion about the determinant's calculation. A counterexample is provided, demonstrating that the determinant does indeed include contributions from elements of L, contradicting earlier assumptions. The participants agree that there is no straightforward formula for calculating this determinant. The complexity of the determinant arises from the interactions between the matrices involved. Overall, the discussion emphasizes the need for careful analysis when determining the determinant of such composite matrices.
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Hi,

Suppose we have the following matrix:

\begin{center}\begin{pmatrix}\mathbf{L}&\mathbf{A}^T\\\mathbf{A}&\mathbf{0}\end{pmatrix}\end{center}

where L is n-by-n matrix, A is m-by-n matrix. How to find the determinant of this square matrix?

Thanks in advance
 
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det(-AA')
 
Some Pig said:
det(-AA')

Thank you for replying, but can you elaborate more, please?
 
Terms containing elements of L will contains zeroes,
so terms only containing elements of A and A'.
The negative sign indicates orders of the elements.
 
Some Pig said:
Terms containing elements of L will contains zeroes...

Why is that?
 
Some Pig said:
det(-AA')

This is wrong.

Counterexample:

L = \begin{pmatrix} 2 &amp; 0 \cr 0 &amp; 2 \end{pmatrix} \quad <br /> A = \begin{pmatrix} 1 \cr 0\end{pmatrix}

Working out the 3x3 determinant shows the mistake in the "proof" that it was right. The only non-zero product in the determinant does contain an element of L.

I don't think there is any "simple" formula for this.
 
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