Let $A,\ B,\ C,\ D \in \mathcal{M}_n(\mathbb{C})$. If $\operatorname{rank}\left( \begin{bmatrix} A &B \\ C &D \end{bmatrix}\right)=n$, prove that $\det(AD)=\det(BC)$.


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Assume first that the first $n$ columns are independent. Then the last $n$ columns are linear combinations of the first $n$. Therefore we have the equality

$$\begin{bmatrix} A \\ C \end{bmatrix} \cdot V = \begin{bmatrix} B \\ D \end{bmatrix}$$

so we're done.

Now, if the first $n$ columns are not linearly independent, both sides are $0$.

Obs: to understand what $V$ is: the first column of $V$ consists of the coefficients in the writing of the first column of $\begin{bmatrix} B \\ D \end{bmatrix}$ as a linear combination of the columns of $\begin{bmatrix} A \\ C \end{bmatrix}$.

  • $\begingroup$ Can you elaborate on what is $V$ ? $\endgroup$ – James S. Cook Jan 16 '15 at 1:06
  • 3
    $\begingroup$ @JamesS.Cook: note that $AV=B$ and $CV=D$ then compute $\det(AD)=\det(ACV)$ and $\det(BC)=\det(AVC)$ $\endgroup$ – robjohn Jan 16 '15 at 1:10
  • $\begingroup$ @James S. Cook: Just did, good call. $\endgroup$ – Orest Bucicovschi Jan 16 '15 at 1:10
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    $\begingroup$ @robjohn thanks. I should have seen this, that said, his answer is improved by your comment. $\endgroup$ – James S. Cook Jan 16 '15 at 2:39

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