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This is a follow up question to this.

I want to solve the following problem:

Let $n \in \Bbb N \space \text{/{0}} \space \text{and} \space n_1,n_2 \in \Bbb N \space \text{such that} \space n_1+n_2=n$

$$M=\begin{pmatrix}E_{n_1}&B\\O&C\end{pmatrix}$$

where $E_{n_1} \in \Bbb R^{n_1 \times n_1}=\text{identity matrix}, \space B \in \Bbb R^{n_1 \times n_2}, \space O \in \Bbb R^{n_2 \times n_1}= \text{zero matrix}, C\in \Bbb R^{n_2 \times n_2}$

Show that $\det(M)=\det(C)$

On Wikipedia it says:

Suppose A, B, C, and D are matrices of dimension n × n, n × m, m × n, and m × m, respectively. Then,

$\det \begin{pmatrix}A&0\\C&D\end{pmatrix}=\det A \det{C}=\det \begin{pmatrix}A&B\\0&D\end{pmatrix} $

Since I don't want to just say "it says so on wikipedia" I would like to know how to derive the formula. As someone in my previous question suggested, I tried using Laplace's formula but I get a slightly wrong answer. Here is what I did:

$$\det(M) =\sum_{j=1}^n (-1)^{i+j} \space m_{ij} \space \lvert M_{ij} \rvert$$

Expanding the first column:

$$\begin{align} \det(M) & = \sum_{j=1}^2 (-1)^{i+1} \space m_{i1} \space \lvert M_{i1} \rvert \\ & = m_{11} \lvert M_{11} \rvert-m_{21} \lvert M_{22} \rvert \\ & = E_{n_1} \lvert M_{11}\rvert-O \lvert M_{21} \rvert \\ & = E_{n_1} \lvert C\rvert \\ & \not= \det(C) \end{align}$$

Can anybody tell me what I am doing wrong? Do I need to use a different approach like Leibniz's formula?

$$\det(A)=\sum_{\sigma \in S_n}\text{sgn}(\sigma)\prod_{i=1}^{n} a_{\sigma(i),i}$$

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You're applying Laplace's formula as if it applied to blocks, but it applies to elements. (This is also reflected in the notation; usually lowercase letters are used for matrix entries and uppercase letters for matrices.)

If Laplace's formula held for blocks (with the determinant of the block where you're using the block itself), then what you're trying to prove would be just a special case of it. However, while it works in this special case, for arbitrary blocks as in

$$ \pmatrix{A&B\\C&D} $$

with all four blocks non-zero, it doesn't work, because there are cross-terms involving entries in all four blocks whereas the expansion would contain only terms with entries from two blocks at a time.

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  • $\begingroup$ Thanks! That would explain why I kept getting nonsense. Is there some "easy" way to show $\det(M)=\det(C)$ without using complicated proofs? Maybe I approached this problem in the wrong way and there is a really straightforward way to do this. $\endgroup$ – qmd Jul 4 '15 at 20:02
  • $\begingroup$ @SuH: Have you checked out the question I linked to in the comment under the question? $\endgroup$ – joriki Jul 4 '15 at 20:04
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    $\begingroup$ Oops. Sorry! I missed that. That is exactly what I was looking for! Thanks! $\endgroup$ – qmd Jul 4 '15 at 20:06
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It can be easily proved with the generalisation of Laplace's formula to the expansion of a determinant along a group of columns – here along the $n_1$-first columns.

For any minor made up of elements in the first $n_1$ columns and in a subset $H$ of $n_1$ rows, one associates the complementary minor, made up of elements in the remaining $n_2$ columns and rows. Now $\det M$ is the sum of the products of these minors, with a relevant $+$ or $-$ sign. If the matrix is block upper-triangular, only the first minor of order $n_1$ ($\det A$) can be non-zero, and its complementary minor is $\det B$.

You can find details in Bourbaki, Algebra, Ch. III, Tensor algebras, Exterior algebras, Symmetric algebras, § 8 Determinants.

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  • $\begingroup$ Thanks for the reference! Will take me a bit to digest. $\endgroup$ – qmd Jul 4 '15 at 20:02
  • $\begingroup$ I admit Bourbaki is not for self-study… I'm sorry I couldn't find another reference. $\endgroup$ – Bernard Jul 4 '15 at 20:07
  • $\begingroup$ I hope that wasn't serious ;) I am extremely grateful for every response I get even though the answer I get might be hard to understand. After all, this is basically personal tutoring for free. $\endgroup$ – qmd Jul 4 '15 at 20:10

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