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

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

where $I_{n_1} \in \Bbb R^{n_1 \times n_1}$ is the identity matrix, $B \in \Bbb R^{n_1 \times n_2}$, $O \in \Bbb R^{n_2 \times n_1}$ is the zero matrix, $C\in \Bbb R^{n_2 \times n_2}$

I want to show that $\det(M)=\det(C)$

Are there some special rules when dealing with matrices inside a matrix? Can I just compute the determinant like this:

$$\det(M)=\begin{vmatrix}I_{n_1}&B\\O& C\end{vmatrix}=I_{n_1}C-BO=I_{n_1}C \space ?$$

However $I_{n_1}C$ is not defined because they have different dimensions ($n_1 \times n_1 \space \text{and} \space n_2 \times n_2$). What am I doing wrong here?

Rodrigo de Azevedo
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qmd
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  • In your case you could just do a row expansion. Take a $3\times 3$ example to see how it works. In general (when $B=0$) use the fact that the determinant is multiplicative. – user 1987 Jul 04 '15 at 14:29
  • Try Laplace's formula and induction over $n_1$. – user251257 Jul 04 '15 at 14:31
  • @user251257 Sorry for the late response. Do you mean $$\sum_{j=1}^n (-1)^{j+k} a_{jk} \lvert A_{jk}\rvert$$ ? – qmd Jul 04 '15 at 15:47
  • yiep. you could also first try a 3x3 example with $n_1=1$ like @xhimi suggested. – user251257 Jul 04 '15 at 15:52
  • @user251257 So could't I just write: $$ \sum_{j=1}^2 (-1)^{j+2} m_{j2} \lvert M_{j2} \rvert=-m_{12} \lvert M_{12} \rvert+m_{22} \lvert M_{22}\rvert= -B \lvert M_{12}\rvert+C \lvert C_{22}\rvert$$

    However, I don't know what $\lvert C_{22}\rvert$ and $\lvert M_{12} \rvert$ are.

     – qmd Jul 04 '15 at 15:57
  • Oh, you should expand the first column, not the $k$-th row. – user251257 Jul 04 '15 at 16:03
  • @user251257 I always thought it doesn't matter which column/row I choose to expland from. – qmd Jul 04 '15 at 16:11
  • you expand the column/row with many zeros to avoid extensive calculation :D – user251257 Jul 04 '15 at 16:12
  • @user251257 That makes sense :). Expanding the first column I get:

    $$E_{n_1}\lvert M_{11}\rvert-O \lvert M_{21} \rvert=E_{n_1}\lvert M_{11}\rvert$$

    However, I haver the same problem as before. What is $\lvert M_{11} \rvert$

     – qmd Jul 04 '15 at 16:16
  • @user251257 Would it be just the determinant of the remaining entries when you cross out the 1st row and 1st column so $det(C)$? – qmd Jul 04 '15 at 16:27
  • @SuH exactly :D – user251257 Jul 04 '15 at 16:30
  • @user251257 But then my answer would be $E_{n_1} \det(C)$ which $\not= \det(C)$ or am I missing something here? – qmd Jul 04 '15 at 16:34
  • how does $E_{n_1}$ look like? – user251257 Jul 04 '15 at 16:38
  • @user251257 $$E_{n_1}=\begin{pmatrix} 1&0& \cdots &0\ 0&1& \cdots & 0 \ \vdots& \vdots& \ddots&\vdots\0&0&\cdots&1 \end{pmatrix}$$? – qmd Jul 04 '15 at 16:54
  • @SuH for $n_1=1$ it is just 1. the determinant of the identity is always 1. – user251257 Jul 04 '15 at 17:23
  • @user251257 I don't get it. Didn't we just compute the determinant with laplace and as far as I understand the answer was : $$E_{n_1}\lvert M_{11}\rvert$$ and NOT

    $$\lvert E_{n_1} \rvert \lvert M_{11}\rvert$$

     – qmd Jul 04 '15 at 18:00
  • @SuH: Sorry, i was confused. You are right. You $\det(M) = E_{n_1}\det(M_{11})$ for the case $n_1=1$. Now, here $E_{n_1} = 1$ and $M_{11} = C$ as you removed the first row and first column – user251257 Jul 05 '15 at 01:52

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If you write a matrix so: $$M=\begin{pmatrix}A&B\\C&D\end{pmatrix}$$ you can't say, in general, that $$\det M=\det A\det D-\det B\det C$$ Just note that even $A$ and $D$ are square, $B$ and $C$ needn't be.

But if $B$ or $C$ are zero, then we have $$\det M=\det A\det D$$

To show it, note that if you consider the entries of $M$ lying in an algebraically closed field, it is clear that the eigenvalues of $M$ are the ones of $A$ and $D$ together.

ajotatxe
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  • Thanks! I am a bit confused. Suppose I have some matrix A. $$A=\begin{pmatrix}a&b\c&d\end{pmatrix}$$ then I always thought the determinant is $ac-bd$ and not $det(a)det(c)-det(b)det(d).$ Why is the determinant of matrix $M$ $det A det D$ and not $AD$? – qmd Jul 04 '15 at 14:22
  • you don't need eigenvalues for that. just apply Laplace's formula. – user251257 Jul 04 '15 at 14:30
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    @SuH $A$ and $D$ are matrices and the determinant is a scalar. Moreover, if $A$ and $D$ are square matrices of different sizes, can not even be multiplied. – ajotatxe Jul 04 '15 at 14:41
  • @ajotatxe I have another question that just came up. As another user suggested I tried calculating the determinant by usinge Laplace's forumla. Howerver I end up with a wrong answer. Can you maybe see where I went wrong? $\det(M)=\sum_{j=1}^{n} (-1)^{i+j} m_{ij} \lvert M_{ij} \rvert$. Expanding the first column yields

    $\det(M)=\sum_{j=1}^{2} (-1)^{i+1} m_{i1} \lvert M_{i1} \rvert$

    $=E_{n_1} \lvert M_{11} \rvert-O \lvert M_{21}\rvert=E_{n_1} \lvert M_{11}\rvert$

    $=E_{n_1} \lver C \rvert\ \not = \lvert C \rvert$

     – qmd Jul 04 '15 at 18:32
  • @ajotatxe the last line should read:

    $$=E_{n_1} \lvert C \rvert \not= \lvert C \rvert$$

     – qmd Jul 04 '15 at 18:38