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


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?

  • $\begingroup$ 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. $\endgroup$ – user 1987 Jul 4 '15 at 14:29
  • $\begingroup$ Try Laplace's formula and induction over $n_1$. $\endgroup$ – user251257 Jul 4 '15 at 14:31
  • $\begingroup$ @user251257 Sorry for the late response. Do you mean $$\sum_{j=1}^n (-1)^{j+k} a_{jk} \lvert A_{jk}\rvert$$ ? $\endgroup$ – qmd Jul 4 '15 at 15:47
  • $\begingroup$ yiep. you could also first try a 3x3 example with $n_1=1$ like @xhimi suggested. $\endgroup$ – user251257 Jul 4 '15 at 15:52
  • $\begingroup$ @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. $\endgroup$ – qmd Jul 4 '15 at 15:57

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.

  • $\begingroup$ 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 $AD$? $\endgroup$ – qmd Jul 4 '15 at 14:22
  • $\begingroup$ you don't need eigenvalues for that. just apply Laplace's formula. $\endgroup$ – user251257 Jul 4 '15 at 14:30
  • 1
    $\begingroup$ @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. $\endgroup$ – ajotatxe Jul 4 '15 at 14:41
  • $\begingroup$ @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\$ $\endgroup$ – qmd Jul 4 '15 at 18:32
  • $\begingroup$ @ajotatxe the last line should read: $$=E_{n_1} \lvert C \rvert \not= \lvert C \rvert$$ $\endgroup$ – qmd Jul 4 '15 at 18:38

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