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?
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.