Originally Posted by

**bubble86** Let M be a nxn matrix and A is a mxm and C is kxk square matrix.

If $\displaystyle \[

M =

\left[ {\begin{array}{cc}

A & B \\

O & C \\

\end{array} } \right]

\]$

where O is made of all entries equal to zero and B is any matrix.

Prove det(M) = det(A) det(C)

the only way i can think of it is to expand about the first column but it seems way too tedious. and we only have learnt the definition of determinant and Cramer's rule.

Originally Posted by

**FernandoRevilla** That is not a trivial problem, requires a large proof based on linear maps. Is it homework?.

Regards.

Fernando Revilla I wonder...

This is trivial if $\displaystyle A,B$ are triangular since if $\displaystyle D(\cdot)$ deonts the diagonal entries

$\displaystyle \displaystyle \det(M)=\prod_{x\in D(M)}x=\prod_{x\in D(A)\cup D(C)}x=\prod_{x\in D(A)}x\cdot \prod_{x\in D(C)}x=\det(A)\det(C)$

Recall though that $\displaystyle \det:\text{Mat}_n(\mathbb(R)}\to\text{Mat}_n\left( \mathbb{R}\right)$ is continuous (where $\displaystyle \text{Mat}_n(\mathbb{R})$ is given the topology as a subspace of $\displaystyle \mathbb{R}^{n^2}$) and thus continuous on $\displaystyle \text{TriMat}_n\left(\mathbb{R}\right)=\left\{A\in \text{Mat}_n(\mathbb{R}):A\text{ is upper triangular}\right\}$. That said, I'm fairly positive (but not absolutely positive) that

$\displaystyle \displaystyle \mathcal{D}=\left\{M=\begin{pmatrix}A & B\\ 0 & C\end{pmatrix}:A\text{ and }C\text{ are triangular}\right\}$

is dense in $\displaystyle \text{TriMat}_n\left(\mathbb{R}\right)$.

Thoughts?