determinant of block matrix

**bubble86** · December 1st 2010, 09:06 AM

Let M be a nxn matrix and A is a mxm and C is kxk square matrix.
If $ 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.

**FernandoRevilla** · December 1st 2010, 09:31 AM

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

Regards.

Fernando Revilla

**bubble86** · December 1st 2010, 10:52 AM

yes it was a hw question for last week. is it possible to give me a hint in the direction of the proof using linear maps? it will be much appreciated.

**Drexel28** · December 1st 2010, 03:31 PM

Originally Posted by bubble86

Let M be a nxn matrix and A is a mxm and C is kxk square matrix.
If $ 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 $A,B$ are triangular since if $D(\cdot)$ deonts the diagonal entries

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

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

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

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

Thoughts?

**FernandoRevilla** · December 1st 2010, 09:17 PM

Originally Posted by bubble86

yes it was a hw question for last week. is it possible to give me a hint in the direction of the proof using linear maps? it will be much appreciated.

An easier alternative:

(i) Prove:

$\det \begin{bmatrix}{I}&{Q}\\{O}&{P}\end{bmatrix}=\det P$

(ii) Use:

$\begin{bmatrix}{A}&{C}\\{O}&{B}\end{bmatrix}=\begi n{bmatrix}{I_m}&{C}\\{O}&{B}\end{bmatrix}\begin{bm atrix}{A}&{O}\\{O}&{I_n}\end{bmatrix}$

(iii) Use:

$\det (DE)=\det(D)\det(E)$

Regards.

Fernando Revilla

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