# determinant of block matrix

• Dec 1st 2010, 09:06 AM
bubble86
determinant of block matrix
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.
• Dec 1st 2010, 09:31 AM
FernandoRevilla
That is not a trivial problem, requires a large proof based on linear maps. Is it homework?.

Regards.

Fernando Revilla
• Dec 1st 2010, 10:52 AM
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.
• Dec 1st 2010, 03:31 PM
Drexel28
Quote:

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.

Quote:

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? (Thinking)
• Dec 1st 2010, 09:17 PM
FernandoRevilla
Quote:

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:

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

(ii) Use:

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

(iii) Use:

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

Regards.

Fernando Revilla