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Math Help - determinant of block matrix

  1. #1
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    determinant of block matrix

    Let M be a nxn matrix and A is a mxm and C is kxk square matrix.
    If \[<br />
M =<br />
\left[ {\begin{array}{cc}<br />
 A & B  \\<br />
 O & C  \\<br />
 \end{array} } \right]<br />
\]
    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.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    That is not a trivial problem, requires a large proof based on linear maps. Is it homework?.

    Regards.

    Fernando Revilla
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  3. #3
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    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.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bubble86 View Post
    Let M be a nxn matrix and A is a mxm and C is kxk square matrix.
    If \[<br />
M =<br />
\left[ {\begin{array}{cc}<br />
 A & B  \\<br />
 O & C  \\<br />
 \end{array} } \right]<br />
\]
    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 View Post
    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?
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by bubble86 View Post
    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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