Results 1 to 3 of 3

Thread: Linear Algebra: Determinant question

  1. #1
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6

    Linear Algebra: Determinant question

    Hello,

    Problem:

    If A,B,C,D are matrices such that C and D commute and D is invertible, then


    $\displaystyle \text{det}\begin{pmatrix} A & B \\ C & D \end{pmatrix} = \text{det}(AD - BC)$

    There is a hint that says " multiply on the right by $\displaystyle \begin{pmatrix} 1 & 0 \\ X & 1 \end{pmatrix}$".

    My effort:

    So I observed that:

    $\displaystyle \text{det}\begin{pmatrix} A & B \\ C & D \end{pmatrix} = $ $\displaystyle \text{det}\begin{pmatrix} A & B \\ C & D \end{pmatrix}\cdot\text{det}\begin{pmatrix} 1 & 0 \\ X & 1 \end{pmatrix} = \text{det}\begin{pmatrix} A+BX & B \\ C+DX & D \end{pmatrix}$

    If we choose $\displaystyle X = -D^{-1}C$, then $\displaystyle \text{det}\begin{pmatrix} A+BX & B \\ C+DX & D \end{pmatrix} = \text{det}\begin{pmatrix} A- BD^{-1}C & B \\ 0 & D \end{pmatrix}$

    Now since C and D commute, $\displaystyle D^{-1}C = CD^{-1}$. Thus

    $\displaystyle \text{det}\begin{pmatrix} A-BD^{-1}C & B \\ 0 & D \end{pmatrix} = \text{det}\begin{pmatrix} A-BCD^{-1} & B \\ 0 & D \end{pmatrix} $$\displaystyle = \text{det}(A-BCD^{-1})\text{det}(D) = \text{det}(AD - BC)$

    It looked like a convincing proof until I realised that I have used the hypothesis in the penultimate step. I have also used it while setting $\displaystyle \text{det}\begin{pmatrix} 1 & 0 \\ X & 1 \end{pmatrix} = 1$.

    So how can I fix those steps? Or do we need a different direction?

    Thank you,
    Srikanth
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2006
    Posts
    244
    Quote Originally Posted by Isomorphism View Post
    I have also used it while setting $\displaystyle \text{det}\begin{pmatrix} 1 & 0 \\ X & 1 \end{pmatrix} = 1$.

    So how can I fix those steps? Or do we need a different direction?

    Thank you,
    Srikanth
    The matrix (assume its square):

    $\displaystyle \begin{pmatrix} I_n & 0_{m,n} \\ X & I_m \end{pmatrix}$

    is lower triangular, and so its determinant is the product of the elements on the diagonal, but these are all 1 so the determinant is 1 without having to assume what was to be proven.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2006
    Posts
    244
    Assume $\displaystyle A,\ B,\ C,\ D$ are all square and multiplicativly conformant (I believe that is implied by the conditions in the problem)

    $\displaystyle \det\begin{pmatrix} A-BCD^{-1} & B \\ 0 & D \end{pmatrix} =\det \left[\begin{pmatrix} A-BCD^{-1} & B \\ 0 & D \end{pmatrix}\begin{pmatrix} I & 0 \\ 0 & D^{-1} \end{pmatrix}\right] \det(D)$ $\displaystyle
    =\text{det}\begin{pmatrix} A-BCD^{-1} & BD^{-1} \\ 0 & I \end{pmatrix} \det(D)
    $

    expanding the first determinant in the last expression, from the bottom right gives:

    $\displaystyle \text{det}\begin{pmatrix} A-BCD^{-1} & B \\ 0 & D \end{pmatrix} =\det(A-BCD^{-1})\det(D)$

    etc...
    Last edited by Constatine11; Dec 25th 2008 at 02:31 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear Algebra: Linear Independence question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 3rd 2011, 05:28 AM
  2. Linear Algebra Question - Linear Operators
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Mar 3rd 2009, 04:33 AM
  3. linear algebra question..
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Dec 15th 2008, 11:23 PM
  4. Determinant - Linear Algebra
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 23rd 2007, 04:46 PM
  5. Replies: 3
    Last Post: Oct 9th 2007, 08:28 AM

Search Tags


/mathhelpforum @mathhelpforum