Results 1 to 5 of 5

Math Help - matrixes

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    56

    matrixes

    I've got two questions which im struggling with:

    1. Let A and B be n\times n matrixes such that A^2=I, B^2=I and (AB)^2=I. Prove that AB=BA

    2. Let A be some 2\times2 matrix such that AX=XA for all real 2\times2 matrxes. Prove that A=\alphaI for \alpha \in R
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,306
    Thanks
    1282
    Quote Originally Posted by vuze88 View Post
    I've got two questions which im struggling with:

    1. Let A and B be n\times n matrixes such that A^2=I, B^2=I and (AB)^2=I. Prove that AB=BA
    Use A^2= I and B^2= I to show that (AB)(BA)= I. Then you have (AB)(BA)= (AB)(AB). Multiply both sides by (AB)^{-1} (after showing that AB is invertible, of course).

    2. Let A be some 2\times2 matrix such that AX=XA for all real 2\times2 matrxes. Prove that A=\alpha I for \alpha \in R
    Let X be \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}, and \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    56
    The word is "matrices" not "matrixes"
    Last edited by satx; March 4th 2010 at 05:56 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2009
    Posts
    56
    for the first one, how do i prove AB is invertible? do i need to prove anythin else other than the fact that its square?

    for the second one, how did you know to choose those 4 "matrices"
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,306
    Thanks
    1282
    A matrix is invertible if and only if its determinant is non-zero. If A had determinant 0, then det(A^2)= det(A)det(A)= 0. but A^2= I which has determinant 1, not 0. Therefore det(A) is not 0. Same thing for B. Then det(AB)= det(A)det(B) is non-zero.

    I chose those four matrices because they are the "standard basis" for the vector space of 2 by 2 matrices. Every such matrix can be written as a linear combination of them:
    \begin{barray}a & b \\ c & d\end{bmatrix}= a\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}+ b\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}+ c\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}+ d\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}
    so they "represent" all 2 by 2 matrices.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inverse of Matrixes
    Posted in the Business Math Forum
    Replies: 1
    Last Post: February 20th 2010, 11:18 PM
  2. Matrixes Arithmetic
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 16th 2010, 11:28 AM
  3. Matrixes
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 25th 2009, 11:58 PM
  4. Matrixes
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: June 11th 2008, 03:17 PM
  5. MATLAB Matrixes
    Posted in the Math Software Forum
    Replies: 1
    Last Post: February 26th 2008, 12:36 PM

Search Tags


/mathhelpforum @mathhelpforum