Results 1 to 3 of 3

Math Help - A^2+I=o

  1. #1
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318

    A^2+I=o

    If A is a n-rowed square matrices with real entries and satisfies the equation A^2+I=O,then compute value of det(A).


    I tried it out like this in two ways:

    1.If given equation in A is its characterstic equation then det(A)=(-1)^n(Product of eigenvalues) which gives the value of det(A)=1,which is the answer.


    2.But is there some elementary explanation which does not use the concept of characterstic equation.I tried to do like this:
    A^2=-I


    det(A^2)=det(-I)


    (det(A))^2=(-1)^n


    Thus, n must be even since A is a square matrix with real entries


    (det(A))^2=1


    (det(A))^2-1=0


    (det(A)-1)(det(A)+1)=0


    det(A)-1=0 or det(A)+1=0


    det(A)=1 or -1


    How do I rule out the possibility of det(A)=-1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,696
    Thanks
    620

    Re: A^2+I=o

    Hey pankaj.

    Could you consider the properties of the inverse for this question? Since A^2 = -I then (A^2)^(-1) = (-I)^(-1). Since the matrix is invertible, then this might add an extra constraint which eliminates one value (this is just a brain-storm session trying to look for answers).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,904
    Thanks
    764

    Re: A^2+I=o

    Also spitballing, but maybe prove that for n=2, the determinant cannot be -1, then use induction?
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum