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Thread: A^2+I=o

  1. #1
    Senior Member pankaj's Avatar
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    A^2+I=o

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


    I tried it out like this in two ways:

    1.If given equation in $\displaystyle A$ is its characterstic equation then $\displaystyle det(A)=(-1)^n$(Product of eigenvalues) which gives the value of $\displaystyle det(A)=1$,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:
    $\displaystyle A^2=-I$


    $\displaystyle det(A^2)=det(-I)$


    $\displaystyle (det(A))^2=(-1)^n$


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


    $\displaystyle (det(A))^2=1$


    $\displaystyle (det(A))^2-1=0$


    $\displaystyle (det(A)-1)(det(A)+1)=0$


    $\displaystyle det(A)-1=0$ or $\displaystyle det(A)+1=0$


    $\displaystyle det(A)=1$ or $\displaystyle -1$


    How do I rule out the possibility of $\displaystyle det(A)=-1$
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  2. #2
    MHF Contributor
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    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).
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  3. #3
    MHF Contributor
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    Re: A^2+I=o

    Also spitballing, but maybe prove that for n=2, the determinant cannot be -1, then use induction?
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