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Math Help - Characteristic equation proof

  1. #1
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    Characteristic equation proof

    Hi

    Im confused on where to start in proving this question

    Let A = \left(\begin{array}{cc}a&b\\c&d\end{array}\right) where a, b, c and d are real numbers. Show that the characteristic equation of A is \lambda^2 - tr(A)\lambda + det(A) = 0
    where tr(A) = trace of A.

    Thanks
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  2. #2
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    [quote=chuckienz;312692]Hi

    Im confused on where to start in proving this question

    Let A = \left(\begin{array}{cc}a&b\\c&d\end{array}\right) where a, b, c and d are real numbers. Show that the characteristic equation of A is \lambda^2 - tr(A)\lambda + det(A) = 0
    where tr(A) = trace of A.


    Well lets start with what we know...

    \det (A)=ac-bd

    tr(a)=a+d

    to compute the char poly we evaluate the determinate of

    \left(\begin{array}{cc}a -\lambda &b\\c&d -\lambda \end{array}\right)=(a-\lambda)(d-\lambda)-cb

    Expanding we get

    ad -a\lambda -b\lambda +\lambda^2-cb

    \lambda^2-(a+b)\lambda +ad-cb=

    Now just sub we what we know from above and we are done
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