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 May 7th, 2013, 06:07 AM #1 Newbie   Joined: May 2013 Posts: 3 Thanks: 0 With characteristic polynomial show that A is invertible Show that square matrix [itex]A[/latex] is reversible if [itex]p_{A}(0)\neq 0[/latex] where [itex]p_{A}[/latex] is characteristic polynomial. Prove similar statement for minimal polynomial. The first part should be very similar to this (if not completely correct?): If [itex]A[/latex] is reversible than [itex]detA\neq 0[/latex]. Characteristic polynomial is by it's definition calculated as $p_{A}(\lambda )=det(A-\lambda I)$. for $\lambda=0$ than: $p_{A}(0)=det(A)$ but $A$ is reversible and therefore $detA\neq 0$ so $p_{A}(0)\neq 0$. End of part one (the other direction is similar). Part two: Minimal polynomial: For minimal polynomial $p(A)=0$ but I am not sure how to continue.. BTW, why would this differ from the first part at all? $p(A)$ could easily also be minimal? If in characteristic polynomial I insert $A$ instead of $\lambda$: $p(A)=det(A-AI)=0$ but I don't see how this proves that $A$ is reversible. Could somebody give ma hint what's the whole idea here? :/ Thanks for all the help! This is actually Home Work I have to do so if there is a separated topic for this kind of questions I sincerely apologize!
May 7th, 2013, 06:09 AM   #2
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Re: With characteristic polynomial show that A is invertible

SORRY for the mistakes in first post:
Quote:
 Originally Posted by skrat Show that square matrix $A$ is reversible if $p_{A}(0)\neq 0$ where $p_{A}$ is characteristic polynomial. Prove similar statement for minimal polynomial. The first part should be very similar to this (if not completely correct?): If $A$ is reversible than $detA\neq 0$. Characteristic polynomial is by it's definition calculated as $p_{A}(\lambda )=det(A-\lambda I)$. for $\lambda=0$ than: $p_{A}(0)=det(A)$ but $A$ is reversible and therefore $detA\neq 0$ so $p_{A}(0)\neq 0$. End of part one (the other direction is similar). Part two: Minimal polynomial: For minimal polynomial $p(A)=0$ but I am not sure how to continue.. BTW, why would this differ from the first part at all? $p(A)$ could easily also be minimal? If in characteristic polynomial I insert $A$ instead of $\lambda$: $p(A)=det(A-AI)=0$ but I don't see how this proves that $A$ is reversible. Could somebody give ma hint what's the whole idea here? :/ Thanks for all the help! This is actually Home Work I have to do so if there is a separated topic for this kind of questions I sincerely apologize!

 May 17th, 2013, 04:04 AM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: With characteristic polynomial show that A is invertible Don't insert "A" or $\lambda$ (that doesn't really make sense, $\lambda$ is a number, not a matrix), insert $\lambd= 0$. In that case $p(\lambda)= |A- \lambda I|$ becomes $p(0)= |A|$, exactly what you want.

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