Proof Expla.: $lambda$ is an eigenvalue of $Ain M_ntimes n(F)$ iff $det(A-lambda I)=0$?

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I don't understand how Theorem 2.5 works in the following proof



theorem5.2



And my idea of this proof is that: If $large(A-lambda I_n)$ is invertible then $v$ has to be $mathbf0,$ which contradicts the assumption that ["]there exists a nonzero vector $v$[..."]. Is this correct?



Here is theorem 2.5



theorem2.5




linear-algebra eigenvalues-eigenvectors linear-transformations proof-explanation




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    Yes, you are correct.
    – Fimpellizieri
    Jul 18 at 16:40










  • @Fimpellizieri: Thank you sir but I also want to know how theorem 2.5 applies here.
    – Nong
    Jul 18 at 16:41














up vote
1
down vote

favorite












I don't understand how Theorem 2.5 works in the following proof



theorem5.2



And my idea of this proof is that: If $large(A-lambda I_n)$ is invertible then $v$ has to be $mathbf0,$ which contradicts the assumption that ["]there exists a nonzero vector $v$[..."]. Is this correct?



Here is theorem 2.5



theorem2.5




linear-algebra eigenvalues-eigenvectors linear-transformations proof-explanation




share|cite|improve this question

















  • 1




    Yes, you are correct.
    – Fimpellizieri
    Jul 18 at 16:40










  • @Fimpellizieri: Thank you sir but I also want to know how theorem 2.5 applies here.
    – Nong
    Jul 18 at 16:41












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I don't understand how Theorem 2.5 works in the following proof



theorem5.2



And my idea of this proof is that: If $large(A-lambda I_n)$ is invertible then $v$ has to be $mathbf0,$ which contradicts the assumption that ["]there exists a nonzero vector $v$[..."]. Is this correct?



Here is theorem 2.5



theorem2.5




linear-algebra eigenvalues-eigenvectors linear-transformations proof-explanation




share|cite|improve this question













I don't understand how Theorem 2.5 works in the following proof



theorem5.2



And my idea of this proof is that: If $large(A-lambda I_n)$ is invertible then $v$ has to be $mathbf0,$ which contradicts the assumption that ["]there exists a nonzero vector $v$[..."]. Is this correct?



Here is theorem 2.5



theorem2.5





linear-algebra eigenvalues-eigenvectors linear-transformations proof-explanation





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 18 at 16:49









José Carlos Santos

114k1698177




114k1698177









asked Jul 18 at 16:39









Nong

1,1471520




1,1471520







  • 1




    Yes, you are correct.
    – Fimpellizieri
    Jul 18 at 16:40










  • @Fimpellizieri: Thank you sir but I also want to know how theorem 2.5 applies here.
    – Nong
    Jul 18 at 16:41












  • 1




    Yes, you are correct.
    – Fimpellizieri
    Jul 18 at 16:40










  • @Fimpellizieri: Thank you sir but I also want to know how theorem 2.5 applies here.
    – Nong
    Jul 18 at 16:41







1




1




Yes, you are correct.
– Fimpellizieri
Jul 18 at 16:40




Yes, you are correct.
– Fimpellizieri
Jul 18 at 16:40












@Fimpellizieri: Thank you sir but I also want to know how theorem 2.5 applies here.
– Nong
Jul 18 at 16:41




@Fimpellizieri: Thank you sir but I also want to know how theorem 2.5 applies here.
– Nong
Jul 18 at 16:41










1 Answer
1






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oldest

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1
down vote



accepted










The idea is:beginaligntextexists vneq0text such that Av=lambda v&ifftextexists vneq0text such that (A-lambdaoperatornameId)v=0\&iff A-lambdaoperatornameIdtext is not one-to-one\&iff A-lambdaoperatornameIdtext is not invertibleendalignand it is in this last equivalence that theorem 2.5 is used, since it says that a linear map between vector spaces of the same (finite) dimension is one-to-one if and only if it is onto and that therefore it is one-to-one if and only if it is invertible.






share|cite|improve this answer





















  • Oh so it's the existence of $vnot=0$ causes the "not one-to-one" happens?
    – Nong
    Jul 18 at 16:51






  • 1




    Yes, of course, since we always have $(A-lambdaoperatornameId)0=0$.
    – José Carlos Santos
    Jul 18 at 16:52










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










The idea is:beginaligntextexists vneq0text such that Av=lambda v&ifftextexists vneq0text such that (A-lambdaoperatornameId)v=0\&iff A-lambdaoperatornameIdtext is not one-to-one\&iff A-lambdaoperatornameIdtext is not invertibleendalignand it is in this last equivalence that theorem 2.5 is used, since it says that a linear map between vector spaces of the same (finite) dimension is one-to-one if and only if it is onto and that therefore it is one-to-one if and only if it is invertible.






share|cite|improve this answer





















  • Oh so it's the existence of $vnot=0$ causes the "not one-to-one" happens?
    – Nong
    Jul 18 at 16:51






  • 1




    Yes, of course, since we always have $(A-lambdaoperatornameId)0=0$.
    – José Carlos Santos
    Jul 18 at 16:52














up vote
1
down vote



accepted










The idea is:beginaligntextexists vneq0text such that Av=lambda v&ifftextexists vneq0text such that (A-lambdaoperatornameId)v=0\&iff A-lambdaoperatornameIdtext is not one-to-one\&iff A-lambdaoperatornameIdtext is not invertibleendalignand it is in this last equivalence that theorem 2.5 is used, since it says that a linear map between vector spaces of the same (finite) dimension is one-to-one if and only if it is onto and that therefore it is one-to-one if and only if it is invertible.






share|cite|improve this answer





















  • Oh so it's the existence of $vnot=0$ causes the "not one-to-one" happens?
    – Nong
    Jul 18 at 16:51






  • 1




    Yes, of course, since we always have $(A-lambdaoperatornameId)0=0$.
    – José Carlos Santos
    Jul 18 at 16:52












up vote
1
down vote



accepted







up vote
1
down vote



accepted






The idea is:beginaligntextexists vneq0text such that Av=lambda v&ifftextexists vneq0text such that (A-lambdaoperatornameId)v=0\&iff A-lambdaoperatornameIdtext is not one-to-one\&iff A-lambdaoperatornameIdtext is not invertibleendalignand it is in this last equivalence that theorem 2.5 is used, since it says that a linear map between vector spaces of the same (finite) dimension is one-to-one if and only if it is onto and that therefore it is one-to-one if and only if it is invertible.






share|cite|improve this answer













The idea is:beginaligntextexists vneq0text such that Av=lambda v&ifftextexists vneq0text such that (A-lambdaoperatornameId)v=0\&iff A-lambdaoperatornameIdtext is not one-to-one\&iff A-lambdaoperatornameIdtext is not invertibleendalignand it is in this last equivalence that theorem 2.5 is used, since it says that a linear map between vector spaces of the same (finite) dimension is one-to-one if and only if it is onto and that therefore it is one-to-one if and only if it is invertible.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 18 at 16:49









José Carlos Santos

114k1698177




114k1698177











  • Oh so it's the existence of $vnot=0$ causes the "not one-to-one" happens?
    – Nong
    Jul 18 at 16:51






  • 1




    Yes, of course, since we always have $(A-lambdaoperatornameId)0=0$.
    – José Carlos Santos
    Jul 18 at 16:52
















  • Oh so it's the existence of $vnot=0$ causes the "not one-to-one" happens?
    – Nong
    Jul 18 at 16:51






  • 1




    Yes, of course, since we always have $(A-lambdaoperatornameId)0=0$.
    – José Carlos Santos
    Jul 18 at 16:52















Oh so it's the existence of $vnot=0$ causes the "not one-to-one" happens?
– Nong
Jul 18 at 16:51




Oh so it's the existence of $vnot=0$ causes the "not one-to-one" happens?
– Nong
Jul 18 at 16:51




1




1




Yes, of course, since we always have $(A-lambdaoperatornameId)0=0$.
– José Carlos Santos
Jul 18 at 16:52




Yes, of course, since we always have $(A-lambdaoperatornameId)0=0$.
– José Carlos Santos
Jul 18 at 16:52












 

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