If `v` is an eigenvector of `A` with corresponding eigenvalue `lambda` and `c` is a scalar, show that `v` is an eigenvector of `A-cI` with corresponding eigenvalue `lambda-c.` I'm not sure where to...
If `v` is an eigenvector of `A` with corresponding eigenvalue `lambda` and `c` is a scalar, show that `v` is an eigenvector of `A-cI` with corresponding eigenvalue `lambda-c.`
I'm not sure where to start here.
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Expert Answers
calendarEducator since 2012
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Just start from what you're given and the definitions. Since `v` is an eigenvector of `A` with eigenvalue `lambda` , we have
`Av=lambda v.`
For `v` to be an eigenvector of `A-cI` with eigenvalue `lambda-c,` this would mean that
`(A-cI)v=(lambda-c)v.` This isn't too hard to show:
`(A-cI)v=Av-cIv=lambda v-cv=(lambda-c)v,`
where the first equality is true by linearity and the second is true by the definition of `I` and because we're told that ```Av=lambda v.`
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