# 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.

### 1 Answer | Add Yours

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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