Question

Suppose that v is an eigenvector of matrix A with eigenvalue λA, and it is also an eigenvector of matrix B with eigenvalue λB. (a) Show that v is an eigenvector of A + B and find its associated eigenvalue. (b) Show that v is an eigenvector of AB and find its associated eigenvalue.

Answer 1

(a) Yes,
`λ_(A)+λ_(B)`

(b) Yes,
`λ_(A)λ_(B)`

Explanation:

First, lets understand what are eigenvectors and eigenvalues?

Note: I am using the notation
`λ_(A)` to denote Lambda(A) sign.

`v` is an eigenvector of matrix A with eigenvalue
`λ_(A)`

`v` is also eigenvector of matrix B with eigenvalue
`λ_(B)`

So we can write this in equation form as

`Av=λ_(A)v`

So what does this equation say?

When you multiply any vector by A they do change their direction. any vector that is in the same direction as of
`Av`, then this
`v` is called the eigenvector of
`A`.
`Av` is
`λ_(A)` times the original
`v`. The number
`λ_(A)` is the eigenvalue of A.

`λ_(A)` this number is very important and tells us what is happening when we multiply
`Av`. Is it shrinking or expanding or reversed or something else?

It tells us everything we need to know!

Bonus:

By the way you can find out the eigenvalue of
`Av` by using the following equation:

`det(A-λI)=0`

where I is identity matrix of the size of same as A.

Now lets come to the solution!

(a) Show that
`v` is an eigenvector of
`A + B` and find its associated eigenvalue.

The eigenvalues of
`A` and
`B` are
`λ_(A)` and
`λ_(B)`, then

`(A+B)(v)=Av+Bv=(λ_(A))v + (λ_(B))v=(λ_(A)+λ_(B))(v)`

so,
`(A+B)(v)=(λ_(A)+λ_(B))(v)`

which means that
`v` is also an eigenvector of
`A+B` and the associated eigenvalues are
`λ_(A)+λ_(B)`

(b) Show that
`v` is an eigenvector of
`AB` and find its associated eigenvalue.

The eigenvalues of
`A` and
`B` are
`λ_(A)` and
`λ_(B)`, then

`(AB)(v)=A(Bv)=A(λ_(B))=λ_(B)(Av)=λ_(B)λ_(A)(v)=λ_(A)λ_(B)(v)`

so,

`(AB)(v)=λ_(A)λ_(B)(v)`

which means that
`v` is also an eigenvector of
`AB` and the associated eigenvalues are
`λ_(A)λ_(B)`