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

Answer:

(a) Yes, $λ_{A}+λ_{B}$

(b) Yes, $λ_{A}λ_{B}$

Step-by-step 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 $λ_{A}λ_{B}$