Is 2 an Eigenvalue of the Matrix Product AB?

[sayan2009]

please solve this eigen value problem

A nd B are matrices of order n*n.now it is given that sum of each row of A is 2 nd that of B is 1...then show that 2 is an eige value of the product matrix AB

 

[statdad]

Let $$\vec v$$ be an $$n \times 1$$ vector like this:

$$\vec v' = \left[\frac 1 n \frac 1 n \dots \frac 1 n \right]$$

Then compute all of

$$\begin{align*}<br /> & A \vec v \\<br /> & B \vec v\\<br /> & (AB) \vec v<br /> \end{align*}$$

and remember that for any matrix $$W$$ and vector $$\vec z$$, if there is a scalar $$k$$ such that

$$W \vec z = k \vec z$$

then $$k$$ is an eigenvalue of the matrix $$W$$.

 

[sayan2009]

how to compute (AB)v

 

[statdad]

Here is a small example (note: the rows in this matrix do not sum to either 1 or 2, as doing that would be solving the problem for you. However, precisely the same steps work)

$$\begin{align*}<br /> A &= \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\\<br /> \vec v & = \begin{bmatrix} \frac 1 2 & \frac 1 2 \end{bmatrix}'<br /> \end{align*}$$

Then
$$A \vec v = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1/2 \\ 1/2 \end{bmatrix} = \begin{bmatrix} {(1+2)}/2 \\ {(3+4)}/2 \end{bmatrix}$$

so in this case, and in every case, the product $$A \vec v$$ has as its entries the means of the rows of $$A$$.

 

[sayan2009]

but why should i try to get mean here?i mean i can take v (transpose) as[1 1 1 ... 1](n times)
why r u taking [1/n 1/n ... 1/n]?

 

[Dick]

but why should i try to get mean here?i mean i can take v (transpose) as[1 1 1 ... 1](n times)
why r u taking [1/n 1/n ... 1/n]?

Go ahead. Just use [1,1,1...]. (Not that there's anything wrong with using [1/n,1/n,...], you'll get the same result in the end).

 

[statdad]

Just try the multiplication (or use Dick's suggestion) and notice how the result compares to the vector $$\vec v$$.

Remember that if $$A \vec v = k \vec b$$ then $$k$$ is an eigenvalue of the matrix $$A$$.

 

[sayan2009]

so the solution is using v=[1 1 1 1 ... 1]
we can easily get Av=2v & Bv=1v
then(AB)v=A(Bv)=A(1v)=Av=2v
so 2 is an eigen value of AB...
is this right solution??

 

[Dick]

so the solution is using v=[1 1 1 1 ... 1]
we can easily get Av=2v & Bv=1v
then(AB)v=A(Bv)=A(1v)=Av=2v
so 2 is an eigen value of AB...
is this right solution??

(AB)v=2v. That sure looks like it says 2 is an eigenvalue to me.

 

[sayan2009]

is that solution correct man??

 

[Dick]

Do you have any doubts?? You don't need me to approve your solution. If you believe in it go for it.

 

[HallsofIvy]

Just try the multiplication (or use Dick's suggestion) and notice how the result compares to the vector $$\vec v$$.

Remember that if $$A \vec v = k \vec b$$ then $$k$$ is an eigenvalue of the matrix $$A$$.

Surely you meant $$A\vec v= k \vec v$$?
Which is the definition of "eigenvalue".