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:

<br /> \vec v&#039; = \left[\frac 1 n \frac 1 n \dots \frac 1 n \right]<br />

Then compute all of

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

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

<br /> W \vec z = k \vec z<br />

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)

<br /> \begin{align*}<br /> A &amp;= \begin{bmatrix} 1 &amp; 2 \\ 3 &amp; 4 \end{bmatrix}\\<br /> \vec v &amp; = \begin{bmatrix} \frac 1 2 &amp; \frac 1 2 \end{bmatrix}&#039;<br /> \end{align*}<br />

Then
<br /> A \vec v = \begin{bmatrix} 1 &amp; 2 \\ 3 &amp; 4 \end{bmatrix} \begin{bmatrix} 1/2 \\ 1/2 \end{bmatrix} = \begin{bmatrix} {(1+2)}/2 \\ {(3+4)}/2 \end{bmatrix}<br />

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".