Proving \delta as Eigenvalue of Matrix A with Constant Column Sum

[seang]

eigenvalue "show that"

Homework Statement

Let A be a matrix whose columns all add up to a fixed constant [tex]\delta[/tex]. Show that [tex]\delta[/tex] is an eigenvalue of A

Homework Equations

The Attempt at a Solution

My solution manual's hint is: If the columns of A each add up to a fixed constant [tex]\delta[/tex], then the row vectors of [tex]A - \delta I[/tex] all add up to (0,0...0).

I don't even understand the hint.

 

[antonantal]

First of all do you understand why "If the columns of A each add up to a fixed constant [tex]\delta[/tex], then the row vectors of [tex]A - \delta I[/tex] all add up to (0,0...0)."?

If yes, then

What is the equation that [tex]delta[/tex] has to fit in order to be an eigenvalue of [tex]A[/tex]?

What is the relation between the determinant of matrix [tex]A[/tex] and the determinant of the matrix obtained by adding to one of the rows of matrix [tex]A[/tex] all the others?

What is the determinant of a matrix that has a row of 0's?

 

[seang]

1. Ax = lambda*x ?

2. det(A)

3. 0.

Yes?

 

[antonantal]

1. Ax = lambda*x ?

Yes but more helpful det([tex]A - \delta I[/tex]) = 0

 

[seang]

so obviously I see the answer IF i can show that somehow I can get A to include a row of all zeroes.

 

[antonantal]

Ok. Because each column of A adds up to a fixed constant [tex]\delta[/tex], it means that the rows (and the columns) of A add up to a constant of n*[tex]\delta[/tex], which means that the rows of [tex]A - \delta I[/tex] add up to 0.
So the matrix formed by, say, adding to the first row of [tex]A - \delta I[/tex] all the other rows will have the first row all 0's, and the same determinant as [tex]A - \delta I[/tex], which means that det([tex]A - \delta I[/tex]) = 0 and [tex]\delta[/tex] is an eigenvalue of A