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 \delta. Show that \delta 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 \delta, then the row vectors of A - \delta I 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 \delta, then the row vectors of A - \delta I all add up to (0,0...0)."?

If yes, then

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

What is the relation between the determinant of matrix A and the determinant of the matrix obtained by adding to one of the rows of matrix A 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(A - \delta I) = 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 \delta, it means that the rows (and the columns) of A add up to a constant of n*\delta, which means that the rows of A - \delta I add up to 0.
So the matrix formed by, say, adding to the first row of A - \delta I all the other rows will have the first row all 0's, and the same determinant as A - \delta I, which means that det(A - \delta I) = 0 and \delta is an eigenvalue of A