# Eigenvalue show that

1. Dec 6, 2006

### seang

eigenvalue "show that"

1. The problem statement, all variables and given/known data
Let A be a matrix whose columns all add up to a fixed constant $$\delta$$. Show that $$\delta$$ is an eigenvalue of A

2. Relevant equations

3. 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.

2. Dec 6, 2006

### 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?

Last edited: Dec 6, 2006
3. Dec 6, 2006

### seang

1. Ax = lambda*x ?

2. det(A)

3. 0.

Yes?

4. Dec 6, 2006

### antonantal

Yes but more helpful det($$A - \delta I$$) = 0

5. Dec 6, 2006

### seang

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

6. Dec 6, 2006

### 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