Consider an nxn matrix A with the property that the row sums

Consider an nxn matrix A with the property that the row sums all equal the same number s. Show that s is eigenvalue of A. (Hint:Find an eigenvector) Please explain well.

Solution

We note that s is an eigenvalue of A iff (A - sI) is singular (where I is the identity matrix). Note that if every column of A adds up to s, then when you subtract s from one entry, the sum of the resulting vector is zero. Thus, (A - sI) is a matrix such that each column vector adds to zero.

Claim: the set of vectors whose entries add up to zero forms a proper subspace of R^n of dimension n-1 (i.e. the set of such vectors is closed under linear combination).
I leave the proof of this claim to you.

We know that the image of a matrix is the span of its column vectors. Since all the column vectors are members of the above-mentioned subspace, the image of the transformation is contained in this subspace. Thus, the rank of our n x n matrix is less than n. Thus, (A - sI) is singular.

Thus, s is an eigenvalue of A.