Question

If every row of A adds up to zero, and x is a column vector of ones, i.e. [1, 1,…, 1]^T, what is Ax? How do you know that det A=0?

Answer 1

Since we want to prove that det(A) = 0, it is apparent that A is a square matrix.

Let A be a nxn matrix as under:

a11	a12	…	a1n
a21	a22		a2n
…	…	…	…
an1	an2	…	ann

Since every row of A adds up to zero, hence (a_i1 +a_i2+…+a_in) = 0 for 1 ≤ i ≤ n.

Now, if X is the n-vector (1, 1,…, 1)^T , then AX is a n x 1 column vector i.e. a n-vector. Further, the ith entry in AX is a_i1 +a_i2+…+a_in = 0. Since this is true for every 1 ≤ i ≤ n, hence AX is a zero n-vector.

Also, since (a_i1 +a_i2+…+a_in) = 0 for 1 ≤ i ≤ n, hence a_in = -( a_i1 +a_i2+…+a_in-1) for 1 ≤ i ≤ n.

Now, if the columns of A are denoted by c₁,c₂,…c_n, then this implies that c_n = -(c₁+c₂+…+c_n-1) i.e. the nth column of A is a linear combination of its preceding (n-1) columns.

Hence det(A) = 0.