Question

Suppose A²=A. On the left side A multiplies each columns of A.

1. Show that the column space contains the eigenvectors with λ=1.

2. Show that the nullspace contains the eigenvectors with λ=0.

3.From previous two answers, you may have found the dimension of the column space and null space. Is the matrix A diagonalizable?

Answer 1

A matrix A is idempotent if and only if all its eigenvalues are either 0 or 1.

1. If A²=A, then A²-A = 0 or, A(A-I) = 0. Now, if A ≠ I , then A -I ≠ 0 so that A-I contains at least one non-=zero column, say u. Then Au = 0 .

For any u ∈ E₀, A u = 0 so that u ∈ Null(A). Also, for any v ∈ null(A), Av =0 = 0v. Hence v ∈ E₀. Thus, E₀ = Null(A).

2. Let A = [c₁,c₂,…,c_n] where c₁,c₂,…,c_n are the columns of A. Now, since A²=A, we have A[c₁,c₂,…,c_n] = A = [c₁,c₂,…,c_n]. Then Ac_i = c_i for 1 ≤ i ≤ n.

This implies that 1 is an eigenvalue of A and E₁ = span { c₁,c₂,…,c_n } = Col(A).

3. Further, rank(A) = dim (Col(A)) = dim E₁ and nullity(A)= dim (Null(A)= dim E₀.

Now, by the dimension theorem, rank(A) + nullity(A) = n = dim E₁ + dim E₀.

This implies that A has n distinct and linearly independent eigenvectors so that A is diagonalizable.