Question

7) Let B be a matrix with a repeated zero eigenvalues. Then show that B2 = 0 (the 2 × 2 zero matrix). Use this to show: if A has a repeated eigenvalue λ0, then (A − λ0I) 2 = 0. (Hint: Use the fact that Bv = 0 for some nonzero vector v)

Answer 1

1. Let B =

a	b
c	d

where a,b,c,d are arbitrary real numbers. The characteristic polynomial of B is det(B-ʎI₂) or, (a-ʎ)(d-ʎ) -bc or, ʎ²-(a+d)+(ad-bc) .

Now, if 0 is a repeated eigenvalue of B, then its characteristic polynomial is (ʎ-0)(ʎ-0)= ʎ² so that a+d = 0 or, d = -a . Also, ad-bc = 0 or, a(-a)- bc = 0 or, bc = -a² so that c = - a²/b . Then B =

a	b
- a2/b	-a

so that B² =

0	0
0	0

Thus, B is the 2 × 2 zero matrix.

2. Now, let A be a 2 × 2 zero matrix with a repeated eigenvalue λ₀. Then there is a non-zero vector v (say) such that Av = λ₀v. Then (A − λ₀I₂)v =0 = 0.v . This means that 0 is a repeated eigenvalue of A − λ₀I₂. Therefore, as per part 1 above, ( A − λ₀I₂)² = 0.