Let A=I₂-2 M Mᵀ, where M is a real matrix of order 2 × 1 such that the relation Mᵀ M=I₁ holds. If λ is a real number such that the relation A X=λ X holds for some non-zero real matrix X of order 2 × 1, then the sum of squares of all possible values of λ is equal to __________.

Solution

Maths sir · 4 steps

Square A first: since MᵀM = I₁, you get A² = I, and that pins every λ to ± 1.

1. Square the matrix. Expand A² = (I - 2MMᵀ)(I - 2MMᵀ). The last term contains M(MᵀM)Mᵀ, and MᵀM = I₁ is just the number 1.

   A² = I - 4MMᵀ + 4M(MᵀM)Mᵀ = I - 4MMᵀ + 4MMᵀ = I

2. Apply to X. From AX = λ X, applying A again gives A²X = λ² X. Since A² = I, this is X = λ² X. As X ≠ 0, λ² = 1, so λ = 1 or λ = -1.
3. Both values occur. Take X = M: AM = M - 2M(MᵀM) = -M, so λ = -1. Take any non-zero X with MᵀX = 0 (perpendicular to M): AX = X, so λ = 1.
4. So. The possible values are 1 and -1, and the sum of their squares is 1 + 1 = 2.

The common slip: 1. Keeping only λ = -1 (from X = M) gives 1. λ = 1 also works for any non-zero X perpendicular to M, so both values count.

The idea to keep. When a matrix is built from a unit vector like I - 2MMᵀ, square it first. If A² = I, every λ satisfies λ² = 1; then check which of ± 1 actually occur.

Answer from the NTA final answer key. Solution by PrepWiser.