Let A = [3 1; 1 2]. What is the maximum value of xᵀAx, where the maximum is…
2007
Let A = [3 1; 1 2]. What is the maximum value of xᵀAx, where the maximum is taken over all x that are unit eigenvectors of A?
- A.
5
- B.
(5 + √5)/2
- C.
3
- D.
(5 - √5)/2
Attempted by 4 students.
Show answer & explanation
Correct answer: B
For any unit eigenvector x of A with eigenvalue λ, we have
Ax = λx.
Therefore,
xᵀAx = xᵀ(λx) = λ(xᵀx) = λ,
since x is a unit vector.
So the required maximum is the largest eigenvalue of A.
For A = [3 1; 1 2],
det(A - λI) = (3 - λ)(2 - λ) - 1
= λ² - 5λ + 5.
Solving λ² - 5λ + 5 = 0 gives
λ = (5 ± √5)/2.
The maximum value is the larger eigenvalue, (5 + √5)/2.