Question

Consider a 3 × 3 real symmetric matrix S such that two of its eigenvalues are a ≠ 0, b ≠ 0 with respective eigen vectors \(\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right],\left[ {\begin{array}{*{20}{c}} {{y_1}}\\ {{y_2}}\\ {{y_3}} \end{array}} \right]\).

If a ≠ b then x₁y₁ + x₂y₂ + x₃y₃ equals 

1. a
2. b
3. ab
4. 0

Answer 1

Correct Answer - Option 4 : 0

Concept:

We know that eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix are also orthogonal.

⇒ X₁.X₂^T = 0

Where X₁ and X₂ are the eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix and X₂^T is the transpose of X2

Calculation:

Given:

\(X_{1}=\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right]\),\(X_{2}=\left[ {\begin{array}{*{20}{c}} {{y_1}}\\ {{y_2}}\\ {{y_3}} \end{array}} \right]\)

Using equation (1),

\(⇒ \left[ {\begin{array}{*{20}{c}} {{x_1}}&{{x_2}}&{{x_3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{y_1}}\\ {{y_2}}\\ {{y_3}} \end{array}} \right]=0\)

= x₁y₁ + x₂y₂ + x₃y₃ = 0 

Note:

we know that dot product of two vectors is |a||b|cos(α) where |a| and |b| is the magnitude and α is the angle between them, hence its zero when α is 90 degree.