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Let $\alpha, \beta$ be two non-zero real numbers and $v_{1}, v_{2}$ be two non-zero real vectors of size $3 \times 1.$ Suppose that $v_{1}$ and $v_{2}$ satisfy $v_{1}^{T} v_{2} = 0, v_{1}^{T} v_{1} = 1,$ and $v_{2}^{T} v_{2} = 1.$ Let $A$ be the $3 \times 3$ matrix given by:

$$A = \alpha  v_{1} v_{1}^{T} + \beta v_{2} v_{2}^{T} $$

The eigenvalues of $A$ are ________________.

  1. $0, \alpha, \beta$
  2. $0, \alpha+\beta, \alpha-\beta$
  3. $0, \frac{\alpha+\beta}{2}, \sqrt{\alpha \beta}$
  4. $0, 0, \sqrt{\alpha^{2} + \beta^{2}}$

1 Answer

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Multiply $v_1$ to both sides above, and take advantage of orthogonality & normality.

$$A v_1 = \alpha v_1 (v_1^{T}v_1) + \beta v_2 (v_2^{T}v_1)$$

$$ A v_1 = \alpha v_1 (1) + \beta v_2 (0)$$

$$ A v_1 = \alpha v_1 $$

Do the same for $v_2$. You'll find $\alpha$ & $\beta$ are eigenvalues. So, A is correct.
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