Question

Let $\alpha, \beta$ be non-zero real numbers and $v_1, v_2$ be two non-zero vectors of size $3 \times 1$. Suppose $v_1^T v_2 = 0$, $v_1^T v_1 = 1$, $v_2^T v_2 = 1$. Let \[ A = \alpha v_1 v_1^T + \beta v_2 v_2^T. \] The eigenvalues of $A$ are ____________.

• A. $0,\ \alpha,\ \beta$
• B. $0,\ \alpha+\beta,\ \alpha-\beta$
• C. $0,\ \dfrac{\alpha+\beta}{2},\ \sqrt{\alpha\beta}$
• D. $0,\ 0,\ \sqrt{\alpha^2 + \beta^2}$

Hint:

For matrices of the form $\alpha v_1 v_1^T + \beta v_2 v_2^T$ with orthonormal vectors, each term contributes exactly one non-zero eigenvalue.

Answer 1

The Correct Option is A

The matrix \[ A = \alpha v_1 v_1^T + \beta v_2 v_2^T \] is a sum of two rank-1 matrices. Since $v_1$ and $v_2$ are orthonormal, \[ v_1^T v_1 = 1,\qquad v_2^T v_2 = 1,\qquad v_1^T v_2 = 0. \] Step 1: Evaluate $A v_1$.
\[ A v_1 = \alpha v_1 (v_1^T v_1) + \beta v_2 (v_2^T v_1) = \alpha v_1 + 0 = \alpha v_1. \] So $v_1$ is an eigenvector with eigenvalue $\alpha$.
Step 2: Evaluate $A v_2$.
\[ A v_2 = \alpha v_1 (v_1^T v_2) + \beta v_2 (v_2^T v_2) = 0 + \beta v_2 = \beta v_2. \] Thus $v_2$ is an eigenvector with eigenvalue $\beta$.
Step 3: Remaining eigenvalue.
$A$ is at most rank 2 (sum of two rank-1 matrices), so the third eigenvalue is: \[ 0. \] Thus eigenvalues are: \[ 0,\ \alpha,\ \beta. \] Final Answer: $0,\ \alpha,\ \beta$