Let's solve the problem step by step: --- **Step 1: State the assumption for linear dependence** Assume, for contradiction, that \( v_1 \) and \( v_2 \) are linearly dependent. This means there exist scalars \( a_1 \) and \( a_2 \), **not both zero**, such that \[ a_1 v_1 + a_2 v_2 = 0. \] **Explanation:** If we can show that this equation only has the trivial solution \( a_1 = a_2 = 0 \), it will prove that \( v_1 \) and \( v_2 \) are linearly independent. --- **Step 2: Apply the linear transformation \( T \) to both sides** Apply \( T \) to the above equation: \[ T(a_1 v_1 + a_2 v_2) = T(0) \implies a_1 T(v_1) + a_2 T(v_2) = 0. \] But \( T(v_1) = \lambda_1 v_1 \) and \( T(v_2) = \lambda_2 v_2 \), so: \[ a_1 \lambda_1 v_1 + a_2 \lambda_2 v_2 = 0. \] **Explanation:** By the definition of eigenvectors and eigenvalues, the effect of \( T \) on each eigenvector is simply to scale it by its eigenvalue. --- **Step 3: Use the two equations to isolate the coefficients** Now subtract \( \lambda_2 \) times the first equation from the second equation: \[ (a_1 \lambda_1 v_1 + a_2 \lambda_2 v_2) - \lambda_2 (a_1 v_1 + a_2 v_2) = 0 \] \[ a_1 (\lambda_1 - \lambda_2) v_1 = 0 \] Since \( v_1 \neq 0 \) and \( \lambda_1 \neq \lambda_2 \) (they are distinct), it follows that \( a_1 = 0 \). Plug back into the original equation: \[ a_1 v_1 + a_2 v_2 = 0 \implies 0 + a_2 v_2 = 0 \implies a_2 = 0 \] **Explanation:** Because the eigenvalues are distinct and the eigenvectors are nonzero, the only solution is the trivial one. --- ### **Final Answer:** The only solution to \( a_1 v_1 + a_2 v_2 = 0 \) is \( a_1 = a_2 = 0 \), so the eigenvectors \( v_1 \) and \( v_2 \) corresponding to **distinct eigenvalues** are **linearly independent**.