Let's break down the solution into **3 steps**, with explanations after each step, and a final answer at the end. --- ### **Step 1: Setup the Linear Combination** Suppose \(v_1\) and \(v_2\) are eigenvectors of \(T\) corresponding to distinct eigenvalues \(\lambda_1\) and \(\lambda_2\), respectively. Assume a linear combination equals zero: \[ a_1 v_1 + a_2 v_2 = 0 \] where \(a_1, a_2\) are scalars. **Explanation:** This is the definition of linear independence. If \(a_1\) and \(a_2\) must both be zero for this equation to hold, then \(v_1\) and \(v_2\) are linearly independent. --- ### **Step 2: Apply the Linear Transformation** Apply \(T\) to both sides: \[ T(a_1 v_1 + a_2 v_2) = T(0) = 0 \] Using linearity and the eigenvector property (\(T(v_i) = \lambda_i v_i\)): \[ a_1 T(v_1) + a_2 T(v_2) = a_1 \lambda_1 v_1 + a_2 \lambda_2 v_2 = 0 \] **Explanation:** We use the fact that eigenvectors satisfy \(T(v_i) = \lambda_i v_i\), and \(T\) is linear. --- ### **Step 3: Form a System and Solve** Now you have two equations: 1. \(a_1 v_1 + a_2 v_2 = 0\) 2. \(a_1 \lambda_1 v_1 + a_2 \lambda_2 v_2 = 0\) Multiply the first equation by \(\lambda_2\): \[ a_1 \lambda_2 v_1 + a_2 \lambda_2 v_2 = 0 \] Subtract from the second equation: \[ (a_1 \lambda_1 v_1 + a_2 \lambda_2 v_2) - (a_1 \lambda_2 v_1 + a_2 \lambda_2 v_2) = 0 \] \[ a_1 (\lambda_1 - \lambda_2) v_1 = 0 \] Since \(v_1 \neq 0\) and \(\lambda_1 \neq \lambda_2\), it must be that \(a_1 = 0\). Plug \(a_1 = 0\) into the first equation: \[ a_2 v_2 = 0 \implies a_2 = 0 \] **Explanation:** Distinct eigenvalues guarantee that the only solution is \(a_1 = a_2 = 0\). --- ## **Final Answer** **Thus, the eigenvectors \(v_1\) and \(v_2\) corresponding to distinct eigenvalues are linearly independent.**

Let's approach this problem step-by-step to demonstrate that eigenvectors corresponding to distinct eigenvalues are linearly independent. --- ### **Step 1: Assume a Linear Combination Equals Zero** Suppose \( v_1 \) and \( v_2 \) are eigenvectors of \( T \) corresponding to the distinct eigenvalues \( \lambda_1 \) and \( \lambda_2 \), respectively. Assume a linear combination: \[ a_1 v_1 + a_2 v_2 = 0 \] where \( a_1, a_2 \) are scalars. **Explanation:** To prove linear independence, we start by assuming that a linear combination of the vectors equals the zero vector. If the only solution is \( a_1 = a_2 = 0 \), then the vectors are linearly independent. --- ### **Step 2: Apply the Transformation \( T \) to the Equation** Apply \( T \) to both sides: \[ T(a_1 v_1 + a_2 v_2) = T(0) = 0 \] Using linearity: \[ a_1 T(v_1) + a_2 T(v_2) = 0 \] Since \( v_1 \) and \( v_2 \) are eigenvectors: \[ a_1 \lambda_1 v_1 + a_2 \lambda_2 v_2 = 0 \] **Explanation:** Applying the linear transformation preserves linear combinations, and because \( v_1, v_2 \) are eigenvectors, we replace \( T(v_i) \) with \( \lambda_i v_i \). This step simplifies the problem to a relation involving eigenvalues and eigenvectors. --- ### **Step 3: Derive the System and Conclude** Now, you have: \[ a_1 v_1 + a_2 v_2 = 0 \quad \text{(from Step 1)} \] \[ a_1 \lambda_1 v_1 + a_2 \lambda_2 v_2 = 0 \] Subtract \( \lambda_2 \) times the first equation from the second: \[ a_1 (\lambda_1 - \lambda_2) v_1 + a_2 (\lambda_2 - \lambda_2) v_2 = 0 \] \[ a_1 (\lambda_1 - \lambda_2) v_1 = 0 \] Since \( v_1 \neq 0 \), and \( \lambda_1 \neq \lambda_2 \), it follows that: \[ a_1 (\lambda_1 - \lambda_2) = 0 \implies a_1 = 0 \] Plugging \( a_1 = 0 \) into the first equation: \[ a_2 v_2 = 0 \implies a_2 = 0 \] **Explanation:** Because the eigenvalues are distinct (\( \lambda_1 \neq \lambda_2 \)), the only solution to the linear combination equaling zero is \( a_1 = a_2 = 0 \). Thus, \( v_1 \) and \( v_2 \) are linearly independent. --- ## **Final Answer:** **Eigenvectors corresponding to distinct eigenvalues are linearly independent because any linear combination that sums to zero necessarily has all coefficients zero, confirming their independence.**