Let's break down the proof into **2 steps** as requested: --- ### **Step 1: If T is positive definite, then all its eigenvalues are positive.** **Proof:** - Recall: \( T \) is positive definite if \(\langle Tv, v \rangle > 0\) for all nonzero \( v \in V \). - Let \( \lambda \) be an eigenvalue of \( T \) with eigenvector \( v \neq 0 \), so \( T v = \lambda v \). - Then, \( \langle T v, v \rangle = \langle \lambda v, v \rangle = \lambda \langle v, v \rangle \). - Since \( v \neq 0 \), \( \langle v, v \rangle > 0 \). - Since \( T \) is positive definite, \( \langle T v, v \rangle > 0 \). - Therefore, \( \lambda \langle v, v \rangle > 0 \implies \lambda > 0 \). **Explanation:** If \( T \) is positive definite, any eigenvalue \( \lambda \) must be positive, because otherwise \( \langle T v, v \rangle \) would not be positive for the corresponding eigenvector. --- ### **Step 2: If all eigenvalues of T are positive, then T is positive definite.** **Proof:** - Since \( T \) is self-adjoint (Hermitian), there exists an orthonormal basis of eigenvectors \( \{e_i\} \) for \( V \), with corresponding eigenvalues \( \lambda_i > 0 \). - Any \( v \in V \) can be written as \( v = \sum_i c_i e_i \). - Then, \[ \langle T v, v \rangle = \left\langle T \left(\sum_i c_i e_i\right), \sum_j c_j e_j \right\rangle = \sum_{i,j} c_i \overline{c_j} \langle T e_i, e_j \rangle \] But \( T e_i = \lambda_i e_i \), and the basis is orthonormal, so \( \langle e_i, e_j \rangle = \delta_{ij} \), giving: \[ \langle T v, v \rangle = \sum_i \lambda_i |c_i|^2 \] - Since all \( \lambda_i > 0 \) and not all \( c_i = 0 \) (since \( v \neq 0 \)), \( \langle T v, v \rangle > 0 \). **Explanation:** If all eigenvalues are positive, any vector \( v \) (except zero) gives a positive value for \( \langle T v, v \rangle \), so \( T \) is positive definite. --- ## **Final Answer:** **T is positive definite if and only if all of its eigenvalues are positive.** This is true because positive definiteness ensures all eigenvalues are positive (step 1), and having all positive eigenvalues ensures positive definiteness (step 2).