91影视

The 2-norm of a matrix A, denoted 鈥朅鈥栤倐, is equal to the square root of the largest eigenvalue of A岬�A. Since A is symmetric, we have A岬�A = A虏. Therefore, 鈥朅鈥栤倐 = \(\sqrt{\max _{1 \leq i \leq n}\left(\lambda_{i}^2\right)}\).

To compute the 2-norm of the inverse of A, 鈥朅鈦宦光�栤倐, we need to find the eigenvalues of (A鈦宦�)岬�A鈦宦� = (A鈦宦�)虏. Notice that if 位_i is an eigenvalue of A, with eigenvector x, then Ax = 位_i x. Since A is nonsingular, we can multiply both sides of the equation by A鈦宦� to get A鈦宦笰x = A鈦宦刮籣i x, which implies A鈦宦箈 = (1/位_i) x. Thus, 1/位_i is an eigenvalue of A鈦宦�. Therefore, the 2-norm of the inverse of A is equal to the square root of the largest eigenvalue of (A鈦宦�)虏: 鈥朅鈦宦光�栤倐 = \(\sqrt{\max _{1 \leq i \leq n}\left(\frac{1}{\lambda_{i}^2}\right)}\).

Now that we have the 2-norm of A and its inverse, we can find the condition number using the formula cond(A) = 鈥朅鈥柭封�朅鈦宦光��, which takes the following form in this case: \[ \operatorname{cond}_{2}(A)=\sqrt{\max _{1 \leq i \leq n}\left(\lambda_{i}^2\right)}\cdot \sqrt{\max _{1 \leq i \leq n}\left(\frac{1}{\lambda_{i}^2}\right)} \] Simplify the expression by taking the product of square roots: \[ \operatorname{cond}_{2}(A)=\sqrt{\max _{1 \leq i \leq n}\left(\lambda_{i}^2\right) \cdot \max _{1 \leq i \leq n}\left(\frac{1}{\lambda_{i}^2}\right)} \] Then, notice that the product of the two max values is equivalent to: \[ \frac{\max _{1 \leq i \leq n}\left|\lambda_{i}\right|^2}{\min _{1 \leq i \leq n}\left|\lambda_{i}\right|^2} \] Thus, we have: \[ \operatorname{cond}_{2}(A)= \sqrt{\frac{\max _{1 \leq i \leq n}\left|\lambda_{i}\right|^2}{\min _{1 \leq i \leq n}\left|\lambda_{i}\right|^2}} \] Finally, we simplify by taking the square root, leading to the result: \[ \operatorname{cond}_{2}(A)=\frac{\max _{1 \leq i \leq n}\left|\lambda_{i}\right|}{\min _{1 \leq i \leq n}\left|\lambda_{i}\right|} \] This completes the proof.