Let π = \(\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ \end{bmatrix} \) , and letβ¦
2024
Let π =Β \(\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ \end{bmatrix} \) , and let π1 , π2 ,π3 , π4 , π5 be the singular values of the matrix
π΄= πππ» (where ππ» is the transpose of π). The value ofΒ \(\sum_{i=1}^{5} \sigma_i \) is ______.
Attempted by 4 students.
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Correct answer: 55
Answer: 55
Explanation:
Compute the squared norm of u:
||u||^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55.
Reason via eigenvalues: The matrix M = u u^T is symmetric, positive semidefinite, and has rank 1. M u = (u^T u) u, so there is a single nonzero eigenvalue equal to ||u||^2 = 55, and the other four eigenvalues are 0. For a symmetric positive semidefinite matrix, singular values equal eigenvalues, so the singular values are 55, 0, 0, 0, 0.
Alternative check via trace: trace(M) = trace(u u^T) = u^T u = 55. The trace equals the sum of eigenvalues, which for this matrix equals the sum of singular values, confirming the sum is 55.