 Solution

ID: past-exam-of-the-numerical-linear-algebra-and-approximation-course-of-the-university-of-oxford/2008/1/c/solution

A = U Σ V^{T}

gives

A^{T} A = V Σ^{2} V^{T} .

Therefore

σ_{i} (A) = λ_{i} (A^{T} A)

, counting multiplicities. These are the nonnegative square roots: zero singular values must be included if

A

is singular. Strict positivity requires full rank, which the question does not assume.

Write the lower bidiagonal matrix entries as

B_{ii} = d_{i}

and

B_{i + 1, i} = e_{i}

. Its columns

i

and

j

have disjoint supports if

∣ i - j ∣ > 1

, so

T = B^{T} B

has no entries outside the three central diagonals. More explicitly,

T_{ii} = d_{i}^{2} + e_{i}^{2} (i < n), T_{nn} = d_{n}^{2}, T_{i, i + 1} = T_{i + 1, i} = e_{i} d_{i + 1} .

This proves both symmetry and the tridiagonal matrix structure.

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