det (λ I - A^{T}) = det [(λ I - A)^{T}] = det (λ I - A) .

Thus they have the same eigenvalues, including algebraic multiplicities.

The proposed sum identity is false. For

A = (0010),

λ_{i}

are zero, whereas

A^{T} A = (0001)

0, 1

. Hence

i \sum μ_{i} = 1 \neq = 0 = i \sum λ_{i}^{2} .

Equivalently, the two sides are generally

tr (A^{T} A)

and

tr (A^{2})

, which need not agree.

ii

Solution

The product identity always holds, since

\prod_{i} μ_{i} = det (A^{T} A) = det (A^{T}) det A = (det A)^{2} = (\prod_{i} λ_{i})^{2} .

Therefore

i = 1 \prod n μ_{i} = i = 1 \prod n λ_{i}^{2} .

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