= Solution
The <singular value decomposition> $A=U\Sigma V^T$ gives
$$
A^TA=V\Sigma^2V^T.
$$
Therefore $\boxed{\sigma_i(A)=\sqrt{\lambda_i(A^TA)}}$, 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 $T=B^TB$ has no entries outside the three central diagonals. More explicitly,
$$
\boxed{T_{ii}=d_i^2+e_i^2\ (i<n),\quad T_{nn}=d_n^2,\quad T_{i,i+1}=T_{i+1,i}=e_id_{i+1}.}
$$
This proves both symmetry and the <tridiagonal matrix> structure.
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