= Solution
A matrix and its <matrix transpose> have the same <characteristic polynomial>, because
$$
\det(\lambda I-A^T)=\det[(\lambda I-A)^T]=\det(\lambda I-A).
$$
Thus they have the same <eigenvalues>, including algebraic multiplicities.
The proposed sum identity is false. For
$$
A=\begin{pmatrix}0&1\\0&0\end{pmatrix},
$$
both eigenvalues $\lambda_i$ are zero, whereas
$$
A^TA=\begin{pmatrix}0&0\\0&1\end{pmatrix}
$$
has eigenvalues $0,1$. Hence
$$
\boxed{\sum_i\mu_i=1\ne0=\sum_i\lambda_i^2}.
$$
Equivalently, the two sides are generally $\operatorname{tr}(A^TA)$ and $\operatorname{tr}(A^2)$, which need not agree.
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