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How do I show that the Schur Theorem ($A\in \mathbb{C}^{\ n\ x\ n}: A=UTU^*\ (T,U\in\mathbb{C}^{\ n\ x\ n})$) allows to put, in the diagonal of $T$, the eigenvalues in the order that I want.

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I don't have an elegant answer, but since noone answered so far I'll share it anyways. It only works if $A$ is diagonalizable (i.e., for each eigenvalue, its algebraic and geometric multiplicity coincide).

If $A$ is diagonalizable, it has an eigenvalue decomposition of the form $$A=E \cdot \Lambda \cdot E^{-1},$$ where $E$ is the matrix of eigenvalues and $\Lambda$ is a diagonal matrix containing the eigenvalues on its main diagonal. It is clear that in this decomposition, the order of the eigenvalues in $\Lambda$ is not unique since $$A=E\cdot\Lambda\cdot E^{-1} = \bar{E} \cdot \bar{\Lambda} \cdot \bar{E}$$ for $\bar{E} = E \cdot \Pi$ and $\bar{\Lambda} = \Pi \cdot\Lambda\cdot\Pi$ where $\Pi$ is an arbitrary permutation matrix.

Now, we can convert the eigenvalue decomposition into a Schur decomposition by employing the QRD decomposition of $E$, i.e., $$E=Q\cdot R \cdot D,$$ where $Q$ is a unitary matrix, $R$ is a "unitary" upper triangular matrix (i.e., having ones on its main diagonal) and $D$ is a diagonal matrix. Inserting $E$ into the EVD, we have $$A = Q \cdot R \cdot D \cdot \Lambda \cdot D^{-1} \cdot R^{-1} \cdot Q^* = Q \cdot R \cdot \Lambda \cdot R^{-1} \cdot Q^*.$$ The last equation can be identified with the Schur decomposition: $Q \rightarrow U$ (since it is unitary) and $R \cdot \Lambda \cdot R^{-1} \rightarrow T$: for a unitary upper triangular matrix $R$, its inverse is also unitary upper triangular and therefore, $R \cdot \Lambda \cdot R^{-1}$ is upper triangular with $\Lambda$ on its main diagonal.

Since this "conversion" works regardless of the order of the eigenvalues we started with, it is in fact arbitrary.

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