Schur decomposition

Remarks

The proof is by induction on $n$. Over $\C$, the matrix $A$ has an eigenvector $v$ of norm $1$. Extend $v$ to an orthonormal basis; in this basis the first column of the matrix of $A$ is $(\lambda, 0, \dots, 0)^T$. Apply the induction hypothesis to the lower right $(n-1) \times (n-1)$ block.

The order of the eigenvalues on the diagonal of $T$ can be chosen freely. If $A$ is a normal matrix, then $T$ is normal and triangular, hence diagonal; this gives the spectral theorem for normal matrices. The complex field matters. A real matrix with non-real eigenvalues has no factorisation $A = QTQ^T$ with $Q$ real orthogonal and $T$ real upper triangular. Indeed, the diagonal of such a $T$ would hold the eigenvalues of $A$. The real Schur form uses an orthogonal $Q$ and a block upper triangular $T$ with diagonal blocks of size $1$ or $2$ instead.

Going deeper

For a student

The Schur decomposition is the unitary form of triangularisation. It is stable to compute, and the QR algorithm for eigenvalues computes it. It gives quick proofs that the trace is the sum and the determinant is the product of the eigenvalues. It is also the standard route to the spectral theorem for normal matrices.