Schur Decomposition - Proof

Proof

A constructive proof for the Schur decomposition is as follows: every operator A on a complex finite-dimensional vector space has an eigenvalue λ, corresponding to some eigenspace Vλ. Let Vλ⊥ be its orthogonal complement. It is clear that, with respect to this orthogonal decomposition, A has matrix representation (one can pick here any orthonormal bases spanning Vλ and Vλ⊥ respectively)

A = \begin{bmatrix} \lambda \, I_{\lambda} & A_{12} \\ 0 & A_{22} \end{bmatrix}: \begin{matrix} V_{\lambda} \\ \oplus \\ V_{\lambda}^{\perp} \end{matrix} \rightarrow \begin{matrix} V_{\lambda} \\ \oplus \\ V_{\lambda}^{\perp} \end{matrix}

where Iλ is the identity operator on Vλ. The above matrix would be upper-triangular except for the A22 block. But exactly the same procedure can be applied to the sub-matrix A22, viewed as an operator on Vλ⊥, and its submatrices. Continue this way n times. Thus the space Cn will be exhausted and the procedure has yielded the desired result.

The above argument can be slightly restated as follows: let λ be an eigenvalue of A, corresponding to some eigenspace Vλ. A induces an operator T on the quotient space Cn modulo Vλ. This operator is precisely the A22 submatrix from above. As before, T would have an eigenspace, say Wμ ⊂ Cn modulo Vλ. Notice the preimage of Wμ under the quotient map is an invariant subspace of A that contains Vλ. Continue this way until the resulting quotient space has dimension 0. Then the successive preimages of the eigenspaces found at each step form a flag that A stabilizes.

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