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 V_λ. Let V_λ⊥ 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 V_λ and V_λ⊥ respectively)

where I_λ is the identity operator on V_λ. 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 V_λ⊥, 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 V_λ. A induces an operator T on the quotient space Cn modulo V_λ. This operator is precisely the A₂₂ submatrix from above. As before, T would have an eigenspace, say W_μ ⊂ Cn modulo V_λ. Notice the preimage of W_μ under the quotient map is an invariant subspace of A that contains V_λ. 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.