Editing Schur decomposition (section)

== Proof ==
A constructive proof for the Schur decomposition is as follows: every operator ''A'' on a complex finite-dimensional vector space has an eigenvalue ''&lambda;'', corresponding to some eigenspace ''V<sub>&lambda;</sub>''. Let ''V<sub>&lambda;</sub>''<sup>⊥</sup> 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 ''Z''<sub>1</sub> and ''Z''<sub>2</sub> spanning ''V<sub>&lambda;</sub>'' and ''V<sub>&lambda;</sub>''<sup>⊥</sup> respectively)
<math display="block">\begin{bmatrix} Z_1 & Z_2 \end{bmatrix}^{*} A \begin{bmatrix}Z_1 & Z_2\end{bmatrix} = \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}
</math>
where ''I<sub>&lambda;</sub>'' is the identity operator on ''V<sub>&lambda;</sub>''. The above matrix would be upper-triangular except for the ''A''<sub>22</sub> block. But exactly the same procedure can be applied to the sub-matrix ''A''<sub>22</sub>, viewed as an operator on ''V<sub>&lambda;</sub>''<sup>⊥</sup>, and its submatrices. Continue this way until the resulting matrix is upper triangular. Since each conjugation increases the dimension of the upper-triangular block by at least one, this process takes at most ''n'' steps. Thus the space '''C'''<sup>''n''</sup> will be exhausted and the procedure has yielded the desired result.<ref>{{cite web |last1=Wagner |first1=David |title=Proof of Schur's Theorem |url=https://math.mit.edu/~gs/linearalgebra/ila5/lafe_schur03.pdf |website=Notes on Linear Algebra}}</ref>

The above argument can be slightly restated as follows: let ''&lambda;'' be an eigenvalue of ''A'', corresponding to some eigenspace ''V<sub>&lambda;</sub>''. ''A'' induces an operator ''T'' on the [[quotient space (linear algebra)|quotient space]] '''C'''<sup>''n''</sup>/''V<sub>&lambda;</sub>''. This operator is precisely the ''A''<sub>22</sub> submatrix from above. As before, ''T'' would have an eigenspace, say ''W<sub>&mu;</sub>'' ⊂ '''C'''<sup>''n''</sup> modulo ''V<sub>&lambda;</sub>''. Notice the preimage of ''W<sub>&mu;</sub>'' under the quotient map is an [[invariant subspace]] of ''A'' that contains ''V<sub>&lambda;</sub>''. 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.