Editing Invariant subspace (section)

== Diagonalization via projections ==
Determining whether a given subspace ''W'' is invariant under ''T'' is ostensibly a problem of geometric nature. Matrix representation allows one to phrase this problem algebraically. 

Write {{Mvar|V}} as the [[direct sum]] {{Math|''W''&nbsp;&oplus;&nbsp;''W''&prime;}}; a suitable {{Math|''W''&prime;}} can always be chosen by extending a basis of {{mvar|W}}.  The associated [[projection operator]] ''P'' onto ''W'' has matrix representation
:<math> 
P = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} : \begin{matrix}W \\ \oplus \\ W' \end{matrix} \rightarrow \begin{matrix}W \\ \oplus \\ W' \end{matrix}.
</math>

A straightforward calculation shows that ''W'' is {{Mvar|T}}-invariant if and only if ''PTP'' =&thinsp;''TP''.   

If 1 is the [[identity operator]], then {{Math|1-''P''}} is projection onto {{Math|''W''&prime;}}.  The equation {{math|''TP'' {{=}} ''PT''}} holds if and only if both im(''P'') and im(1&thinsp;−&nbsp;''P'') are invariant under ''T''. In that case, ''T'' has matrix representation <math display=block> 
T = \begin{bmatrix} T_{11} & 0 \\ 0 & T_{22} \end{bmatrix} : \begin{matrix} \operatorname{im}(P) \\ \oplus \\ \operatorname{im}(1-P) \end{matrix} \rightarrow  \begin{matrix} \operatorname{im}(P) \\ \oplus \\ \operatorname{im}(1-P) \end{matrix} \;.
</math>

Colloquially, a projection that commutes with ''T'' "diagonalizes" ''T''.