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(Added {{Template:Functional analysis}}) 
(→Putnam's generalization: explained where normality is used.) 

This is equal to
:<math>e^{\lambda M^*} \left[e^{\bar\lambda M}T e^{\bar\lambda N}\right] e^{\lambda N^*} = U(\lambda) T V(\lambda)^{1}</math>,
where <math>U(\lambda) = e^{\lambda M^*  \bar\lambda M}</math> because <math>M</math> is normal, and similarly <math>V(\lambda) = e^{\lambda N^*  \bar\lambda N}</math>. However we have
:<math>U(\lambda)^* = e^{\bar\lambda M  \lambda M^*} = U(\lambda)^{1}</math>
so U is unitary, and hence has norm 1 for all λ; the same is true for ''V''(λ), so
