Sa se arate ca pentru orice operator \( x\in B(\mathbb{H}) \) are loc
\( Ker(x^{*})=Ran(x)^{ \perp } \), unde \( x^{*} \) este adjunctul operatorului \( x \).
Pentru orice operator din B(H) avem Ker(x*)=Ran(x)^{_|_}
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Pentru orice operator din B(H) avem Ker(x*)=Ran(x)^{_|_}
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Fie \( a\in ker(x^*) \), atunci \( (x^*a, u)=0, \forall u\in H \), de unde\( (a, xu)=0 \), deci \( a\in Ran(x)^{\perp} \).
Reciproc, fie \( a\in Ran(x)^{\perp} \), atunci \( (a, xu)=0, \forall u\in H \), deci\( (x^*a,u)=0 \forall u\in H \) (de unde e gata, sau...) luam \( u=x^*a \) si rezulta \( ||x^*a||=0 \), deci \( a\in Ker(x^*) \).
Reciproc, fie \( a\in Ran(x)^{\perp} \), atunci \( (a, xu)=0, \forall u\in H \), deci\( (x^*a,u)=0 \forall u\in H \) (de unde e gata, sau...) luam \( u=x^*a \) si rezulta \( ||x^*a||=0 \), deci \( a\in Ker(x^*) \).
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