If $T$ is a normal linear map from vector space $V$ to itself, then image of $T$ is the same as image of $T^*$.
Can anyone help me to prove this?
Asked
Active
Viewed 481 times
2
-
1From $|Tx|^2=(Tx,Tx)=(T^Tx,x)=(TT^x,x)=(T^x,T^x)=|T^x|^2$ you get that $\operatorname{Ker}(T)=\operatorname{Ker}(T^)$. Then prove that the image is the orthogonal complement of the kernel. – plop May 09 '21 at 14:34
-
@plop In general, the orthogonal complement of the kernel is the closure of the image, so this argument only shows that the closures of the images coincide. – J. De Ro May 09 '21 at 21:00
-
@QuantumSpace This argument shows that they are equal in the finite-dimensional case. If you want it for a general Hilbert, quotient out the kernel and consider the extension to the closure of their ranges of operator $U=A^A^{-1}$. You get that $A^U=A$ and $A^*=AU$, which implies that their ranges are themselves equal. – plop May 09 '21 at 21:38
-
Thank you very much @plop – Ehsan Poursaeed May 12 '21 at 14:21
-
Am I supposed to write an answer myself now? @plop – Ehsan Poursaeed May 12 '21 at 14:22