Proof regarding the image and kernel of a normal operator

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If T is a normal operator, it is established that T and its adjoint T* share the same kernel and image. The discussion emphasizes the importance of unitarily diagonalizing T, which allows for a clearer understanding of the relationship between the kernels and images of T and T*. The approach involves examining the right and left nullspaces of both operators, utilizing the properties of orthonormal eigenvectors. The challenge lies in demonstrating the equality of the kernels and images through these decompositions. Ultimately, the relationship between T and T* is rooted in their shared structural properties as normal operators.
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Homework Statement


Show that if T is normal, then T and T* have the same kernel and the same image.

Homework Equations


N/A

The Attempt at a Solution


At first I tried proving that Ker T ⊆ Ker T* and Ker T* ⊆ Ker, thus proving Ker T = Ker T* and doing the same thing with I am T, but could not find a way of doing so. I am relatively certain this has to do with discomposing T into direct sum subspace or doing something with the orthonormal basis of V comprised of eigenvalues of T but I cannot seem to figure it out.
I would love some assistance on the matter.
 
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You understand that ##T## is unitarily diagonalizable here, right? You really can't ask for much more as the mutually orthonormal eigenvectors form a partition (and indeed are a proper, length preserving, coordinate system).

##T = U D U^*##

##T^* = \big(U D U^*\big)^* = U D^* U^*##

with respect to the the right nullspace, how would you identify it in ##T##, above? What would it look like in ##T^*##? If you are so interested, what would the left nullspace look like for both of them?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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