Proof regarding the image and kernel of a normal operator

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Adgorn
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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?