# Unitary Operator apply to Entangled vector

I am trying of resolve this exercise: Show that if $|\psi \rangle$ is an entangled state of two Qbits, then the application of a unitary operator of the form $U_1 \otimes U_2$ necessarily generates an entangled state.

Suppose that $(U_1 \otimes U_2)|\psi\rangle$ is not entangled then it must have the form

$$(U_1 \otimes U_2)|\psi\rangle = (a|0\rangle+b|1\rangle)\otimes (c|0\rangle+d|1\rangle),$$ but the unique way this will be true is when $|\psi\rangle$ is not entangled due to definition of linear operator.

My proof is bad? Help me please

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Your language is not completely clear to me. If I understand your arguments correctly, your proof seems fine. – Siva Apr 8 '13 at 5:19
Exist any way without use contradiction? – juaninf Apr 9 '13 at 10:08
your proof sounds alright, though, I am not the best person to verify it. Any unitary operation on the individual Qbits should not disentangle the system. You should work on expressing your questions clearly. – Prathyush Apr 10 '13 at 0:05
Is there any mathematic way to solve this? – juaninf May 12 '13 at 1:36