- #1
Shackleford
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Here's the definition.
Let T be a linear operator on a finite-dimensional inner product space V. If [itex] \|\vec{T(x)}\| = \|\vec{x}\| \\[/itex] for all x in V, we call T a unitary operator.
The question is asking about for all x in some orthornormal basis for V. Isn't that the same as for all x in V?
Let T be a linear operator on a finite-dimensional inner product space V. If [itex] \|\vec{T(x)}\| = \|\vec{x}\| \\[/itex] for all x in V, we call T a unitary operator.
Let U be a linear operator on a finite-dimensional inner product space V. If [itex] \|\vec{U(x)}\| = \|\vec{x}\| \\[/itex] for all x in some orthonormal basis for V, must U be unitary? Justify your answer with a proof or a counterexample.
The question is asking about for all x in some orthornormal basis for V. Isn't that the same as for all x in V?