Answer to Question #121754 in Functional Analysis for Henry

Question #121754
Please, this is urgent , solve it as soon as possible.

Suppose V is finite-dimensional and phi is a linear functional on V. Then
there is a unique vector u in V such that
phi(v)=< v, u> for every v in V.
1
Expert's answer
2020-06-15T18:57:28-0400

The vector "u"  is unique. Let "u_1"  and "u_2"  be such that "\\phi(v) = \\langle v, u_1\\rangle"  and "\\phi(v) = \\langle v, u_2\\rangle"  for every "v\\in V". Then "\\langle v, u_1-u_2\\rangle = \\langle v, u_1\\rangle - \\langle v, u_2\\rangle = \\phi(v) - \\phi(v) = 0"  for every "v\\in V" . Substituting "u_1-u_2"  for "v" , we get "\\langle u_1-u_2, u_1-u_2\\rangle = 0" , hence "u_1-u_2 = 0" .




The vector "u"  exists. Since "V" is finite-dimensional, the Gram-Schmidt procedure gives an orthonormal basis "\\{e_i\\}_i"  of "V". Choose "u = \\sum_i \\overline{\\phi(e_i)} e_i" .


Let "v\\in V". Since "\\{e_i\\}_i"  is a basis, there are scalars "\\{a_i\\}_i"  such that "v = \\sum_i a_i e_i" . Hence


"\\phi(v) = \\phi\\left(\\sum_i a_i e_i\\right) = \\sum_i a_i \\phi(e_i)."

On the other hand,


"\\langle v, u\\rangle = \\left\\langle \\sum_i a_i e_i, \\sum_i \\overline{\\phi(e_i)} e_i\\right\\rangle \\\\= \\sum_i \\sum_j a_i \\phi(e_j) \\langle e_i, e_j\\rangle = \\sum_i a_i \\phi(e_i)"

where the last equality follows from the orthonormality of "\\{e_i\\}_i" . Therefore, "\\phi(v) = \\langle v, u\\rangle" for every "v\\in V".


The expression "\\overline{a}"  denotes the complex conjugate of a complex scalar "a"  and is used for complex vector spaces. For real vector spaces, the proof is the same as above except that complex conjugation should be eliminated.


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Comments

Assignment Expert
11.06.20, 23:30

Dear Henry, please kindly wait for a solution of the question. If you need meeting specific requirements, then you can submit an order.

Henry
11.06.20, 19:29

Can you please solve at the earliest as it's urgent.

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