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complex conjugate

complex conjugation, matrix complex conjugate
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Mathematics Subject Classification

12D99 no label found30-00 no label found32-00 no label found



Does the $\overline{Av} = \overline{A}\overline{v}$ result rely upon $A$ being finite dimensional, or will it hold for a general linear operator?


Hi Rhys,

In my opinion the answer is yes, but it is necessary to define bases in the spaces between which the linear operator maps. What does this mean?
Let A,B be spaces of dimension n, m resp., and let L be a linear operator such that
L: A \mapsto B.
So for every element x in A we have y = Lx, where y is some element in B.
Next let's choose bases e_j (j=1,...,n), f_i (i=1,...,m) in A,B resp. Thus, if x=(x_1,...,x_n), y=(y_1,...,y_m), we may writing out the linear system
y_i = L_{ij} x_j.
A,B are defined over the field F where the entries L_{ij} belong.
Then we say that the matrix ||L_{ij}|| is associated to the operator L. Obviously, if we change any basis, then another matrix will represent the operator L.

What do you mean by \overline{Av}, \overline{A}, and \overline{v}? Are these representations of the vector Av, the matrix A, and the vector v with respect to some basis for your space? What do you mean by "$A$ being finite dimensional?" Do you mean that the space under considered (on which A acts) is finite dimensional?

- Keenan

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