Two questions on linear transformations.

  • #1
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Main Question or Discussion Point

1) there's given a transformation f:C^n->C^n (C is the complex field), it's known that f is linear on R (real numbers) and its rank on R equals 3 i.e, dim_R Imf=3. now is f linear on C?
2) there's a function f:C->C and its known that f is linear on R, and det_R f<0, is f linear on C?

im kind of stuck on those questions, obviously in the second question, if the determinant is different than zero then the matrix is invertible (i.e has an inverse) and so it has an f^-1, so it's isomorphism on R, but im not sure if its linear on C.

now about the first question if dim_R Imf=3<n then function isnt onto C^n and thus isnt injective, but i dont know how to deduce from that about its linearity on C.

your help is apprecited.
 

Answers and Replies

  • #2
HallsofIvy
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1) there's given a transformation f:C^n->C^n (C is the complex field), it's known that f is linear on R (real numbers) and its rank on R equals 3 i.e, dim_R Imf=3. now is f linear on C?
2) there's a function f:C->C and its known that f is linear on R, and det_R f<0, is f linear on C?

im kind of stuck on those questions, obviously in the second question, if the determinant is different than zero then the matrix is invertible (i.e has an inverse) and so it has an f^-1, so it's isomorphism on R, but im not sure if its linear on C.

now about the first question if dim_R Imf=3<n then function isnt onto C^n and thus isnt injective, but i dont know how to deduce from that about its linearity on C.

your help is apprecited.
I am confused by this. If f is defined on Cn, what do you mean by "f is linear on R"? Do you mean on (x+ 0i,0,0...)? And it has to have rank 3? How can a function defined on a one-dimensional vector space have rank 3?
 
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  • #3
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i think it means:
for example, that the basis of C^n is (1,0,...,0),...(0,0,...0,1),(i,0,...,0),...(0,...,i)
where on C it would only be the first n vectors.
 
  • #4
matt grime
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Hmm. There are several ways to interpret this question.

R^n is an obvious real subspace of C^n, but that is an artificial way of thinking about things - and the canonical counter example is to take the transformation that conjugates in each variable.

No, you really have to think about this as C^n being a 2n dimensional real vector space.

Anyway, buried in there is the hint: as a real vector space, C^m is 2m dimensional, and the image of f is a complex vector space, so it certainly can't be a 3 dimensional real vector space, can it?
 
  • #5
HallsofIvy
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i think it means:
for example, that the basis of C^n is (1,0,...,0),...(0,0,...0,1),(i,0,...,0),...(0,...,i)
where on C it would only be the first n vectors.
No, a basis for C^n is (1, 0, ...,0)... (0, 0, ...., 1). Since you are multiplying by complex numbers. Do you mean to say a basis for C^n as a vector space over R?

matt, my problem is that the problem did not say "linear on R^n", it said "linear on R". Exactly how are you thinking of a function from C^n to C^n as restricted to R?
 
  • #6
matt grime
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I was taking 'linear on R' to mean R-linear (linear *over* R), the normal meaning we'd take for field extensions K over k, which is almost certainly what should have been written.
 
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