Semilinear Transformation, Kernel

[LHeiner]

Hello

I'm trying to proof the following: f is a semilinear transformation between the vectorspaces $$V \rightarrow W,c^\ast \in W^\ast , G:=ker \ c^\ast$$. Show that $$f^{-1}(G)=ker(f^T(c^\ast ))$$ and that the f-preimage of a hyperplane of W a hyperplane of V or V as a whole is.

Can you help me?

 

[ilia1987]

If I understand correctly: $$f^T(c^\ast ) = c^\ast \circ f$$
so that: $$ker(f^T(c^\ast))=ker(c^\ast \circ f)$$
which is all the vectors of V that are mapped by f into the kernel of c*.
Or: $$f^{-1}(ker(c\ast))$$ in other words...
for the second part, you should show that $$f^{-1}(ker (c\ast))$$ is a linear subspace of V, and then remember that every linear subspace of W can be represented as an intersection of kernels of some number of linear functionals, hence every inverse image is an intersection of some number of subspaces of V.

 

[LHeiner]

If I understand correctly: $$f^T(c^\ast ) = c^\ast \circ f$$
so that: $$ker(f^T(c^\ast))=ker(c^\ast \circ f)$$
which is all the vectors of V that are mapped by f into the kernel of c*.
Or: $$f^{-1}(ker(c\ast))$$ in other words...
for the second part, you should show that $$f^{-1}(ker (c\ast))$$ is a linear subspace of V, and then remember that every linear subspace of W can be represented as an intersection of kernels of some number of linear functionals, hence every inverse image is an intersection of some number of subspaces of V.

thank you for your quick answer, but I'm not sure if i got it!

I mean if you say that $$ker(c^\ast \circ f)$$ are all Vectors of V that are mapped by f into the kernel of c* it is already $$=f^{-1}(ker(c\ast))$$.

And for the second party I am confused, i didn't know that every linear subspace of W can be represented as an intersection of kernels of some number of linear functionals.

 

[ilia1987]

$$ker(c^\ast \circ f) = \{v|v\in V,c^\ast (f(v))=0\} = \{v|v\in V,f(v)\in ker(c^\ast)\}$$ the last identity comes from the definition of kernel.

As for the second part, I'm not sure if it can be applied to any space W, but if W is finite dimensional with dimension n, then every linear functional's kernel has dimension n-1 or n, and for every n-1 dimensional subspace of W a linear functional can be found that has that subspace as its kernel (that defines that linear functional up to a multiplication by a scalar).

Try thinking of a linear functional acting on the space of nx1 column vectors as a row vector (1xn matrix), that matrix clearly has kernel>=n-1. And every matrix A nx(n-1) which is of full column rank has left kernel of dimension 1. The row vector that spans this kernel is a linear functional, the kernel of which is the column space of A. The same is true if the matrix A is nxm of full column rank, and the column space of A is the kernel of n-m linear functionals which span the left kernel of A, or, in other words, the intersection of the kernels of these n-m linear functionals.

But as for infinite dimensional linear spaces I don't have enough knowledge to give an answer. Basically each subspace of W would have to be the intersection of an infinite amount of kernels of linear functionals, and infinity tends to be strange, so you'd have to read about it yourself in http://en.wikipedia.org/wiki/Dual_space" .

Now, all you have to do is prove that $$f^{-1}(ker(c^\ast))$$ is a linear subspace of V and that's it.

 

[blue_raver22]

Sure, I would be happy to help you with this proof. First, let's define some terms for clarity. A semilinear transformation is a function between two vector spaces that satisfies the following conditions:
1. f(ax) = af(x) for any scalar a and any vector x in V
2. f(x+y) = f(x) + f(y) for any vectors x and y in V

A kernel is the set of all vectors in the domain of a linear transformation that map to the zero vector in the codomain. In this case, G is the kernel of the linear transformation c* from W to its dual space W*.

Now, let's prove the given statement. We want to show that f^-1(G) = ker(f^T(c*)), where f^T is the transpose of f.

First, let's take an element x in f^-1(G). This means that f(x) is in G, which means that c*(f(x)) = 0. Using the properties of linear transformations, we can rewrite this as (c* o f)(x) = 0. Since c* o f is a linear transformation, this means that x is in the kernel of this linear transformation, which is exactly what we wanted to show. Therefore, f^-1(G) is a subset of ker(f^T(c*)).

Next, let's take an element x in ker(f^T(c*)). This means that (f^T(c*) o f)(x) = 0. Using the properties of linear transformations and the transpose, we can rewrite this as (c* o f)(x) = 0. This means that f(x) is in G, so x is in f^-1(G). Therefore, ker(f^T(c*)) is a subset of f^-1(G).

Since we have shown that each set is a subset of the other, they must be equal. Thus, we have proven that f^-1(G) = ker(f^T(c*)).

Finally, let's consider the f-preimage of a hyperplane in W. A hyperplane in W is defined as the set of all vectors that satisfy the equation c*(w) = b for some fixed vector w and scalar b. The f-preimage of this hyperplane is the set of all vectors in V that map to this hyperplane under f. Using the definition of a semilinear transformation