Dual basis and kernel intersection

In summary: That is to say, we have \psi_1, \dots, \psi_n \in V^{*} such that \psi_i(\phi_j) = \delta_{ij}.Then we can show that v = \sum_i \psi_i \in \bigcap_i \ker(\phi_i) \setminus \{0\}.In summary, we can prove the other implication by assuming that \Phi is not a basis and constructing a non-zero vector v \in \bigcap_i \ker(\phi_i).
  • #1
mahler1
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The problem statement, all variable
Let ##\phi_1,...,\phi_n \in V^*## all different from the zero functional. Prove that

##\{\phi_1,...,\phi_n\}## is basis of ##V^*## if and only if ##\bigcap_{i=1}^n Nu(\phi_i)={0}##.

The attempt at a solution.

For ##→##: Let ##\{v_1,...,v_n\}## be a basis of ##V## such that ##\phi_i(v_j)=δ_{ij}##. Now let ##x \in \bigcap_{i=1}^n Nu(\phi_i)={0}##, as ##x \in V## then ##x=α_1v_1+...+α_nv_n##.
By hypothesis ##\phi_i(x)=0## for all ##1\leq i \leq n##. But this means ##0=\phi_i(x)=\phi_i(α_1v_1+...+α_nv_n)=α_i## for all ##1\leq i \leq n##, it follows ##x=0##.

I don't know how to show the other implication. For ←, in the exercise is given the suggestion: if ##\{ψ_1,...,ψ_n\}## is a basis of ##V^*##, one can try to show that the matrix obtained from writing ##\phi_1,...,\phi_n## in coordinates with respect to the basis ##\{ψ_1,...,ψ_n\}##, is an invertible matrix, I am stuck and I haven't got a clue on how to use this idea, I would like any hints or another idea to prove the remaining implication.
 
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  • #2
mahler1 said:
The problem statement, all variable
Let ##\phi_1,...,\phi_n \in V^*## all different from the zero functional. Prove that

##\{\phi_1,...,\phi_n\}## is basis of ##V^*## if and only if ##\bigcap_{i=1}^n Nu(\phi_i)={0}##.

The attempt at a solution.

For ##→##: Let ##\{v_1,...,v_n\}## be a basis of ##V## such that ##\phi_i(v_j)=δ_{ij}##. Now let ##x \in \bigcap_{i=1}^n Nu(\phi_i)={0}##, as ##x \in V## then ##x=α_1v_1+...+α_nv_n##.
By hypothesis ##\phi_i(x)=0## for all ##1\leq i \leq n##. But this means ##0=\phi_i(x)=\phi_i(α_1v_1+...+α_nv_n)=α_i## for all ##1\leq i \leq n##, it follows ##x=0##.

I don't know how to show the other implication. For ←, in the exercise is given the suggestion: if ##\{ψ_1,...,ψ_n\}## is a basis of ##V^*##, one can try to show that the matrix obtained from writing ##\phi_1,...,\phi_n## in coordinates with respect to the basis ##\{ψ_1,...,ψ_n\}##, is an invertible matrix, I am stuck and I haven't got a clue on how to use this idea, I would like any hints or another idea to prove the remaining implication.
An alternative approach is to assume that [itex]\Phi = \{\phi_1, \dots, \phi_n\}[/itex] is not a basis for [itex]V^{*}[/itex], and exhibit a non-zero [itex]v \in \bigcap_i \ker(\phi_i)[/itex].

To that end: let [itex]\Phi' \subset \Phi[/itex] be a linearly-independent subset of [itex]\Phi[/itex] of maximum dimension. This [itex]\Phi'[/itex] can then be extended to a basis [itex]B[/itex] of [itex]V^{*}[/itex]. Now consider the dual basis of [itex]B[/itex].
 
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Related to Dual basis and kernel intersection

1. What is a dual basis?

A dual basis is a set of vectors that form a basis for the dual space of a given vector space. It is defined as the set of linear functionals that map each basis vector to 1 and every other vector to 0.

2. How is a dual basis related to a given basis?

A dual basis is closely related to a given basis as it is constructed using the same vectors as the given basis. However, the dual basis is used to define linear functionals that map the given basis vectors to specific values, whereas the given basis is used to define the vectors in the vector space.

3. What is the intersection of a dual basis and a kernel?

The intersection of a dual basis and a kernel is the set of vectors that are mapped to 0 by all the linear functionals in the dual basis. This intersection is also known as the annihilator of the dual basis.

4. Can a dual basis and a kernel intersection be empty?

Yes, it is possible for a dual basis and a kernel intersection to be empty. This occurs when there are no linear functionals that map the given basis vectors to 0, meaning that the kernel is equal to the entire vector space and there is no intersection with the dual basis.

5. How is the dual basis and kernel intersection used in mathematics?

The dual basis and kernel intersection is used in mathematics to define the orthogonal complement of a subspace. It is also used in linear algebra and functional analysis to prove theorems and solve problems related to vector spaces and linear transformations.

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