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I 1+121+\frac{1}{2} 22\frac{2}{2} กั 48\frac{4}{8} 올 ≓\risingdotseq Q5. Bo...
Apr 28, 2024
I 1+121+\frac{1}{2} 22\frac{2}{2} กั 48\frac{4}{8} 올 ≓\risingdotseq Q5. Bonus marks. Challenge question - this question is optional. Some aspects of the puestion ary beyond the scope of examinable content in this course. It inll only be trorth at mast a cowell marks, and you need to have this question almost complefely correct to obtain any marks at all. The marks that you do earn on this question seill only serve to make up for marks lost on the other question in this assigmment So far in this course, you have soen an abstract definition of a vector space. If you have not already, you will soon see the definition of linear transformations which are the functions of vector spaces that preserve their linear structure? In mathematics, this idea of studying objects (eg. vector spaces) and the functions between them (eg. linear transformations) that preserve a certain structure is ubiquitous. For example, the study of category theory studies, in a very abstract way, these types of mathematical structures. In this course, you have seen the definition of a basis for a wetor space as a linearly independent spanning set. This definition uses the terminology in linear algebra does not lend itself to further abstraction. In category theory, constructions like that of a basis are better described of basis. To answer the following question, you don't need to know what category theory, of a univenal property is. The background above is intended to provide students with an opportunity to do some research and light reading to see one direction where the kind of abetraction in this course can lead. The question: Let BB be a subet of a vector space VV over an arbitrary field FF, Prove that BB is a basis for VV if and only if for every F\mathrm{F}-vector space WW, and every (ect) function f:B→Wf: B \rightarrow W, there exists a unique extension of ff to a linear transformation fE:V→W.  f_{E}: V \rightarrow W \text {. } Being an extension means that fS(v)=f(v)f_{S}(\mathrm{v})=f(\mathrm{v}) for every v∈B\mathrm{v} \in B. Remark: This property of extending set functions to linear transformations is an example of a universal property, and gives another way of defining a basis of a vector space without Ecettioning linear independence or spansing sets. It also formalises the idea that to define a limear transformation, it is enough to say where the basis vectors go. haven't sen it yet, look ahoud in your notes. 13 of 22
Solution by Steps
step 1
To prove that BB is a basis for VV if and only if for every FF-vector space WW, and every function f:B→Wf: B \rightarrow W, there exists a unique extension of ff to a linear transformation fE:V→Wf_{E}: V \rightarrow W, we first assume that BB is a basis for VV
step 2
Since BB is a basis, it is a linearly independent set that spans VV. This means any vector in VV can be uniquely expressed as a linear combination of vectors in BB
step 3
Given a function f:B→Wf: B \rightarrow W, we can define an extension fE:V→Wf_{E}: V \rightarrow W by setting fE(v)=f(b1)c1+f(b2)c2+…+f(bn)cnf_{E}(v) = f(b_1)c_1 + f(b_2)c_2 + \ldots + f(b_n)c_n for any v∈Vv \in V where v=b1c1+b2c2+…+bncnv = b_1c_1 + b_2c_2 + \ldots + b_nc_n and bi∈Bb_i \in B, ci∈Fc_i \in F
step 4
This extension is unique because if there were another extension g:V→Wg: V \rightarrow W such that g(b)=f(b)g(b) = f(b) for all b∈Bb \in B, then for any v∈Vv \in V, g(v)g(v) must equal fE(v)f_{E}(v) due to the uniqueness of the representation of vv in terms of the basis BB
step 5
Conversely, assume that for every FF-vector space WW, and every function f:B→Wf: B \rightarrow W, there exists a unique extension of ff to a linear transformation fE:V→Wf_{E}: V \rightarrow W. We need to show that BB is a basis for VV
step 6
If BB were not linearly independent, then there would exist a non-trivial linear combination of vectors in BB that equals the zero vector, which would contradict the uniqueness of the extension of ff to fEf_{E}
step 7
If BB did not span VV, then there would exist a vector v∈Vv \in V not expressible as a linear combination of vectors in BB, which would contradict the existence of the extension of ff to fEf_{E} for any function f:B→Wf: B \rightarrow W
step 8
Therefore, BB must be a linearly independent set that spans VV, which means BB is a basis for VV
Answer
The statement is proven; BB is a basis for VV if and only if for every FF-vector space WW, and every function f:B→Wf: B \rightarrow W, there exists a unique extension of ff to a linear transformation fE:V→Wf_{E}: V \rightarrow W.
Key Concept
Universal Property of a Basis
Explanation
The universal property of a basis in a vector space is that any function defined on the basis can be uniquely extended to a linear transformation on the entire vector space. This property characterizes a basis without explicitly mentioning linear independence or spanning sets.
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