Vector Space Problems

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Exercise Show that form a linearly independent set of vectors in , viewed as a vector space over .

Proof:
Recall that the set of vectors in a vector space (over a field ) are said to be linearly independent if whenever are scalars in such that then . So for this problem, since we’re considering the complex numbers as a vector space over , we must show that whenever and then . Rearranging the above equation, we obtain Now, a complex number is equal to if and only if its real and imaginary parts are both . So in this case, we conclude that This implies , so that , which yields . Thus we conclude the vectors are linearly independent in (over ).


Exercise Show that form a linearly independent set of vectors in , viewed as a vector space over .

Proof:
Recall that a set of vectors in a vector space (over a field ) is said to be linearly dependent if they are not linearly independent. More concretely, these vectors are linearly dependent if we can find scalars not all equal to zero such that

So for this problem, to show that and are not linearly dependent over , all we need to do is exhibit two complex scalars and that are not both zero such that There are many choices for and , but one such example is and .


Exercise Let be a vector space over a field . If are a linearly independent set of vectors, then show that also form a linearly independent set of vectors in .

Proof:
Recall that the set of vectors in a vector space (over a field ) are said to be linearly independent if whenever are scalars in such that then .

So for this problem, we must show that whenever and we have that After rearranging terms in the above equation, we have that Now since the vectors are linearly independent in by assumption, we have that

In other words, , so that form a linearly independent set as desired.


Exercise Prove that a vector space over a field is infinite-dimensional if and only if there is a sequence in such that is linearly independent for every .

Proof:
Recall that a vector space is said to be finite dimensional if it is spanned by a finite list of vectors In other words, has finite dimension if every vector in may be written as a linear combination of some list of vectors . On the other hand, a vector space is infinite dimensional if it is not finite dimensional, i.e., cannot be spanned by a finite list of vectors. Now before we proceed in the proof, we will need the following fact:

Lemma Suppose is a vector space over a field , and are vectors that span . If in are linearly independent, then .

We are ready now to proceed with the proof:
’: Suppose that is an infinite dimensional vector space. Then, in particular, , so that there is some in . Then is a linearly independent vector in . By way of induction now, suppose that for some , we have produced vectors such that are linearly independent. Since is infinite-dimensional, it cannot be spanned by the (finite!) list of vectors . Thus we have that there is some such that We claim that now that form a linearly independent set in . To see this, suppose that Now if , then we may re-write the above equation as contradicting the fact that is not in the span of . So we conclude , and thus we have that Now by induction hypothesis, since are linearly independent, we must have are all zero. Thus we’ve shown that also form a linearly independent set, completing the induction. Thus we have constructed a sequence of vectors in so that is linearly independent for each .

’: On the other hand, suppose that contains a sequence of vectors so that is linearly independent for each . By way of contradiction, let’s suppose is not infinite dimensional, i.e. is finite dimensional. Then can be spanned by a finite list of vectors .
Now, since contains a linearly independent set of .


Exercise Suppose that and are subspaces of a vector space . Prove that is a subspace of if and only if or .

Proof:
Recall that a subset of a vector space is a subspace of if itself is a vector space with the same addition and scalar multiplication operations as .

’: Instead of proving that is a subspace of implies or , we’ll show the contrapositive of this statement. That is, if and , then is not a subspace of . So suppose there is some that is not in , and likewise that there is some that is not in . We claim that . For if it were, then would lie in either or . If , then since is a subspace, this would imply contradicting our choice of . Likewise, if , this would yield , which is again a contradiction. So we conclude that , and thus fails to be closed under addition, so cannot be a subspace of .
’: Suppose now that or . Then is equal to either or respectively, which, by assumption are subspaces of .


Before we begin the next exercise, we will need the following notation: for an arbitrary non-empty set , let denote the set of all functions . Then is always a vector space, with addition and scalar multiplication defined pointwise.

Exercise Let and consider the set Show that is a subspace of if and only if .

Proof:
Recall that a subset of a vector space is a subspace of if itself is a vector space with the same addition and scalar multiplication operations as . There is a very convenient test that determines if is a subspace of , sometimes called the subspace test. It says the following:

(Subspace Test) Suppose that , where is a vector space over a field . Then is a subspace of if and only if the following conditions are met:

  1. ,
  2. .

We are now ready to proceed with the proof:

’: Suppose is a subspace of . Then by condition of the subspace test, contains the zero vector of , which is just the function that maps to for all . We will write this zero vector as . Now since , by definition of being in , we must have that On the other hand, when we actually integrate , we find the integral must be zero. Thus as desired.
’: Say . We will show that is a subspace of by showing that it passes all three conditions of the subspace test above. For condition (1), just note that by our previous remark, , and since is continuous, we have that . For condition (2), suppose that and . We must show that . Since a continuous function multiplied by a constant is still continuous, is still a continuous function. Now, so that we conclude . Lastly, for condition (3), we must show that if , then . The addition of two continuous functions is always continuous, so that is continuous. Now since , we have that so that , and thus satisfies all three conditions of the subspace test.


Exercise Prove or give a counterexample to the following statement: If are subspaces of a vector space with then .

Proof:
Let’s think about what it means for two subspaces of a vector space to satisfy . This means that and that . In other words, for any , we may write uniquely in the form , where and .

It turns out the statement of the problem is false, so that we must provide a counterexample to this statement: Let and consider its subspaces (one should check that they actually form subspaces first):

Then and are not the same subspaces of , so that all we need to check is Suppose that . Then , where and , so that . Now if , then for some , and thus . Likewise, for some , so that . This shows , and hence .
Now again say . Then , where and so that . Now suppose . Then for some , so that . Likewise, for some , thus , so that we conclude , and thus Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V = U_{2} \oplus W} .