A bag contains one red pen, four black pens, and three blue pens. two pens are randomly chosen from the bag and are not replaced
2 answers:
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Answer:
1. Yes.
2. No.
3. Yes.
Step-by-step explanation:
Consider the following subsets of Pn given by
1.Let W1 be the set of all polynomials of the form , where a is in ℝ.
2.Let W2 be the set of all polynomials of the form , where a is in ℝ.
3. Let W3 be the set of all polynomials of the form , where a is in ℝ.
Recall that given a vector space V, a subset W of V is a subspace if the following criteria hold:
- The 0 vector of V is in W.
- Given v,w in W then v+w is in W.
- Given v in W and a a real number, then av is in W.
So, for us to check if the three subsets are a subset of Pn, we must check the three criteria.
- First property:
Note that for W2, for any value of a, the polynomial we get is not the zero polynomial. Hence the first criteria is not met. Then, W2 is not a subspace of Pn.
For W1 and W3, note that if a= 0, then we have p(t) =0, so the zero polynomial is in W1 and W3.
- Second property:
W1. Consider two elements in W1, say, consider a,b different non-zero real numbers and consider the polynomials
.
We must check that p_1+p_2(t) is in W1.
Note that
Since a+b is another real number, we have that p1(t)+p2(t) is in W1.
W3. Consider two elements in W3. Say . Then
So, again, p1(t)+p2(t) is in W3.
- Third property.
W1. Consider an element in W1 and a real scalar b. Then
.
Since (ba) is another real scalar, we have that bp(t) is in W1.
W3. Consider an element in W3 and a real scalar b. Then
.
Since (ba) is another real scalar, we have that bp(t) is in W3.
After all,
W1 and W3 are subspaces of Pn for n= 2
and W2 is not a subspace of Pn.