User:J D Bowen/Math725 HW2
1) In $\mathbb{R}^4 \ $, let $U_1=\left\{{ (x,y,z,w) : x+y+z+w=x-y+z-w=0 }\right\}, U_2=\left\{{(x,y,z,w):x+z=y-w=0}\right\} \ $.
Then the vectors $\vec{a}=(1,0,-1,0) \ $ and $\vec{b}=(-1,0,1,0) \ $ have the property that $\vec{a}\in U_1, \vec{b}\in U_2, \vec{a}\neq 0, \vec{b}\neq 0, \vec{a}+\vec{b} =\vec{0} \ $. This means that $U_1 + U_2 \ $ is not an internal direct sum.
Consider the set $U_3 = \left\{{ (x,y,z,w): x=z=0, y=w }\right\} \ $ and let $\vec{a}\in U_1, \vec{b}\in U_3 \ $ be any vectors satisfying $\vec{a}+\vec{b} =\vec{0} \ $. Observe that $U_3 \ $ is a subspace, since it is closed under addition.
Now suppose that $\vec{b} \neq \vec{0} \ $. Then $\vec{b}=(0,b,0,b), b\neq 0 $. Then since $\vec{a}+\vec{b}=0, \vec{a}=(0,-b,0,-b) \ $. But then $(x+y+z+w)_{\vec{a}} = -2b \neq 0 \ $, so $\vec{a}\notin U_1 \ $, a contradiction. Hence $\vec{b}=\vec{0} \ $. Since we assumed $\vec{a}+\vec{b}=\vec{0}, \vec{b}=\vec{0}\implies\vec{a}=\vec{0} \ $. Hence $U_1+U_3 \ $ is an internal direct sum.
2) Let $S \ $ be a subset of $V \ $, and let $U \ $ be a subspace of $V \ $. Suppose $S\subset U \ $. Let $\vec{a}_1, \dots, \vec{a}_n \ $ be any finite collection of vectors in $S \ $. Since $S \subset U \ $, each vector $\vec{a}_1, \dots, \vec{a}_n \in U \ $. Since $U \ $ is a subspace, any sum $c_1\vec{a}_1 + \dots c_n\vec{a}_n \in U \ $. Since $\text{span}(S) \ $ is composed exclusively of such vectors, $\text{span}(S)\subset U \ $.
3) Let $S_1, S_2 \subset V, U_1 =\text{span}(S_1), U_2=\text{span}(S_2) \ $.
a) We aim to show that $U_1=U_2 \iff ( S_1\subset U_2 \land S_2 \subset U_1 ) \ $.
($\Rightarrow$)
Let $U_1=U_2 \ $. We have $U_1 =\text{span}(S_1)\implies S_1\subset U_1, U_2 =\text{span}(S_2)\implies S_2\subset U_2 \ $. Since $U_1=U_2, (S_1\subset U_1 \implies S_1\subset U_2) \land (S_2\subset U_2 \implies S_2\subset U_1) \ $.
($\Leftarrow$)
Let $(S_1\subset U_2) \land (S_2 \subset U_1) \ $. Since $U_2 \ $ is a vector space, every linear combination of vectors from $U_2 \ $ is in $U_2 \ $. Therefore, since $S_1\subset U_2 \ $ and since the span of $S_1 \ $ is all linear combinations of vectors from $S_1 \ $, we can be sure that $\text{span}(S_1)\subset U_2 \ $. But $\text{span}(S_1)=U_1 \ $, so $U_1 \subset U_2 \ $.
If we replace all 1s with 2s and all 2s with 1s, this argument shows that $U_2 \subset U_1 \ $. These two facts together imply $U_1=U_2 \ $.
b) Consider the set $\text{span}(S_1\cup S_2) = \left\{{\vec{x}\in V: \vec{x}=\Sigma c_i \vec{a}_i +\Sigma d_j\vec{b}_j }\right\} \ $, where $\vec{a}_i\in S_1, \vec{b}_i\in S_2 \ $ and the c, d are elements of the field V is over. Of course, since $U_1=\text{span}(S_1)=\left\{{\vec{x}\in V:\vec{x}=\Sigma c_i \vec{a}_i}\right\} \ $ and $U_2=\text{span}(S_2)=\left\{{\vec{x}\in V:\vec{x}=\Sigma d_j \vec{b}_j}\right\} \ $, this is just
$\text{span}(S_1\cup S_2) = \left\{{\vec{x}\in V: \vec{x}=\vec{\alpha}+\vec{\beta} }\right\} \ $, where $\vec{\alpha}\in U_1, \vec{\beta}\in U_2 \ $. But that is simply $U_1+U_2 \ $.
c) Let $V=\mathbb{R}^2, S_1=\left\{{ (0,1) }\right\}, S_2 = \left\{{(1,0)}\right\} \ $. Then $U_1 \ $ is the line $x=0 \ $ and $U_2 \ $ is the line $y=0 \ $. Then we have $S_1 \cap S_2 = \varnothing \ $ but $U_1 \cap U_2 = \left\{{ (0,0) }\right\} $.
4) Define $x^{[i]} = \frac{1}{i!}x^i \ $.
a) Note that $x^{[0]} = \frac{1}{0!} x^0 = \frac{1}{1} 1 = 1 \ $ and $x^{[1]} = \frac{1}{1!}x^1=x \ $.
b) Further note that $\frac{d}{dx} x^{[i]} = \frac{d}{dx} \left({ \frac{1}{i!} x^i }\right) = \frac{i}{i!}x^{i-1} = \frac{1}{(i-1)!}x^{i-1} = x^{[i-1]} \ $.
c) Let $f\in\text{span}(1,x, x^2, \dots, x^n) \ $. Then by the definition of span there are constants $a_0, \dots, a_n \ $ such that $f(x)=a_0+a_1x+a_2x^2+\dots+a_nx^n \ $. Define $c_j =j!a_j \ $. Then we have $c_jx^{[j]} = j!a_j \frac{1}{j!}x^j = a_j x^j \ $. The sum $c_0x^{[0]}+\dots+c_nx^{[n]} \ $ is clearly in $\text{span}(1, x^{[1]}, \dots, x^{[n]} ) \ $ and so $f\in\text{span}((1,x, x^2, \dots, x^n) \implies f\in\text{span}(1, x^{[1]}, \dots, x^{[n]} ) \ $.
Similarly, for any function $f=a_0+a_1x+a_2x^{[2]}+\dots+a_nx^{[n]}\in\text{span}(1, x^{[1]}, \dots, x^{[n]} ) $, form the field elements $b_j=\frac{a_j}{j!} \ $. Then we have $b_jx^j = \frac{a_j}{j!}j!x^j \ $. Since the sum $b_0+b_1x+b_2x^2+\dots+b_nx^n \in \text{span}((1,x, x^2, \dots, x^n) \ $, we have $f\in\text{span}(1, x^{[1]}, \dots, x^{[n]} ) \implies f\in\text{span}((1,x, x^2, \dots, x^n) \ $.
Hence $\text{span}(1, x^{[1]}, \dots, x^{[n]} ) = \text{span}((1,x, x^2, \dots, x^n) \ $.
d) Let $P_5 = \left\{{a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5 : a_i\in\mathbb{C} }\right\}, f_i =1+x+\dots+x^{[i]}, U_1=\text{span}(f_0,f_1,f_2), U_2=\text{span}(f_3,f_4,f_5) \ $.
By 3b, we have $U_1+U_2=\text{span}(f_0,f_1,f_2,f_3,f_4,f_5) \ $.
For any function $f=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5\in P_5 \ $, define $b_j=j!a_j \ $.
Now define $c_5= b_5, c_4=b_4-b_5, c_3=b_3-b_4, c_2=b_2-b_3, c_1=b_1-b_2, c_0=b_0-b_1 \ $.
Then $c_0f_0+c_1f_1+c_2f_2+c_3f_3+c_4f_4+c_5f_5= \ $
$(c_0)+(c_1+c_1x)+(c_2+c_2x+c_2x^{[2]})+(c_3+c_3x+c_3x^{[2]}+c_3x^{[3]})+(c_4+c_4x+c_4x^{[2]}+c_4x^{[3]}+c_4x^{[4]})+(c_5+c_5x+c_5x^{[2]}+c_5x^{[3]}+c_5x^{[4]}+c_5x^{[5]}) \ $
$(c_0+c_1+c_2+c_3+c_4+c_5)+(c_1+c_2+c_3+c_4+c_5)x+(c_2+c_3+c_4+c_5)x^{[2]}+(c_3+c_4+c_5)x^{[3]}+(c_4+c_5)x^{[4]}+c_5x^{[5]} \ $
$=b_0+b_1x+b_2x^{[2]}+b_3x^{[3]}+b_4x^{[4]}+b_5x^{[5]}=a_0+a_1x+a_2x^2+a_3x^4+a_5x^5 = f \ $.
Hence $f\in P_5 \implies f\in U_1+U_2 \ $. Since the Us are subspaces, they cannot have a sum larger than their containing space, and so $P_5=U_1+U_2 \ $.
Now let $\vec{a}\in U_1, \vec{b}\in U_2 \ $ be any vectors, and suppose $\vec{a}+\vec{b}=\vec{0} \ $. Since there is no vector in $U_1 \ $ which contains terms of $ x^5 \ $, the only term for this power comes from $\vec{b} \ $. Since the sum $\vec{a}+\vec{b} =\vec{0} \ $, we must have $\vec{b}=c_3f_3+c_4f_4+0f_5 \ $, otherwise we would have a non-zero term for $x^5 \ $ in the sum $\vec{a}+\vec{b} \ $. But of these, the only possible contribution of a $x^4 \ $ term must come from $c_4f_4 \ $, and since there are no $x^4 \ $ terms in the sum $\vec{a}+\vec{b} \ $, we must have $\vec{b}=c_3f_3 \ $. But then the only possible contribution of a $x^3 \ $ term in the sum $\vec{a}+\vec{b} \ $ must come from $c_3f_3 \ $, and since there are no $x^3 \ $ terms, we must have $\vec{b}=\vec{0} \ $. Since $\vec{a}+\vec{b}=\vec{a}+\vec{0}=\vec{0} \ $, we must have $\vec{a}=0 \ $, and so $U_1+U_2 \ $ is an internal direct sum.