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 $$U_1=U_2 \iff ( S_1\subset U_2 \and S_2 \subset U_1 ) \ $$.

($$\Rightarrow$$)

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 $$U_1=U_2, (S_1\subset U_1 \implies S_1\subset U_2) \and (S_2\subset U_2 \implies S_2\subset U_1) \ $$.

($$\Leftarrow$$)

Let $$(S_1\subset U_2) \and (S_2 \subset U_1) \ $$. 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.