User:J D Bowen/Math725 HW3

(1) (a) Recall (4d) from the previous homework:

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 $$f\in P_5 \implies f\in \text{span}(f_0, \dots, f_5) \ $$. So $$P_5 \subset \text{span}(f_0, \dots, f_5) \ $$

We certainly have $$\text{span}(f_0, \dots, f_5) \subset P_5 \ $$, since $$\forall i, f_i \in P_5 \ $$. So $$\text{span}(f_0, \dots, f_5)=P_5 \ $$.

Now suppose we have $$\Sigma a_i f_i(z) = 0 \ \forall z\in\mathbb{C} \ $$. Then we have

$$\Sigma a_if_i(z) \ $$

$$=(a_0+a_1+a_2+a_3+a_4+a_5)+(a_1+a_2+a_3+a_4+a_5)z+(a_2+a_3+a_4+a_5)z^{[2]}+(a_3+a_4+a_5)z^{[3]}+(a_4+a_5)z^{[4]}+a_5z^{[5]}=0 \ $$

Of course, this implies $$a_5=0 \ $$, which causes a chain of implications $$\implies a_4=0 \implies a_3=0 \implies a_2=0 \implies a_1=0 \implies a_0=0 \ $$. Hence, the $$f_i \ $$ are linearly independent. Therefore, they constitute a basis for $$P_5 \ $$, implying $$\text{dim}(P_5)=6 \ $$

(b) Since all the $$f_i \ $$ are linearly independent, some of them certainly are. Since $$\left\{{f_0,f_1, f_2 }\right\} \ $$ is a linearly independent set, setting $$U_1 =\text{span}(\left\{{f_0,f_1, f_2 }\right\}) \ $$ means $$\text{dim}(U_1)=3 \ $$.

We also have $$a_2x^2+a_3x^3+a_4x^4+a_5x^5=0 \implies a_i=0, \ i\in\left\{{2,3,4,5}\right\} \ $$. Hence, this set $$\left\{{x^2, x^3, x^4, x^5 }\right\} \ $$ is linearly independent, and so setting $$U_2 =\text{span}(\left\{{x^2,x^3,x^4,x^5 }\right\}) \ $$ means $$\text{dim}(U_2) = 4 \ $$.

$$\text{dim}(U_1+U_2) = \text{dim}(\text{span}(f_0,f_1, f_2, x^2, x^3, x^4, x^5)) \ $$. We can use a proof similar to part a to demonstrate that we can form any quadratic function from $$f_0,f_1,f_2 \ $$, and of course by adding multiples of $$x^3, x^4, x^5 \ $$ we can them form any function in $$P_5 \ $$. Hence, $$U_1+U_2 = P_5 \ $$, so $$\text{dim}(U_1+U_2) = \text{dim}(P_5) = 6 \ $$.

Let us examine the content of $$U_1 \cap U_2 \ $$ by setting $$g\in U_1, h\in U_2 \ $$. Then $$g(x)=g_0f_0 + g_1f_1+g_2f_2 = g_0 + g_1 + g_1 x+ g_2 +g_2x+g_2\tfrac{x^2}{2} \ $$, and $$h(x)=h_2x^2+h_3x^3+h_4x^4+h_5x^5 \ $$.

Setting $$g=h \implies g_2/2 = h_2 \ $$ and all other constants are zero. Hence, $$U_1\cap U_2 = \left\{{ax^2:a\in\mathbb{C} }\right\} = \text{span}(x^2) \ $$ and so $$\text{dim}(U_1\cap U_2) = 1 \ $$.

We have $$\text{dim}(U_1)+\text{dim}(U_2)= 3+4 =7 \ $$ and $$\text{dim}(U_1+U_2)+\text{dim}(U_1\cap U_2)=6+1=7 \ $$.

(c) In part b we constructed a basis for $$U_1\cap U_2 \ $$, namely, $$\left\{{x^2}\right\} \ $$, and in part a, we demonstrated that $$\left\{{f_0, \dots, f_5 }\right\} \ $$ is a basis for $$U_1+U_2=P_5 \ $$.

(2) (a) This is precisely (3b) from the previous homework. Define $$S_1=\left\{{v_1, \dots, v_s}\right\}, S_2 = \left\{{w_1, \dots, w_t }\right\} \ $$.

Consider the set $$\text{span}(S_1\cup S_2) = \left\{{\vec{x}\in V: \vec{x}=\Sigma c_i \vec{v}_i +\Sigma d_j\vec{w}_j }\right\} \ $$, where $$\vec{v}_i\in S_1, \vec{w}_i\in S_2 \ $$ and the c, d are elements of the field V is over. Of course, since $$U_1=\text{span}(S_1)=\left\{{\vec{x}\in V:\vec{x}=\Sigma c_i \vec{v}_i}\right\} \ $$ and $$U_2=\text{span}(S_2)=\left\{{\vec{x}\in V:\vec{x}=\Sigma d_j \vec{w}_j}\right\} \ $$, 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 $$U_1+U_2 \ $$.

(b) We aim to show that $$\left\{{v_1, \dots, v_s, w_1, \dots, w_t }\right\} \ $$ is a basis for $$U_1+U_2 \ $$ if and only if $$U_1+U_2 \ $$ is an internal direct sum.

(3) Let $$\vec{x}\in\mathbb{C}^n \ $$ be any vector. Then it has a representation $$\vec{x}=(z_1, \dots, z_n) \ $$. Let $$f_i=(0, \dots, 1, \dots, 0) \ $$, where the one occurs in the ith position.

Then $$z_1f_1+\dots+z_nf_n=(z_1, \dots, z_n) \ $$, so any vector in $$\mathbb{C}^n \ $$ can be represented as a linear combination of $$f_i \ $$. Hence $$\text{span}(\left\{{f_i}\right\} \ $$.

Now suppose that $$\Sigma a_i f_i = \vec{0} \ $$. Then $$(a_1, \dots, a_n)=(0,\dots,0) \ $$. This implies $$a_i=0 \ \forall i \ $$. Therefore, $$\left\{{f_i }\right\} \ $$ is linearly independent.

Hence, $$\left\{{f_i }\right\} \ $$ is a basis for $$\mathbb{C}^n \ $$.

(4) Assume that $$P_\infty \ $$, the set of all polynomials over $$\mathbb{C} \ $$, is finite dimensional. Then there exists a spanning set of polynomials $$\left\{{f_1, ..., f_n}\right\} \ $$ for the space. For any polynomial $$f_i(x) \ $$ in the spanning set, define $$\text{deg}:\left\{{f_i}\right\} \to \mathbb{N} \ $$ to be the degree of that polynomial, that is, the largest power of the variable appearing in the polynomial. Then we can form the finite set $$D=\left\{{\text{deg}(f_1), \dots, \text{deg}(f_n) }\right\} \ $$. Since this set is finite, it has a maximum element $$N \ $$. Let $$g(x)=x^{N+1} \ $$. Certainly we have $$g\in P_\infty \ $$, but notice that no linear combination of $$f_i \ $$ can yield this function, since it contains a higher power of $$x \ $$ than any of them. Hence $$\left\{{f_i}\right\} \ $$ does not form a spanning set, contradicting our assumption that such a set exists. Therefore, $$P_\infty \ $$ is not finite dimensional.