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) Suppose $\left\{{v_1, \dots, v_s, w_1, \dots, w_t }\right\} \ $ is a basis for $U_1+U_2 \ $. Then $\text{dim}(U_1+U_2)=s+t \ $. But we also have $\text{dim}(U_1+U_2)=\text{dim}(U_1)+\text{dim}(U_2)-\text{dim}(U_1\cap U_2)=(s)+(t)+\text{dim}(U_1\cap U_2) \ $, so $\text{dim}(U_1\cap U_2) = 0 \ $, implying $U_1\cap U_2=\left\{{0}\right\} \ $ and so $U_1+U_2 \ $ is an internal direct sum.

We could just as easily suppose that $U_1+U_2 \ $ is an internal direct sum; then $U_1\cap U_2=\left\{{0}\right\} \ $ and so $\text{dim}(U_1\cap U_2) = 0 \ $. Then the $v \ $ and the $w \ $ vectors must be linearly independent, and since together they clearly span $U_1+U_2 \ $, the must form a basis.

(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.