User:J D Bowen/Math725 HW4

1) Let $$V, W \ $$ be finite-dimensional vector spaces. We aim to show that the following three statements are equivalent:
 * a) $$\text{dim}(V)\geq \text{dim}(W) \ $$
 * b) There is a surjective linear map $$V \to W \ $$.
 * c) There is an injective linear map $$W \to V \ $$.

(a)$$\implies \ $$ (b):

Suppose we have $$\text{dim}(V)\geq \text{dim}(W) \ $$. There exists basis sets for these spaces $$\left\{{\vec{v_1}, \dots, \vec{v_m}}\right\}\subset V, \left\{{\vec{w_1}, \dots, \vec{w_n} }\right\} \subset W \ $$, where $$m=\text{dim}(V), n=\text{dim}(W) \ $$. We have $$m\geq n \ $$.

Define a linear transformation $$\phi:V\to W \ $$ as $$\sum_{i=1}^m a_i\vec{v_i} \mapsto \sum_{i=1}^n a_i\vec{w_i} \ $$. Since $$m\geq n \ $$, every vector $$\vec{w}\in W \ $$ has a preimage in $$V \ $$. Hence $$\phi \ $$ is a surjection.

(b)$$\implies \ $$ (a):

Suppose there exists at least one surjective linear map $$\phi:V\to W \ $$. If we let $$\left\{{\vec{v_1}, \dots, \vec{v_m}}\right\}\subset V \ $$ be any basis for $$V \ $$, we can be sure that every vector in $$V \ $$ can be represented as $$\vec{v} = \sum_{i=1}^m a_i\vec{v_i} \ $$ for some constants $$a_i \ $$.

Since $$\phi \ $$ is linear, we have

$$\phi(\vec{v}) = \sum_{i=1}^m a_i\phi(\vec{v_i}) \ $$.

Since $$\phi \ $$ is surjective, for every $$\vec{w}\in W \ $$, there is at least one $$\vec{v}\in V \ $$ such that $$\phi(\vec{v})=\vec{w} \ $$. Therefore, for any $$\vec{w}\in W \ $$, there are constants $$a_i \ $$ such that

$$\vec{w}=\sum_{i=1}^m a_i\phi(\vec{v_i}) \ $$.

Hence the $$\phi(\vec{v_i}) \ $$ form a spanning set for $$W \ $$, implying that $$\text{dim}(W)\leq \text{dim}(V) \ $$.

(a)$$\implies \ $$ (c):

Suppose we have $$\text{dim}(V)\geq \text{dim}(W) \ $$. There exists basis sets for these spaces $$\left\{{\vec{v_1}, \dots, \vec{v_m}}\right\}\subset V, \left\{{\vec{w_1}, \dots, \vec{w_n} }\right\} \subset W \ $$, where $$m=\text{dim}(V), n=\text{dim}(W) \ $$. We have $$m\geq n \ $$.

Define a linear transformation $$\phi:W\to V \ $$ as $$\sum_{i=1}^n a_i\vec{w_i} \mapsto \sum_{i=1}^n a_i\vec{v_i} \ $$. Since $$n\leq m \ $$, every $$\vec{v}\in V \ $$ has at most one pre-image (1 or 0, depending on whether or not $$\vec{v} \ $$ is in the subspace $$a_{n+1}=\dots=a_m =0 \ $$ or not). Hence $$\phi \ $$ is an injection.

(c)$$\implies \ $$(a):

Suppose there exists at least one injective linear map $$\phi:W\to V \ $$. Let a basis for $$W \ $$ be $$\left\{{\vec{w_1}, \dots, \vec{w_n} }\right\} \subset W \ $$.

We have $$\sum_{i=1}^n a_i\vec{w_i}=\vec{0} \implies (\forall i, \ a_i=0) \ $$.

Since $$\phi \ $$ is linear, we have two important facts:


 * 1) $$\phi(\vec{0_W})=\vec{0_V} \ $$


 * 2) $$\phi \left({\sum_{i=1}^n a_i\vec{w_i} }\right) = \sum_{i=1}^n a_i\phi(\vec{w_i}) \ $$

Now, since $$\phi \ $$ is an injection, for $$\vec{u},\vec{v}\in V, \ \vec{u}\neq\vec{v} \ $$, we have $$\phi(\vec{u})\neq\phi(\vec{v}) \ $$. This, combined with fact 1, means that the  only  vector which maps to $$\vec{0_V} \ $$ is $$\vec{0_W} \ $$.

Now we can use fact 2 to investigate if the image of the basis vectors in $$W \ $$ is linearly independent in $$V \ $$:

$$\sum_{i=1}^n a_i\phi(\vec{w_i}) = \vec{0_V} \implies \phi \left({\sum_{i=1}^n a_i\vec{w_i} }\right) = \vec{0_V} \implies \sum_{i=1}^n a_i\vec{w_i}=\vec{0_W} \implies (\forall i, \ a_i=0) $$

Therefore, the $$\phi(\vec{w_i}) \ $$ are linearly independent in $$V \ $$, which means $$\text{dim}(V)\geq n = \text{dim}(W) \ $$.

3) Let $$S,T \ $$ be linear operators on a vector space $$V \ $$.

a) Let $$\vec{x}\in\text{ker}(T) \ $$, that is, $$T(\vec{x})=\vec{0} \ $$. Then $$ST(\vec{x})=S(T(\vec{x})) = S(\vec{0}) \ $$.

But $$S \ $$ is linear, so $$S(\vec{y})=S(\vec{0}+\vec{y})=S(\vec{0})+S(\vec{y}) \implies S(\vec{0})=\vec{0} \ $$.

Hence $$T(\vec{x})=\vec{0}\implies ST(\vec{x})=\vec{0} \ $$, so $$\text{ker}(T)\subseteq \text{ker}(ST) \ $$.

b) Let $$T:\mathbb{R}^2 \to \mathbb{R}^2 \ $$ be the linear map defined $$x\vec{e}_1+y\vec{e_2}\mapsto -y\vec{e}_1+x\vec{e_2} \ $$, and let $$S:\mathbb{R}^2\to\mathbb{R}^2 \ $$ be defined $$x\vec{e}_1+y\vec{e_2}\mapsto x\vec{e}_1 \ $$. Then $$\text{ker}(S)=\left\{{ c\vec{e_2}:c\in\mathbb{R} }\right\}, \text{ker}(ST)=\left\{{c\vec{e_1}:c\in\mathbb{R}}\right\} \ $$, and so $$\text{ker}(S)\not\subset\text{ker}(ST) \  $$.

c) Suppose $$\exists \vec{y}\in\text{Im}(ST) \ $$.  Then $$\exists \vec{x}:ST(\vec{x})=\vec{y} \ $$, observe that

$$ST(\vec{x})=S( T(\vec{x})) \ $$

So there exists a $$T(\vec{x}) \ $$ such that $$\vec{y}=S(T(\vec{x})) \ $$. Hence, $$y\in\text{Im}(S) \ $$.

We have $$\text{Im}(ST)\subseteq\text{Im}(S) \ $$.

d) Let $$T:\mathbb{R}^2 \to \mathbb{R}^2 \ $$ be the linear map defined $$x\vec{e}_1+y\vec{e_2}\mapsto x\vec{e}_1 \ $$, and let $$S:\mathbb{R}^2\to\mathbb{R}^2 \ $$ be defined $$x\vec{e}_1+y\vec{e_2}\mapsto -y\vec{e}_1+x\vec{e_2} \ $$. Then $$\text{Im}(ST) = \left\{{ c\vec{e_2}:c\in\mathbb{R}}\right\}, \text{Im}(T)=\left\{{ c\vec{e_1}:c\in\mathbb{R}}\right\} \ $$, and we note $$\text{Im}(ST)\not\subset\text{Im}(T) \ $$

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4) Let $$\mathbf{P} \ $$ be the space of polynomials over a field $$\mathbf{F} \ $$.

a) Let $$P,Q \in \mathbf{P} \ $$ be defined $$P=p_m x^m + \dots +p_0, Q = q_n x^n + \dots + 0 \ $$.

If we let the constants be zero for subscripts not summed over in the relevant polynomial, then observe that for the differential operator,

$$d_x(aP+bQ)= d_x \left({ \sum_{i=0}^{\text{max}(m,n)} (ap_i+bq_i)x^i }\right) =  \sum_{i=0}^{\text{max}(m,n)}d_x\left({  (ap_i+bq_i)x^i }\right) =  \sum_{i=1}^{\text{max}(m,n)} i\left({  (ap_i+bq_i)x^{i-1} }\right)$$

$$=\sum_{i=1}^{\text{max}(m,n)} ap_iix^{i-1}+bq_iix^{i-1} = \sum_{i=1}^m ap_iix^{i-1} + \sum_{i=1} bq_iix^{i-1} = a\sum_{i=1}^m p_iix^{i-1} + b\sum_{i=1} q_iix^{i-1}= ad_x(P)+bd_x(Q) \ $$

So, the differential operator is linear. Note that the two polynomials $$x\mapsto 1, x\mapsto 2 \ $$ both have derivative $$x\mapsto 0 \ $$, and so the differential operator is not injective. Note also that for any polynomial $$P = p_m x^m + \dots +p_0 \ $$, there is another polynomial $$Q= (p_m/(m+1))x^{m+1} + \dots p_0 x \ $$ such that $$d_x Q = P \ $$. Since $$\mathbf{F}= \mathbb{R} \ $$ or $$\mathbb{C} \ $$, we can be sure that $$p_k/(k+1) \ $$ is an element of $$\mathbf{F} \ $$. Hence the differential operator is surjective.

b) Fix $$a\in\mathbf{F} \ $$ and let $$I_a:\mathbf{P}\to\mathbf{P} \ $$ be

$$f\mapsto \int_a^x f \ $$

Observe that for field elements $$r,s \ $$, and polynomials $$P, Q \ $$, we have

$$I_a(rP+sQ)=\int_a^x (rP+sQ)dx= \int_a^x rPdx+\int_a^x sQdx = r\int_a^xPdx+s\int_a^xQdx = rI_a(P)+sI_a(Q) \ $$,

so $$I_a \ $$ is linear.

c) If we let $$P=\sum_{i=0}^n p_i x^i \ $$ be any polynomial, then note that

$$\frac{d}{dx} \int_a^x P = \frac{d}{dx} \int_a^x \left({ \sum_{i=0}^n p_i x^i }\right) dx= \frac{d}{dx} \left({ \sum_{i=0}^n \frac{p_i}{i+1} x^{i+1} }\right) = \sum_{i=0}^n p_i x^i= P \ $$

and so $$I_a \ $$ is a right inverse for $$d_x \ $$. Note, however, that for that same polynomial,

$$\int_a^x \frac{d}{dt} P = \int_a^x\frac{d}{dt}  \left({ p_0+p_1t+\dots+p_nt^n }\right) dt= \int_a^x \left({ p_1+p_2t + \dots +np_nt^{n-1} }\right) $$

$$=\left[{ p_1t + p_2t^2+\dots+p_nt^n }\right]^x_a = p_1x + p_2x^2+\dots+p_nx^n + C \ $$

where $$C=-P(a) \ $$ is some constant. Note that this means for some polynomial $$P \ $$ with constant term $$p_0 \ $$, the expression $$I_a \ $$ is a left inverse of the differential operator if and only if $$P(a)=-p_0 \ $$. Therefore, it is not possible for any one $$I_a \ $$ to be a left inverse of the differential operator for all polynomials.