MATH383: A first course in differential equationsFebruary 7, 2020

Due date: Feb 13, 2020, 11:55PM.

Bibliography: Lesson07.pdf Lesson08.pdf. The first exercise in each problem set is solved for you to use as a model.

1. Consider $𝒱={ℝ}^3$, $𝒮=ℝ$. Establish whether for given operations $\oplus ,\odot$ $(𝒱,𝒮,\oplus ,\odot )$ is a vector space or not.

   Ex 1

   $𝒙=(x_1,x_2,x_3),𝒚=(y_1,y_2,y_3)$, $𝒙,𝒚\in 𝒱={ℝ}^3$, $𝒙\oplus 𝒚=(2x_1+2y_1,2x_2+2y_2,2x_3+2y_3)$, and with $\alpha \in 𝒮=ℝ$, $\alpha \odot 𝒙=(\alpha x_1,\alpha x_2,\alpha x_3)$

   Solution. Check associativity, $(𝒙\oplus 𝒚)\oplus 𝒛=𝒙\oplus (𝒚\oplus 𝒛)$. Compute

   ${\displaystyle \begin{array}{ll}𝒖=𝒙\oplus 𝒚=(2x_1+2y_1,2x_2+2y_2,2x_3+2y_3)& 𝒗=𝒚\oplus 𝒛=(2y_1+2z_1,2y_2+2z_2,2y_3+2z_3)\end{array}}$ ${\displaystyle 𝒖+𝒛=(4x_1+4y_1+2z_1,4x_2+4y_2+2z_2,4x_3+4y_3+2z_2)}$ ${\displaystyle 𝒙+𝒗=(2x_1+4y_1+4z_1,2x_2+4y_2+4z_2,2x_2+4y_2+4z_2)}$

   Since $𝒖+𝒛\ne 𝒙+𝒗$, associativity is not satisified and $(𝒱,𝒮,\oplus ,\odot )$ is not a vector space. Note: it is sufficient to find one unsatisfied property to prove that $(𝒱,𝒮,\oplus ,\odot )$ is not a vector space. However to prove that $(𝒱,𝒮,\oplus ,\odot )$ is indeed a vector space, all properties must be verified/

   Ex 2

   $𝒙\oplus 𝒚=(x_1-y_1,x_2-y_2,x_3-y_3)$, $\alpha \odot 𝒙=(\alpha x_1,\alpha x_2,\alpha x_3)$.

   Ex 3

   $𝒙\oplus 𝒚=(x_1+y_1-1,x_2+y_2-1,x_3+y_3-1_{})$, $\alpha \odot 𝒙=(\alpha x_1,\alpha x_2,\alpha x_3)$.

   Ex 4

   $𝒙\oplus 𝒚=(x_1+y_1,x_2-y_2,x_3+y_3)$, $\alpha \odot 𝒙=(\alpha x_1,\alpha x_2,\alpha x_3)$.

   Ex 5

   $𝒙\oplus 𝒚=(x_1+y_1,x_2+y_2,x_3+y_3)$, $\alpha \odot 𝒙=(\alpha +x_1,\alpha x_2,\alpha x_3)$.

2. Consider $𝒱\subset {ℝ}^{2\times 2}$, a subset of all 2 by 2 real-component matrices with operations

   ${\displaystyle 𝑨,𝑩\in {ℝ}^{2\times 2},𝑨=\left(\begin{array}{ll}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right),𝑩=\left(\begin{array}{ll}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right),𝑨\oplus 𝑩=\left(\begin{array}{ll}{a}_{11}+b_{11}& a_{12}+b_{12}\\ a_{21}+b_{21}& a_{22}+b_{22}\end{array}\right)}$ ${\displaystyle \alpha \odot 𝑨=\left(\begin{array}{ll}\alpha a_{11}& \alpha a_{12}\\ \alpha a_{21}& \alpha a_{22}\end{array}\right).}$

   Determine whether the following are vector spaces

   Ex 1

   $𝒱$ is the set of skew-symmetric matrices, $𝑨\in 𝒱\Rightarrow$ ${𝑨}^T=-𝑨$.

   Solution. From $𝑨=-{𝑨}^T$ deduce that

   ${\displaystyle \left(\begin{array}{ll}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right)=-\left(\begin{array}{ll}{a}_{11}& a_{21}\\ a_{12}& a_{22}\end{array}\right)\Rightarrow 𝑨=\left(\begin{array}{ll}0& a\\ -a& 0\end{array}\right).}$

   Verify vector space properties for $\forall 𝑨,𝑩,𝑪\in 𝒱$, $\forall \alpha ,\beta \in ℝ$:

   Closure

   ${\displaystyle 𝑨+𝑩=\left(\begin{array}{ll}0& a\\ -a& 0\end{array}\right)+\left(\begin{array}{ll}0& b\\ -b& 0\end{array}\right)=\left(\begin{array}{ll}0& a+b\\ -(a+b)& 0\end{array}\right)\in 𝒱?}$

   Associativity

   ${\displaystyle (𝑨+𝑩)+𝑪=\left(\begin{array}{ll}0& a+b\\ -(a+b)& 0\end{array}\right)+\left(\begin{array}{ll}0& c\\ -c& 0\end{array}\right)=\left(\begin{array}{ll}0& a+b+c\\ -(a+b+c)& 0\end{array}\right)}$ ${\displaystyle 𝑨+(𝑩+𝑪)=\left(\begin{array}{ll}0& a\\ -a& 0\end{array}\right)+\left(\begin{array}{ll}0& b+c\\ -(b+c)& 0\end{array}\right)=\left(\begin{array}{ll}0& a+b+c\\ -(a+b+c)& 0\end{array}\right).?}$

   Identity

   ${\displaystyle 𝑨+𝟎=\left(\begin{array}{ll}0& a\\ -a& 0\end{array}\right)+\left(\begin{array}{ll}0& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{ll}0& a\\ -a& 0\end{array}\right)?}$

   Inverse

   ${\displaystyle 𝑨+(-𝑨)=\left(\begin{array}{ll}0& a\\ -a& 0\end{array}\right)+\left(\begin{array}{ll}0& -a\\ a& 0\end{array}\right)=\left(\begin{array}{ll}0& 0\\ 0& 0\end{array}\right)?}$

   Commutativity

   ${\displaystyle 𝑨+𝑩=\left(\begin{array}{ll}0& a+b\\ -(a+b)& 0\end{array}\right)=\left(\begin{array}{ll}0& b+a\\ -(b+a)& 0\end{array}\right)=𝑩+𝑨.?}$

   Distributivity

   ${\displaystyle \alpha (𝑨+𝑩)=\left(\begin{array}{ll}0& \alpha (a+b)\\ -\alpha (a+b)& 0\end{array}\right)=\left(\begin{array}{ll}0& \alpha .a+\alpha b)\\ -\alpha .a-\alpha b)& 0\end{array}\right)=\alpha 𝑨+\alpha 𝑩?}$ ${\displaystyle (\alpha +\beta )𝑨=\left(\begin{array}{ll}0& (\alpha +\beta )a\\ -(\alpha +\beta )a& 0\end{array}\right)=\left(\begin{array}{ll}0& \alpha a+\beta a\\ -\alpha a-\beta a& 0\end{array}\right)=\alpha 𝑨+\beta 𝑨?}$ ${\displaystyle \alpha (\beta 𝑨)=\alpha \left(\begin{array}{ll}0& \beta a\\ -\beta a& 0\end{array}\right)=\left(\begin{array}{ll}0& \alpha \beta a\\ -\alpha \beta a& 0\end{array}\right)=(\alpha \beta )𝑨.?}$

   All properties are verified, hence skew-symmetric matrices form a vector space.

   Ex 2

   $𝒱$ is the set of upper-triangular matrices, $𝑨\in 𝒱\Rightarrow$ $a_{21}=0$.

   Ex 3

   $𝒱$ is the set of symmetric matrices, $𝑨\in 𝒱\Rightarrow$ ${𝑨}^T=𝑨$.

3. Determine whether the set $𝒮$ is linearly dependent or independent within the vector space $𝒱$

   Ex 1

   ${\displaystyle 𝒮=\{{𝒖}_1,{𝒖}_2\}=\{\left(\begin{array}{l}2\\ 1\\ 3\end{array}\right),\left(\begin{array}{l}0\\ -1\\ 1\end{array}\right)\},𝒱={ℝ}^3}$

   Solution. The first equation of the system $a_1{𝒖}_1+a_2{𝒖}_2=𝟎$ is $2a_1+0a_2=0\Rightarrow a_1=0$. The second equation then states $-a_2=0$, hence $a_1=a_2=0$, and ${𝒖}_1,{𝒖}_2$ are linearly independent.

   Ex 2

   ${\displaystyle 𝒮=\{{𝒖}_1,{𝒖}_2,{𝒖}_3\}=\{\left(\begin{array}{l}2\\ 1\end{array}\right),\left(\begin{array}{l}1\\ 0\end{array}\right),\left(\begin{array}{l}8\\ -3\end{array}\right)\},𝒱={ℝ}^2}$

   Ex 3

   ${\displaystyle 𝒮=\{{𝒖}_1,{𝒖}_2,{𝒖}_3\}=\{\left(\begin{array}{l}1\\ 0\\ 1\end{array}\right),\left(\begin{array}{l}-1\\ 1\\ 0\end{array}\right),\left(\begin{array}{l}0\\ 1\\ 1\end{array}\right)\},𝒱={ℝ}^3}$

4. Determine whether the set $𝒮$ is linearly dependent or independent within the vector space $𝒱$. Here ${𝒫}_n$ is the set of polynomials of degree at most $n$.

   Ex 1

   ${\displaystyle 𝒮=\{{𝒑}_1,{𝒑}_2,{𝒑}_3\}=\{1,2x^2+x+2,-x^2+x\},𝒱={𝒫}_2}$

   Solution. Denote $𝒒=a_1{𝒑}_1+a_2{𝒑}_2+a_3{𝒑}_3$, and consider the equality $𝒒=𝟎$. Note that $𝟎$ is the zero polynomial, i.e. $𝒒\left(x\right)=𝟎\left(x\right)\Rightarrow$

   ${\displaystyle a_1+a_2(2x^2+x+2)+a_3(-x^2+x)=0\mathrm{for}\mathrm{all}x}$

   For $x=0$ obtain $a_1+a_2=0.$ Subsequently for $x=1$ obtain $a_1+5a_2=0$. Subtract to obtain $4a_2=0\Rightarrow a_2=0$, and then $a_1=0$. Then for $x=-1$ obtain $-2a_3=0\Rightarrow a_3=0$. The only choice of $a_1,a_2,a_3$ to have $a_1{𝒑}_1+a_2{𝒑}_2+a_3{𝒑}_3=𝟎$ is $a_1=a_2=a_3=0$, hence $𝒮$ is a linearly independent set of vectors.

   Ex 2

   ${\displaystyle 𝒮=\{{𝒑}_1,{𝒑}_2,{𝒑}_3\}=\{2,x,x^3+2x^2-1\},𝒱={𝒫}_3}$

   Ex 3

   ${\displaystyle 𝒮=\{{𝒑}_1,{𝒑}_2,{𝒑}_3,{𝒑}_4\}=\{x,x^2,x^2+2x,x^3-x+1\},𝒱={𝒫}_3}$