MATH383: A first course in differential equationsFebruary 14, 2020

Due date: Feb 20, 2020, 11:55PM.

Bibliography: Lesson09.pdf Lesson10.pdf. The first exercise in each problem set is solved for you to use as a model.

1. Consider $𝒱={ℝ}^n$, $𝒮=ℝ$. Determine whether the column vectors of $𝑨$ form a basis for ${ℝ}^n$.

   Ex 1

   $n=3,$

   ${\displaystyle 𝑨=\left(\begin{array}{lll}-1& 1& 1\\ 2& 0& 1\\ 1& 1& 1\end{array}\right).}$

   Solution. Reduce $𝑨$ to row-echelon form

   ${\displaystyle 𝑨\sim \left(\begin{array}{lll}-1& 1& 1\\ 0& 2& 3\\ 0& 2& 2\end{array}\right)\sim \left(\begin{array}{lll}-1& 1& 1\\ 0& 2& 3\\ 0& 0& -1\end{array}\right)\sim \left(\begin{array}{lll}1& -1& -1\\ 0& 1& \frac{3}{2}\\ 0& 0& 1\end{array}\right)}$

   Since the row-echelon form does not contain a row of zeros, the columns of $𝑨$ form a basis for ${ℝ}^3$. Check in Maxima

   (%i1)

   A: matrix([-1, 1, 1],[2, 0, 1],[1, 1, 1]);

   ${\displaystyle \text{(\%o1)}\left(\begin{array}{ccc}-1& 1& 1\\ 2& 0& 1\\ 1& 1& 1\end{array}\right)}$

   (%i2)

   echelon(A);

   ${\displaystyle \text{(\%o2)}\left(\begin{array}{ccc}1& 0& \frac{1}{2}\\ 0& 1& \frac{3}{2}\\ 0& 0& 1\end{array}\right)}$

   (%i3)

   Note: multiple row-echelon forms are possible (differing by, say, permutation of rows), but the same conclusion on the independence is reached, i.e., no zero rows implies linear independence.

   Ex 2

   $n=3$,

   ${\displaystyle 𝑨=\left(\begin{array}{lll}2& 5& 3\\ -2& 1& 1\\ 1& 2& 1\end{array}\right).}$

   Solution. Reduce $𝑨$ to row-echelon form

   ${\displaystyle 𝑨\sim \left(\begin{array}{lll}2& 5& 3\\ 0& 6& 4\\ 0& -1& -1\end{array}\right)\sim \left(\begin{array}{lll}2& 5& 3\\ 0& 3& 2\\ 0& 0& -1\end{array}\right)\sim \left(\begin{array}{lll}1& \frac{5}{2}& \frac{3}{2}\\ 0& 1& \frac{2}{3}\\ 0& 0& 1\end{array}\right)}$

   Since the row-echelon form does not contain a row of zeros, the columns of $𝑨$ form a basis for ${ℝ}^3$. Check in Maxima

   (%i39)

   A: matrix([2, 5, 3],[-2, 1, 1],[1, 2, 1]);

   ${\displaystyle \text{(\%o39)}\left(\begin{array}{ccc}2& 5& 3\\ -2& 1& 1\\ 1& 2& 1\end{array}\right)}$

   (%i40)

   echelon(A);

   ${\displaystyle \text{(\%o40)}\left(\begin{array}{ccc}1& \frac{5}{2}& \frac{3}{2}\\ 0& 1& \frac{2}{3}\\ 0& 0& 1\end{array}\right)}$

   (%i41)

   Ex 3

   $n=4$,

   ${\displaystyle 𝑨=\left(\begin{array}{llll}1& 2& 2& -1\\ 1& 1& 4& 2\\ -1& 3& 2& 0\\ 1& 1& 5& 3\end{array}\right).}$

   Solution. Reduce $𝑨$ to row-echelon form

   ${\displaystyle 𝑨\sim \left(\begin{array}{llll}1& 2& 2& -1\\ 0& -1& 2& 3\\ 0& 5& 4& -1\\ 0& -1& 3& 4\end{array}\right)\sim \left(\begin{array}{llll}1& 2& 2& -1\\ 0& 1& -2& -3\\ 0& 0& 14& 14\\ 0& 0& 1& 1\end{array}\right)\sim \left(\begin{array}{llll}1& 2& 2& -1\\ 0& 1& -2& -3\\ 0& 0& 1& 1\\ 0& 0& 0& 0\end{array}\right)}$

   The row-echelon form does contain a row of zeros, the columns of $𝑨$ do not form a basis for ${ℝ}^4$. Check in Maxima

   (%i41)

   A: matrix([1, 2, 2, -1],[1, 1, 4, 2],[-1, 3, 2, 0],[1, 1, 5, 3]);

   ${\displaystyle \text{(\%o41)}\left(\begin{array}{cccc}1& 2& 2& -1\\ 1& 1& 4& 2\\ -1& 3& 2& 0\\ 1& 1& 5& 3\end{array}\right)}$

   (%i42)

   echelon(A);

   ${\displaystyle \text{(\%o42)}\left(\begin{array}{cccc}1& -3& -2& 0\\ 0& 1& \frac{3}{2}& \frac{1}{2}\\ 0& 0& 1& 1\\ 0& 0& 0& 0\end{array}\right)}$

   (%i43)

   Ex 4

   $n=4$,

   ${\displaystyle 𝑨=\left(\begin{array}{llll}-1& 2& 1& 2\\ 1& 1& 3& 1\\ 0& -1& 1& 1\\ 1& 2& -1& 2\end{array}\right).}$

   Solution. Having understood the procedure of using row operations to obtain the reduced row-echelon form, resort not to automated computation and obtain in Maxima

   (%i43)

   A: matrix([-1, 2, 1, 2],[1, 1, 3, 1],[0, -1, 1, 1],[1, 2, -1, 2]);

   ${\displaystyle \text{(\%o43)}\left(\begin{array}{cccc}-1& 2& 1& 2\\ 1& 1& 3& 1\\ 0& -1& 1& 1\\ 1& 2& -1& 2\end{array}\right)}$

   (%i44)

   echelon(A);

   ${\displaystyle \text{(\%o44)}\left(\begin{array}{cccc}1& -2& -1& -2\\ 0& 1& 0& 1\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)}$

   (%i45)

   The row-echelon form does not contain any row of zeros, the columns of $𝑨$ form a basis for ${ℝ}^4$.

2. Determine whether the following column vectors of $𝑩$ form a basis for $(𝒱,+,ℝ,\cdot )$

   Ex 1

   $𝑩=\left(\begin{array}{lll}1& 2x^2+x+2& -x^2+x\end{array}\right)$, $𝒱={𝒫}_2$

   Solution. Consider an arbitrary $𝒑\in 𝒱={𝒫}_2$

   ${\displaystyle 𝒑\left(x\right)=a_0+a_{1}x+a_2{x}^2,}$

   and check if $𝒑$ can be expressed as a linear combination of the basis vectors, $𝑩$, i.e., there exists $𝒄\in {ℝ}^3$ such that $𝑩𝒄=𝒑$

   ${\displaystyle 𝑩𝒄=\left(\begin{array}{lll}{𝒃}_1\left(x\right)& {𝒃}_2\left(x\right)& {𝒃}_3\left(x\right)\end{array}\right)\left(\begin{array}{l}{c}_1\\ c_2\\ c_3\end{array}\right)=c_1{𝒃}_1\left(x\right)+c_2{𝒃}_2\left(x\right)+c_3{𝒃}_3\left(x\right)=a_0+a_{1}x+a_2{x}^2=𝒑\left(x\right).}$

   Identify powers of $x$

   ${\displaystyle c_1{𝒃}_1\left(x\right)+c_2{𝒃}_2\left(x\right)+c_3{𝒃}_3\left(x\right)=c_1+c_2(2x^2+x+2)+c_3(-x^2+x)=(c_1+2c_2)\cdot 1+(c_1+c_2)\cdot x+(2c_2-c_3)\cdot x^2}$

   to obtain system $𝑴𝒄=𝒂$

   ${\displaystyle \left(\begin{array}{lll}1& 2& 0\\ 1& 1& 0\\ 0& 2& -1\end{array}\right)\left(\begin{array}{l}{c}_1\\ c_2\\ c_3\end{array}\right)=\left(\begin{array}{l}{a}_0\\ a_1\\ a_2\end{array}\right),}$

   Reduce matrix $𝑴$ to row-echelon form.

   ${\displaystyle 𝑴=\left(\begin{array}{lll}1& 2& 0\\ 1& 1& 0\\ 0& 2& -1\end{array}\right)\sim \left(\begin{array}{lll}1& 2& 0\\ 0& -1& 0\\ 0& 2& -1\end{array}\right)\sim \left(\begin{array}{lll}1& 2& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right)\sim \left(\begin{array}{lll}1& 2& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right).}$

   Since the row-echelon form of $𝑴$ does not have any zero rows, a unique solution of $𝑴𝒄=𝒂$ is found, and $𝑩$ is a basis for ${𝒫}_3$. Check in Maxima

   (%i3)

   M: matrix([1, 2, 0],[1, 1, 0],[0, 2, -1]);

   ${\displaystyle \text{(\%o3)}\left(\begin{array}{ccc}1& 2& 0\\ 1& 1& 0\\ 0& 2& -1\end{array}\right)}$

   (%i4)

   echelon(M);

   ${\displaystyle \text{(\%o4)}\left(\begin{array}{ccc}1& 2& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)}$

   (%i5)

   Ex 2

   $𝒱={ℝ}^{2\times 2}$ the space of real-valued two-by-two matrices,

   ${\displaystyle 𝑩=\left(\begin{array}{llll}\left(\begin{array}{ll}1& 0\\ 0& 0\end{array}\right)& \left(\begin{array}{ll}1& 1\\ 0& 0\end{array}\right)& \left(\begin{array}{ll}-2& 1\\ 1& 1\end{array}\right)& \left(\begin{array}{ll}0& 0\\ 0& 2\end{array}\right)\end{array}\right)}$

   Solution. Here are two approaches:

   1. The basis can be reorganized as a set of ${ℝ}^4$ vectors

      ${\displaystyle \left(\begin{array}{ll}1& 0\\ 0& 0\end{array}\right)\to \left(\begin{array}{l}1\\ 0\\ 0\\ 0\end{array}\right),\mathrm{etc}.;𝑩=\left(\begin{array}{llll}1& 1& -2& 0\\ 0& 0& 1& 0\\ 0& 1& 1& 0\\ 0& 0& 1& 2\end{array}\right)}$

      Reduce $𝑩$ to row echelon form

      (%i45)

      B: matrix([1, 1, -2, 0],[0, 0, 1, 0],[0, 1, 1, 0],[0, 0, 1, 2]);

      ${\displaystyle \text{(\%o45)}\left(\begin{array}{cccc}1& 1& -2& 0\\ 0& 0& 1& 0\\ 0& 1& 1& 0\\ 0& 0& 1& 2\end{array}\right)}$

      (%i47)

      echelon(B);

      ${\displaystyle \text{(\%o47)}\left(\begin{array}{cccc}1& 1& -2& 0\\ 0& 1& 1& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)}$

      (%i48)

      Since the reduced row echelon form has no row of zeros $𝑩$ is a basis

   2. Denote $𝑩=({𝑩}_1,{𝑩}_2,{𝑩}_3,{𝑩}_4)$, and consider whether some general

      ${\displaystyle 𝑨=\left(\begin{array}{ll}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right)\in {ℝ}^{2\times 2}}$

      can be expressed as a linear combination $𝑨=c_1{𝑩}_1+c_2{𝑩}_2+c_3{𝑩}_3+c_4{𝑩}_4$

      ${\displaystyle \left(\begin{array}{ll}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right)=c_1\left(\begin{array}{ll}1& 0\\ 0& 0\end{array}\right)+c_2\left(\begin{array}{ll}1& 1\\ 0& 0\end{array}\right)+c_3\left(\begin{array}{ll}-2& 1\\ 1& 1\end{array}\right)+c_4\left(\begin{array}{ll}0& 0\\ 0& 2\end{array}\right).}$

      Indentify each component to obtain the system

      ${\displaystyle \{\begin{array}{lll}{c}_1+c_2-2c_3& =& a_{11}\\ c_2+c_3& =& a_{12}\\ c_3& =& a_{21}\\ c_3+2c_4& =& a_{22}\end{array}.\Rightarrow \left(\begin{array}{llll}1& 1& -2& 0\\ 0& 1& 1& 0\\ 0& 0& 1& 0\\ 0& 0& 1& 2\end{array}\right)\left(\begin{array}{l}{c}_1\\ c_2\\ c_3\\ c_4\end{array}\right)=\left(\begin{array}{l}{a}_{11}\\ a_{12}\\ a_{21}\\ a_{22}\end{array}\right).}$

      Apart from a permutation of two rows, this is the same as matrix in solution procedure 1, and $𝑩$ does form a basis

   Ex 3

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

   ${\displaystyle 𝑩=\left(\begin{array}{lll}\left(\begin{array}{ll}1& 0\\ 0& 0\end{array}\right)& \left(\begin{array}{ll}0& 0\\ 0& 1\end{array}\right)& \left(\begin{array}{ll}0& 1\\ 0& 0\end{array}\right)\end{array}\right)}$

   Solution. Since

   ${\displaystyle 𝑨=\left(\begin{array}{ll}0& 0\\ 1& 0\end{array}\right)}$

   cannot be expressed as a linear combination of elements in $𝑩$, the system is not a basis. This is verified if we form the matrix

   ${\displaystyle 𝑩=\left(\begin{array}{lll}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right)}$

   which has a row of zeros.

3. Find a basis for the subspace $𝒮$ of vector space $𝒱$. Specify the dimension of $𝒮$.

   Ex 1

   ${\displaystyle 𝒮=\left\{\left(\begin{array}{l}s+2t\\ -s+t\\ t\end{array}\right)|s,t\in ℝ.\right\},𝒱={ℝ}^3}$

   Solution. Recognize that the specification of $𝒮$ gives a linear combination with scalar coefficients $s,t$, and rewrite

   ${\displaystyle \left(\begin{array}{l}s+2t\\ -s+t\\ t\end{array}\right)=s\left(\begin{array}{l}1\\ -1\\ 0\end{array}\right)+t\left(\begin{array}{l}2\\ 1\\ 1\end{array}\right).}$

   Construct

   ${\displaystyle 𝑩=\left(\begin{array}{ll}{𝒃}_1& {𝒃}_2\end{array}\right)=\left(\begin{array}{ll}1& 2\\ -1& 1\\ 0& 1\end{array}\right).}$

   Check if columns of $𝑩$ are linearly independent,

   ${\displaystyle s{𝒃}_1+t{𝒃}_2=𝟎\Rightarrow \left(\begin{array}{l}s+2t\\ -s+t\\ t\end{array}\right)=\left(\begin{array}{l}0\\ 0\\ 0\end{array}\right),}$

   and since $s=t=0$ is the only solution, columns of $𝑩$ are linearly independent and span the subspace, hence are a basis for $𝒮$.

   Ex 2

   ${\displaystyle 𝒮=\left\{\left(\begin{array}{ll}a& a+d\\ a+d& d\end{array}\right)|a,d\in ℝ.\right\},𝒱={ℝ}^{2\times 2}}$

   Solution. Since

   ${\displaystyle \left(\begin{array}{ll}a& a+d\\ a+d& d\end{array}\right)=a\left(\begin{array}{ll}1& 1\\ 1& 0\end{array}\right)+d\left(\begin{array}{ll}0& 1\\ 1& 1\end{array}\right)}$ ${\displaystyle 𝑩=\left(\begin{array}{ll}\left(\begin{array}{ll}1& 1\\ 1& 0\end{array}\right)& \left(\begin{array}{ll}0& 1\\ 1& 1\end{array}\right)\end{array}\right)}$

   is a basis.

   Ex 3

   $𝒮$ is the space of skew-symmetric matrices ($𝑨\in 𝒮\Rightarrow {𝑨}^T=-𝑨$), $𝒱={ℝ}^{2\times 2}$Solution. Since

   ${\displaystyle \left(\begin{array}{ll}0& a\\ -a& 0\end{array}\right)=a\left(\begin{array}{ll}0& 1\\ -1& 0\end{array}\right)}$ ${\displaystyle 𝑩=\left(\begin{array}{l}\left(\begin{array}{ll}0& 1\\ -1& 0\end{array}\right)\end{array}\right)}$

   is a basis.

   Ex 4

   $𝒮=\{.p\left(x\right)\left|p\right(0)=0\}$, $𝒱={𝒫}_2$.

   Solution. Since $𝒑\in 𝒮\Rightarrow 𝒑=a{x}^2+bx$, $𝑩=\{x,x^2\}$ is a basis.

   Ex 5

   $𝒮=\{.p\left(x\right)\left|p\right(0)=0,p\left(1\right)=0\}$, $𝒱={𝒫}_3$.

   Solution. Since $𝒑\in 𝒮\Rightarrow 𝒑=a{x}^3+b{x}^2+cx$, with $𝒑\left(1\right)=a+b+c=0\Rightarrow c=-(a+b)$, hence

   ${\displaystyle 𝒑=a{x}^3+b{x}^2-(a+b)x=a(x^3-x)+b(x^2-x),}$

   such that $𝑩=\{x^3-x,x^2-x\}$ is a basis.

4. Find the coordinates of the vector $𝒗$ in the basis $𝑩$.

   Ex 1

   ${\displaystyle 𝑩=\left(\begin{array}{ll}3& -2\\ 1& 2\end{array}\right),𝒗=\left(\begin{array}{l}8\\ 0\end{array}\right),𝒱={ℝ}^2}$

   Solution. If $𝑩$ is a basis for $𝒱={ℝ}^2$, then the vector $𝒗$ can be expressed as a linear combination of the basis vectors and that scalar coefficients are the coordinates of $𝒗$ in basis $𝑩$

   ${\displaystyle 𝒗=x_1{𝒃}_1+x_2{𝒃}_2=𝑩\left(\begin{array}{l}{x}_1\\ x_2\end{array}\right)\Rightarrow \mathrm{Solve}\left(\begin{array}{ll}3& -2\\ 1& 2\end{array}\right)\left(\begin{array}{l}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{l}8\\ 0\end{array}\right)}$

   to find the coordinates $x_1=2$, $x_2=-1$. Check in Maxima. $$

   (%i5)

   B:matrix([3,-2],[1,2]);

   ${\displaystyle \text{(\%o5)}\left(\begin{array}{cc}3& -2\\ 1& 2\end{array}\right)}$

   (%i6)

   v:matrix([8],[0]);

   ${\displaystyle \text{(\%o6)}\left(\begin{array}{c}8\\ 0\end{array}\right)}$

   (%i7)

   linsolve_by_lu(B,v);

   0 errors, 0 warnings

   ${\displaystyle \text{(\%o7)}[\left(\begin{array}{c}2\\ -1\end{array}\right),\text{false}]}$

   (%i8)

   x: first(%);

   ${\displaystyle \text{(\%o8)}\left(\begin{array}{c}2\\ -1\end{array}\right)}$

   (%i9)

   B.x

   ${\displaystyle \text{(\%o9)}\left(\begin{array}{c}8\\ 0\end{array}\right)}$

   (%i10)

   Ex 2

   ${\displaystyle 𝑩=\left(\begin{array}{ll}-2& -1\\ 4& 1\end{array}\right),𝒗=\left(\begin{array}{l}-2\\ 1\end{array}\right),𝒱={ℝ}^2}$

   Solution. Solve by Gauss elimination (row reduction operations on the extended matrix)

   ${\displaystyle \left(\begin{array}{lll}-2& -1& -2\\ 4& 1& 1\end{array}\right)\sim \left(\begin{array}{lll}-2& -1& -2\\ 0& -1& -3\end{array}\right)\sim \left(\begin{array}{lll}-2& 0& 1\\ 0& 1& 3\end{array}\right)\sim \left(\begin{array}{lll}1& 0& -\frac{1}{2}\\ 0& 1& 3\end{array}\right)}$

   and the coordinates are $(-\frac{1}{2},3)$. Check in Maxima

   (%i48)

   B:matrix([-2,-1],[4,1]);

   ${\displaystyle \text{(\%o48)}\left(\begin{array}{cc}-2& -1\\ 4& 1\end{array}\right)}$

   (%i49)

   v:matrix([-2],[1]);

   ${\displaystyle \text{(\%o49)}\left(\begin{array}{c}-2\\ 1\end{array}\right)}$

   (%i50)

   linsolve_by_lu(B,v);

   0 errors, 0 warnings
   

   ${\displaystyle \text{(\%o50)}[\left(\begin{array}{c}-\frac{1}{2}\\ 3\end{array}\right),\text{false}]}$

   Ex 3

   ${\displaystyle 𝑩=\left(\begin{array}{lll}1& 3& 1\\ -1& -1& 0\\ 2& 1& 2\end{array}\right),𝒗=\left(\begin{array}{l}1\\ 0\\ 2\end{array}\right),𝒱={ℝ}^3}$

   Solution. Since $𝑩=\left(\begin{array}{lll}{𝒃}_1& {𝒃}_2& {𝒃}_3\end{array}\right)$, and $𝒗={𝒃}_3$ the coordinates are $(0,0,1)$.

   Ex 4

   ${\displaystyle 𝑩=\left(\begin{array}{lll}2& 1& 0\\ 2& 0& 0\\ 1& 2& 1\end{array}\right),𝒗=\left(\begin{array}{l}0\\ 1\\ \frac{1}{2}\end{array}\right),𝒱={ℝ}^3}$

   Solution. With $𝑩=\left(\begin{array}{lll}{𝒃}_1& {𝒃}_2& {𝒃}_3\end{array}\right)$, from

   ${\displaystyle 𝒗=c_1{𝒃}_1+c_2{𝒃}_2+c_3{𝒃}_3}$

   observe from second row that $c_1=\frac{1}{2}$. Then

   ${\displaystyle \frac{1}{2}\left(\begin{array}{l}2\\ 2\\ 1\end{array}\right)+c_2\left(\begin{array}{l}1\\ 0\\ 2\end{array}\right)+c_3\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{l}1+c_2\\ 1\\ \frac{1}{2}+2c_2+c_3\end{array}\right)=\left(\begin{array}{l}0\\ 1\\ \frac{1}{2}\end{array}\right)}$

   and the other coordinates are $c_2=-1$, $c_3=2$. Verify in Maxima

   (%i51)

   B:matrix([2,1,0],[2,0,0],[1,2,1]);

   ${\displaystyle \text{(\%o51)}\left(\begin{array}{ccc}2& 1& 0\\ 2& 0& 0\\ 1& 2& 1\end{array}\right)}$

   (%i52)

   v:matrix([0],[1],[1/2]);

   ${\displaystyle \text{(\%o52)}\left(\begin{array}{c}0\\ 1\\ \frac{1}{2}\end{array}\right)}$

   (%i53)

   linsolve_by_lu(B,v);

   ${\displaystyle \text{(\%o53)}[\left(\begin{array}{c}\frac{1}{2}\\ -1\\ 2\end{array}\right),\text{false}]}$

   (%i54)

   Ex 5

   ${\displaystyle 𝑩=\left(\begin{array}{lll}1& x-1& x^2\end{array}\right),𝒗=-2x^2+2x+3,𝒱={𝒫}_2}$

   Solution. With $𝑩=\left(\begin{array}{lll}{𝒃}_1& {𝒃}_2& {𝒃}_3\end{array}\right)$ write

   ${\displaystyle 𝒗=-2{𝒃}_3+2{𝒃}_2+5{𝒃}_1=-2\left(x^2\right)+2(x-1)+5\left(1\right)=-2x^2+2x-2+5=-2x^2+2x+3,}$

   and the coordinates are $(5,2,-2)$.