5

Answer the following problems:

1. Define \(F = \left\{\begin{pmatrix}a & b \\ -b & a\end{pmatrix} : a, b \in \mathbb{R}\right\}\), with the regular matrix addition and multiplication. Show that \((F,+,\cdot)\) is a field.

   Proof

   1. Associativity

      Let \(A=\begin{pmatrix}a & b\\ -b & a\end{pmatrix},B=\begin{pmatrix}c & d\\ -d & c\end{pmatrix},X=\begin{pmatrix}x & y\\ -y & x\end{pmatrix}\)

      \(A+\left(B+C\right)=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}+\left\lbrack\begin{pmatrix}c & d\\ -d & c\end{pmatrix}+\begin{pmatrix}x & y\\ -y & x\end{pmatrix}\right\rbrack=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}+\begin{pmatrix}c+x & d+y\\ -d-y & x+c\end{pmatrix}=\begin{pmatrix}a+c+x & b+d+y\\ -b-d-y & a+x+c\end{pmatrix}\)

      \(\left(A+B\right)+C=\left\lbrack\begin{pmatrix}a & b\\ -b & a\end{pmatrix}+\begin{pmatrix}c & d\\ -d & c\end{pmatrix}\right\rbrack+\begin{pmatrix}x & y\\ -y & x\end{pmatrix}=\begin{pmatrix}a+c & b+d\\ -b-d & a+c\end{pmatrix}+\begin{pmatrix}x & y\\ -y & x\end{pmatrix}=\begin{pmatrix}a+c+x & b+d+y\\ -b-d-y & a+x+c\end{pmatrix}\)

      Thus \(A+(B+C)=(A+B)+C\), \(\forall A,B,C\in F\)

      \(A\cdot(B\cdot C)=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\left\lbrack\begin{pmatrix}c & d\\ -d & c\end{pmatrix}\cdot\begin{pmatrix}x & y\\ -y & x\end{pmatrix}\right\rbrack=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\begin{pmatrix}cx-\mathrm{d}y & cy+\mathrm{d}x\\ -\mathrm{d}x-cy & -\mathrm{d}y+cx\end{pmatrix}=\begin{pmatrix}acx-ady-bdx-bcy & acy+adx-bdy+bcx\\ -bcx+bdy-adx-acy & -bcy-bdx-ady+acx\end{pmatrix}\)

      \((A\cdot B)\cdot C=\left\lbrack\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\begin{pmatrix}c & d\\ -d & c\end{pmatrix}\right\rbrack\cdot\begin{pmatrix}x & y\\ -y & x\end{pmatrix}=\begin{pmatrix}ac-bd & ad+bc\\ -bc-ad & -bd+ac\end{pmatrix}\cdot\begin{pmatrix}x & y\\ -y & x\end{pmatrix}=\begin{pmatrix}acx-ady-bdx-bcy & acy+adx-bdy+bcx\\ -bcx+bdy-adx-acy & -bcy-bdx-ady+acx\end{pmatrix}\)

      Thus \(A\cdot(B\cdot C)=(A\cdot B)\cdot C\), \(\forall A,B,C\in F\) 2. Commutativity

      \(A+B=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}+\begin{pmatrix}c & d\\ -d & c\end{pmatrix}=\begin{pmatrix}a+c & b+d\\ -b-d & a+c\end{pmatrix}\)

      \(B+A=\begin{pmatrix}c & d\\ -d & c\end{pmatrix}+\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=\begin{pmatrix}a+c & b+d\\ -b-d & a+c\end{pmatrix}\)

      Thus \(A+B=B+A\), \(\forall A,B\in F\)

      \(A\cdot B=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\begin{pmatrix}c & d\\ -d & c\end{pmatrix}=\begin{pmatrix}ac-bd & ad+bc\\ -bc-ad & -bd+ac\end{pmatrix}\)

      \(B\cdot A=\begin{pmatrix}c & d\\ -d & c\end{pmatrix}\cdot\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=\begin{pmatrix}ac-bd & ad+bc\\ -bc-ad & -bd+ac\end{pmatrix}\)

      Thus \(A\cdot B=B\cdot A\), \(\forall A,B\in F\) 3. Additive inverse and multiplicative inverse

      \(0+A=\begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}+\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=A\)

      \(A+0=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}+\begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=A\)

      Thus \(\forall A\in F,\exists0\in F~s.t.~0+A=A+0=A\)

      \(1\cdot A=\begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}\cdot\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=A\)

      \(A\cdot1=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=A\)

      Thus \(\forall A\in F,\exists1\in F~s.t.~1\cdot A=A\cdot1=A\) 4. Additive identity and multiplicative identity

      \(A+(-A)=-A+A=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}+\begin{pmatrix}-a & -b\\ b & -a\end{pmatrix}=\begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}\)

      Thus \(\forall A\in F,\exists-A\in F~s.t.~-A+A=A+\left(-A\right)=0\)

      \(A\cdot A^{-1}=A^{-1}\cdot A=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\begin{pmatrix}\frac{a}{a^2+b^2} & \frac{-b}{ad-bc}\\ \frac{-b}{ad-bc} & \frac{a}{ad-bc}\end{pmatrix}=\begin{pmatrix}\frac{a^2+b^2}{a^2+b^2} & \frac{a^2+b^2}{a^2+b^2}\\ \frac{a^2+b^2}{a^2+b^2} & \frac{a^2+b^2}{a^2+b^2}\end{pmatrix}=\begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}\)

      Thus \(\forall A\in F,\exists A^{-1}\in F~s.t.~A^{-1}\cdot A=A\cdot A^{-1}=1\) 5. Distributivity

      \(A\cdot(B+C)=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\left\lbrack\begin{pmatrix}c & d\\ -d & c\end{pmatrix}+\begin{pmatrix}x & y\\ -y & x\end{pmatrix}\right\rbrack=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\begin{pmatrix}c+x & d+y\\ -d-y & c+x\end{pmatrix}=\begin{pmatrix}ac+ax-bd-by & ad+ay+bc+bx\\ -bc-bx-ad-ay & -bd-by+ac+ax\end{pmatrix}\)

      \(AB+AC=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\begin{pmatrix}c & d\\ -d & c\end{pmatrix}+\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\begin{pmatrix}x & y\\ -y & x\end{pmatrix}=\begin{pmatrix}ac-bd & ad+bc\\ -bc-ad & -bd+ac\end{pmatrix}+\begin{pmatrix}ax-by & ay+bx\\ -bx-ay & -by+ax\end{pmatrix}=\begin{pmatrix}ax+ac-bd-by & ad+ay+bc+bx\\ -bc-ad-bx-ay & ac+ax-bd-by\end{pmatrix}\)

      Thus \(A\cdot (B+C)=AB+AC\)\,\(\forall A,B,C\in F\)

2. (a) Let \(A\) be a matrix of order \(2 \times 2\) satisfying \(\text{tr}(A) = 0\). Prove that \(A^2\) is a scalar matrix.

   Since \(\text{tr}(A)=0\Rightarrow a_{11}=-a_{22}\), then \(A = \begin{pmatrix} a & b \\ c & -a \end{pmatrix}\)

   Then \(A^2 = \begin{pmatrix} a & b \\ c & -a \end{pmatrix} \begin{pmatrix} a & b \\ c & -a \end{pmatrix}\)​\(= \begin{pmatrix} a^2 + bc & 0 \\ 0 & a^2 + bc \end{pmatrix}\)​\(= \begin{pmatrix} a^2 + bc \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

   Thus \(A^2\) is a scalar matrix.

   (b) Let \(A\) and \(B\) be matrices of order \(2 \times 2\). Prove that \(\text{tr}(AB - BA) = 0\).

   Since \(A=\begin{pmatrix}a & b\\ c & d\end{pmatrix},B=\begin{pmatrix}w & x\\ y & z\end{pmatrix}\), then \(AB=\begin{pmatrix}a & b\\ c & d\end{pmatrix}\begin{pmatrix}w & x\\ y & z\end{pmatrix}=\begin{pmatrix}aw+by & ax+bz\\ cw+\mathrm{d}y & cx+dz\end{pmatrix}\)

   \(BA=\begin{pmatrix}w & x\\ y & z\end{pmatrix}\begin{pmatrix}a & b\\ c & d\end{pmatrix}=\begin{pmatrix}wa+cx & wb+\mathrm{d}x\\ ay+cz & by+dz\end{pmatrix}\)

   Then \(AB-BA=\begin{pmatrix}aw+by & ax+bz\\ cw+\mathrm{d}y & cx+dz\end{pmatrix}-\begin{pmatrix}wa+cx & wb+\mathrm{d}x\\ ay+cz & by+dz\end{pmatrix}=\begin{pmatrix}by-cx & ax-\mathrm{d}x-bw+bz\\ -ay+\mathrm{d}y+cw-cz & cx-by\end{pmatrix}\)

   Thus \(\text{tr}(AB-BA)=by-cx+cx-by=0\)

   (c) Let \(A\), \(B\), and \(C\) be matrices of order \(2 \times 2\). Prove that \(C(AB - BA)^2 = (AB - BA)^2 C\).

   Let \(C=\begin{pmatrix}e & f\\ g & h\end{pmatrix}\) and we have know that \(AB-BA=P=\begin{pmatrix}p & q\\ r & s\end{pmatrix}\)

   Since \(tr(AB-BA)=0\), then \(P=\begin{pmatrix}p & q\\ r & -p\end{pmatrix}\Rightarrow P^2=\begin{pmatrix}p^2+qr\end{pmatrix}\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\)

   Hence \(CP^2=\begin{pmatrix}e & f\\ g & h\end{pmatrix}\begin{pmatrix}p^2+qr\end{pmatrix}\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}=\left(p^2+qr\right)\begin{pmatrix}e & f\\ g & h\end{pmatrix}\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}=\left(p^2+qr\right)\begin{pmatrix}e & f\\ g & h\end{pmatrix}\)

   \(P^2C=\begin{pmatrix}p^2+qr\end{pmatrix}\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix}e & f\\ g & h\end{pmatrix}=\left(p^2+qr\right)\begin{pmatrix}e & f\\ g & h\end{pmatrix}\)

   Hence \(CP^2=P^2C\), thus \(C(AB - BA)^2 = (AB - BA)^2 C\)

3. Prove or disprove the following claim: Let \(V\) be a vector space over a field \(F\), and let \(U,W\) be vector subspaces of \(V\). If \(U \cup W\) is a vector subspace of \(V\), then \(U \subseteq W\) or \(W \subseteq U\).

   Proof

   \(\Rightarrow\)) Take \(u\in U\) and \(w\in W\), since \(U\subseteq U\cup W,W\subseteq U\cup W\), then \(u\in U\cup W,w\in U\cup W\)

   Thus \(u+w\in U\cup W\), then we have two cases

   If \(u+w\in U\), since \(u\in U\), then \(-u\in U\), then \(u+w+\left(-u\right)\in U\Rightarrow\) \(w\in U\)

   Thus we have \(W\subseteq U\)

   If \(u+w\in W\), since \(w\in W\), then \(-w\in W\), then \(u+w+\left(-w\right)\in W\Rightarrow\) \(u\in W\)

   Thus we have \(U\subseteq W\)

   \(\Leftarrow\)) If \(U_{}\subseteq W\Rightarrow U\cup W=W\) and since \(W\) is a subspace, so is \(U\cup W\).

   If \(W\subseteq U\Rightarrow U\cup W=U\), which is a subspace.

4. Let \(A\) be a \(4 \times 3\) matrix. Denote by \(C_i\) the \(i\)-th column of \(A\), and let \(b = C_1 - 2C_3\).

   (a) Prove that the system \(A x = b\) has a solution.

   Suppose \(A=\begin{pmatrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ a_{41} & a_{42} & a_{43}\end{pmatrix}\), then \(Ax=b\Rightarrow\begin{pmatrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ a_{41} & a_{42} & a_{43}\end{pmatrix}\begin{pmatrix}x_1\\ x_2\\ x_3\end{pmatrix}=\begin{pmatrix}a_{11}-2a_{13}\\ a_{21}-2a_{23}\\ a_{31}-2a_{33}\\ a_{41}-2a_{43}\end{pmatrix}\)

   Then we have \(\begin{cases}a_{11}x_1+a_{12}x_2+a_{13}x_3=a_{11}-2a_{13}\\ a_{21}x_1+a_{22}x_2+a_{23}x_3=a_{21}-2a_{23}\\ a_{31}x_1+a_{32}x_2+a_{33}x_3=a_{31}-2a_{33}\\ a_{41}x_1+a_{42}x_2+a_{43}x_3=a_{41}-2a_{43}\end{cases}\)

   Obviously, there exists a solution which is \(x_0=\begin{pmatrix}1\\ 0\\ -2\end{pmatrix}\)

   Thus the system \(A x = b\) has a solution.

   (b) Assume that \(\text{rank}(A) = 2\), and that \(x_0 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\) is a solution to \(A x = b\). Prove that the system has infinitely many solutions, find the degree of freedom, and write the general solution.

   Since \(\begin{cases}a_{11}x_1+a_{12}x_2+a_{13}x_3=a_{11}-2a_{13}\\ a_{21}x_1+a_{22}x_2+a_{23}x_3=a_{21}-2a_{23}\\ a_{31}x_1+a_{32}x_2+a_{33}x_3=a_{31}-2a_{33}\\ a_{41}x_1+a_{42}x_2+a_{43}x_3=a_{41}-2a_{43}\end{cases}\) and \(x_0 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\), then \(\begin{cases}a_{11}+a_{12}+a_{13}=a_{11}-2a_{13}\\ a_{21}+a_{22}+a_{23}=a_{21}-2a_{23}\\ a_{31}+a_{32}+a_{33}=a_{31}-2a_{33}\\ a_{41}+a_{42}+a_{43}=a_{41}-2a_{43}\end{cases}\)

   Then \(\begin{cases}a_{12}=-3a_{13}\\ a_{22}=-3a_{23}\\ a_{32}=-3a_{33}\\ a_{42}=-3a_{43}\end{cases}\), Let's consider \(A\) be the row-reduced echelon form, thus the first column must have a pivot which is \(a_{11}=1\)

   Then \(a_{i2}=-3a_{i3}\), since \(a_{i2}\) is in the front of \(a_{i3}\), then \(\exists a_{i2}=1\)

   Since \(rank(A)=2\), then \(\exists!a_{i2}=1,a_{i3}=-\frac13\)

   Since \(A\) is the row-reduced echelon form, then \(a_{22}=1,a_{23}=-\frac13\) and \(a_{12},a_{32},a_{42}=0\Rightarrow a_{13},a_{33},a_{43}=0\)

   We have \(A=\begin{pmatrix}1 & 0 & 0\\ 0 & 1 & -\frac13\\ 0 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}\), then \(\begin{cases}x_1=1\\ x_2-\frac13x_3=\frac23\end{cases}\)

   Then \(x_3\) can be any values, thus the system has infinitely many solutions

   The degree of freedom is 1

   The general solution is \(\begin{cases}x_1=1\\ x_2=\frac23+\frac13x_3\\ x_3 \end{cases}\)

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Answer the following questions:

1. Suppose problem 1 is given in the midterm, and is worth 15 points. How many points do you think you gained for your solution? Please explain.

   15, My proof is very detail and structured, and just use the definition and property taught in the lecture.

   Suppose problem 2 is given in the midterm, and is worth 15 points. How many points do you think you gained for your solution? Please explain.

   15, My proof is very detail and structured, and just use the definition and property taught in the lecture. Suppose problem 3 is given in the midterm, and is worth 15 points. How many points do you think you gained for your solution? Please explain.

   15, My proof is totally same with the proof of professor. Because this question is taught in the class

   Suppose problem 4 is given in the midterm, and is worth 15 points. How many points do you think you gained for your solution? Please explain.

   13, I am not sure whether the general solutions should be written in this form. Although i think it is ok, but a bit strange

2. Did you choose to adjust some of your solutions? Please explain.

   No, because every time i submitted my log. I have tried my best to correct and modify it in a better form and accuracy. There is no more things i can do without instruction

3. Looking back at your solutions, which skills do you think you have improved throughout the course?

   Maybe i learn how to write a proof in a good form and let's people understand. I can write a proof in a detail and good way

   I also learn how to slow my pace and think a problem as deep as i can.

4. How many learning logs have you submitted so far? 1 2 3 4 5

5. Why did you choose to participate in this study?

   I am eager to help professors understand our learning state.

   I want to do more exercise to improve my level as well.

   I also really want the points bonus to improve my grade.

   It's a win-win thing

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