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Homework 4

Homework 4.pdf

999027873 Yue Shi

  1. Let \(V\) be a \(F\)-vector space such that \(\dim V=n\) and let \(U \subseteq V\) be subspace of \(V\) such that \(\dim U=n-1\)

    1. Prove that if \(v \notin U\), then \(V = U \oplus \langle v \rangle\)​

      Suppose \(\mathcal{B'}\) is the basis of \(U\) and \(\mathcal{B^{\prime}}=\left\lbrace v_{1},\ldots,v_{n-1}\right\rbrace\) since \(\dim U=n-1\)​

      Then since \(v\notin U\), then we extend the \(\mathcal{B}'\) to the basis of \(V\): \(\mathcal{B}=\{v_1,...,v_{n-1},v\}\)​

      Then for any \(w\in V\), we have \(w=a_{1}v_{1}+...+a_{n-1}v_{n-1}+a_{n}v=\underbrace{a_{1}v_{1}+...+a_{n-1}v_{n-1}} _{\in U}+\underbrace{a_{n}v}_{\in \lang v\rang}\)​

      Thus \(V=U+\langle v\rangle\).

      Then consider \(w\in U\cap \lang v\rang\). NTP \(w=0_V\)​

      Since \(w\in U\), then \(w=a_{1}v_{1}+...+a_{n-1}v_{n-1}\)​

      Since \(w\in\lang v\rang\), then \(w=\lambda v\), then \(a_{1}v_{1}+...+a_{n-1}v_{n-1}=\lambda v\)​

      Then \(a_{1}v_{1}+...+a_{n-1}v_{n-1}-\lambda v=0\) and since \(\mathcal{B}=\{v_1,...,v_{n-1},v\}\)​

      Then \(a_1=...=a_{n-1}=\lambda=0\), then \(w=0\). Thus \(V = U \oplus \langle v \rangle\) 2. Prove that if \(W\) is a subspace of \(V\) such that \(W \nsubseteq U\)​, then \(V = U + W\).

      Since \(W\nsubseteq U\), then \(\exists v\in W:v\notin U\)​

      Suppose \(\mathcal{B_1}\) is the basis of \(U\) and \(\mathcal{B_1}=\left\lbrace v_{1},\ldots,v_{n-1}\right\rbrace\) since \(\dim U=n-1\)

      Suppose \(\mathcal{B_2}\) is the basis of \(W\) and \(\mathcal{B_2}=\left\lbrace u_{1},\ldots,u_{k-1},v\right\rbrace\) where \(\dim W=k\in [1,n]\)

      Then since \(v\notin U\), then we extend the \(\mathcal{B_1}\) to the basis of \(V\): \(\mathcal{B}=\{v_1,...,v_{n-1},v\}\)

      Then for any \(w\in V\), we have \(w=a_{1}v_{1}+...+a_{n-1}v_{n-1}+a_{n}v=\underbrace{a_1v_1+...+a_{n-1}v_{n-1}}_{\in U}+\underbrace{a_{n}v}_{\in W}\)

      Thus \(V = U + W\)​

  2. Let \(U = \{p \in P_4(F) : p(2) = p(5) = p(6)\}\).

    1. Find a basis of \(U\).

      Assume \(p(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}\)​

      We know \(p(2)=p(5)=p(6)\), then

      \(a_{0}+2a_{1}+4a_{2}+8a_{3}+16a_{4}=a_{0}+5a_{1}+25a_{2}+125a_{3}+625a_{4}=a_{0}+ 6a_{1}+36a_{2}+216a_{3}+1296a_{4}\)​

      Then \(3a_1+21a_2+117a_3+609a_4=0\) and \(a_{1}+11a_{2}+91a_{3}+671a_{4}=0\)​

      Then \(12a_2+156a_3+1404a_4=0\), then \(a_{2}=-13a_{3}-117a_{4}\)​

      Also \(a_{1}+11\left(-13a_{3}-117a_{4}\right)+91a_{3}+671a_{4}=0\), then \(a_{1}=52a_{3}+616a_{4}\)​

      Thus \(p(x)=a_{0}+\left(52a_{3}+616a_{4}\right)x-\left(13a_{3}+117a_{4}\right)x^{2}+a_{3} x^{3}+a_{4}x^{4}\)

      Thus the basis is \(\{1,52x-13x^{2}+x^{3},616x-117x^{2}+x^{4}\}\)​ 2. Extend the basis in part (1) to a basis of \(P_4(F)\).

      Since one way is that \(x^3,x^{4}\) is not free in that basis, thus we can add \(x^3,x^{4}\)​

      Then \(\{1,52x-13x^{2}+x^{3},616x-117x^{2}+x^{4},x^{3},x^{4}\}\) is the basis of \(P_4(F)\)​ 3. Find a subspace \(W\) of \(P_4(F)\) such that \(P_4(F) = U \oplus W\).

      Let \(W=\{x^3,x^4\}\). Then we check

      Let \(p(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}\), then

      \(p(x)=a_{0}\cdot1+\left(-\frac{9a_{1}+36a_{2}}{148}-\frac{1}{13}a_{2})(52x-13x^{2} +x^{3}\right)+\frac{a_{1}+4a_{2}}{148}(616x-117x^{2}+x^{4})+\left(a_{3}+\frac{9a_{1}+36a_{2}}{148} +\frac{1}{13}a_{2}\right)x^{3}+\left(a_{4}-\frac{a_{1}+4a_{2}}{148}\right)x^{4}\)

      Then \(p(x)=\underbrace{a_{0}\cdot1+\left(-\frac{9a_{1}+36a_{2}}{148}-\frac{1}{13}a_{2})(52x-13x^{2} +x^{3}\right)+\frac{a_{1}+4a_{2}}{148}(616x-117x^{2}+x^{4})}_{\in U}+\underbrace{\left(a_{3}+\frac{9a_{1}+36a_{2}}{148} +\frac{1}{13}a_{2}\right)x^{3}+\left(a_{4}-\frac{a_{1}+4a_{2}}{148}\right)x^{4}}_{\in W}\)

      Thus \(P_{4}(F) = U +W\). Take \(p(x)\in U\) and \(p(x)\in W\), then

      \(p(x)=a_{0}+a_{1}(52x-13x^{2}+x^{3})+a_{2}(616x-117x^{2}+x^{4})=a_{0}+\left(52a_{1} +616a_{2}\right)x-\left(13a_{1}+117a_{2}\right)x^{2}+a_{1}x^{3}+a_{2}x^{4}\)​

      \(p(x)=a_{3}x^{3}+a_{4}x^{4}\)​

      Then \(a_{0}+\left(52a_{1}+616a_{2}\right)x-\left(13a_{1}+117a_{2}\right)x^{2}+a_{1}x^{3} +a_{2}x^{4}=a_{3}x^{3}+a_{4}x^{4}\)

      Then \(a_{0}=0,\begin{cases} 52a_1+616a_2=0\\ 13a_1+117a_2=0 \end{cases},a_1=a_3,a_2=a_4\)

      Then \(a_1,a_2=0\), thus \(p(x)=0\). Thus \(P_4(F) = U \oplus W\)​

  3. Consider the vector space of \(2 \times 2\) matrices \(M_2(\mathbb{R})\). Let \(W_1\) be the set of matrices of the form \(\begin{pmatrix} x & -x \\ y & z \end{pmatrix}\) and let \(W_2\) be the set of matrices of the form \(\begin{pmatrix} a & b \\ -a & c \end{pmatrix}\).

    1. Prove that \(W_1\) and \(W_2\) are subspaces of \(M_2(\mathbb{R})\).

      Take \(A,B\in W_{1}\), then \(A= \begin{pmatrix} x & -x \\ y & z \end{pmatrix},B= \begin{pmatrix} a & -a \\ b & c \end{pmatrix}\)​

      Then \(A+B= \begin{pmatrix} x+a & -x-a \\ y+b & z+c \end{pmatrix}= \begin{pmatrix} x+a & -(x+a) \\ y+b & z+c \end{pmatrix}\in W_1\)​

      And \(\lambda A=\begin{pmatrix} \lambda x & -\lambda x \\ \lambda y & \lambda z \end{pmatrix}\in W_1\), thus \(W_1\) is subspaces of \(M_2(\mathbb{R})\).

      Take \(A,B\in W_{2}\), then \(A= \begin{pmatrix} x & y \\ -x & z \end{pmatrix},B= \begin{pmatrix} a & b \\ -a & c \end{pmatrix}\)

      Then \(A+B= \begin{pmatrix} x+a & y+b \\ -x-a & z+c \end{pmatrix}= \begin{pmatrix} x+a & y+b \\ -\left(x+a\right) & z+c \end{pmatrix}\in W_{2}\)

      And \(\lambda A= \begin{pmatrix} \lambda x & \lambda y \\ -\lambda x & \lambda z \end{pmatrix}\in W_{2}\), thus \(W_{2}\) is subspaces of \(M_2(\mathbb{R})\). 2. Find the dimensions of \(W_1\), \(W_2\), \(W_1 + W_2\) and \(W_1 \cap W_2\). Give an explicit basis for each of these subspaces.

      Since there are \(3\) free variables in \(W_1,W_2\), then \(\dim W_1=\dim W_2=3\)​

      The basis of \(W_1\) is \(\left\lbrace \begin{pmatrix} 1 & -1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right\rbrace\), the basis of \(W_2\) is \(\left\lbrace \begin{pmatrix} 1 & 0 \\ -1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right\rbrace\)

      Then we get the basis of \(W_{1}+W_{2}\text{ is }\left\lbrace \begin{pmatrix} 1 & -1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right\rbrace+\left\lbrace \begin{pmatrix} 1 & 0 \\ -1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right\rbrace\)​

      Since \(\begin{pmatrix} 1 & -1 \\ 0 & 0 \end{pmatrix}= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}- \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\) and \(\begin{pmatrix} 1 & 0 \\ -1 & 0 \end{pmatrix}= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}- \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\)​

      Thus the basis of \(W_{1}+W_{2}\text{ is }\left\lbrace \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\right\rbrace\)

      Take \(A\in W_1\cap W_2\), then \(\begin{pmatrix} x & -x \\ y & z \end{pmatrix}= \begin{pmatrix} a & b \\ -a & c \end{pmatrix}\), then \(x=a,x=-b,y=-a,z=c\)​

      Then \(A= \begin{pmatrix} a & -a \\ -a & c \end{pmatrix}\), then basis of \(W_1\cap W_2\) is \(\left\lbrace \begin{pmatrix} 1 & -1 \\ -1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right\rbrace\)​

  4. Consider the real vector space \(M_2(F)\). Find a basis \(\{A_1, A_2, A_3, A_4\}\) for \(M_2(F)\) such that \(A_k^2 = A_k\) for each \(k = 1, 2, 3, 4\).

    Clearly, \(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}^{2}= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}^{2}= \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}^{2}= \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}^{2}= \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}\)

    Thus claim: \(\left\lbrace \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\right\rbrace\) is a basis of \(M_2(F)\)​

    1. Linearly independent

      Let \(0=a_{1} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}+a_{2} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}+a_{3} \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}+a_{4} \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\)

      Then \(\begin{cases} a_1+a_4=0\\ a_4=0\\ a_3=0\\ a_2+a_3=0 \end{cases}\), then \(a_1=a_2=a_3=a_4=0\)​ 2. Spanning the \(M_{2}(F)\)​

      Take any \(A= \begin{pmatrix} a & b \\ c & d \end{pmatrix}\in M_{2}(F)\), then \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}=\left(a-b\right) \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}+\left(d-c\right) \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}+c \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}+b \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\)​

      Thus this list can span the \(M_2(F)\)​

    Thus \(\left\lbrace \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\right\rbrace\) is a basis of \(M_2(F)\)

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