Homework 3

999027873 Yue Shi

Homework 3.pdf

1. (30 pts) Find a spanning list of vectors for \(S + T\) as a subspace of the \(F\)-vector space \(V\) for the following subspaces \(S\) and \(T\)​

   1. \(S = \{(x, y, z) \in \mathbb{R}^3 : 2x - 3y + 2z = 0\}\) and \(T = \{(x, y, z) \in \mathbb{R}^3 : x + z = 0\}\), where \(V = \mathbb{R}^3\), \(F = \mathbb{R}\)​

      Solution

      We know \(\forall v\in S+T,\exists s\in S,t\in T:v=s+t\)​

      Let \(s=\left(2x,2y,3y-2x\right)\) and \(t=\left(a,b,-a\right)\)​

      Thus \(S+T=\langle(2x+a,2y+b,3y-2x-a)\rangle\)

      Thus the spanning list is \(\{(2,0,-2),(0,2,3),\left(1,0,-1\right),\left(0,1,0\right)\}\)​ 2. \(S = \langle (2,-i), (1, i), (1 + i,-i) \rangle\) and \(T = \langle (i, 0), (i, 2i), (0,1) \rangle\), where \(V = \mathbb{C}^2\), \(F = \mathbb{R}\).

      Solution

      We know \(\forall v\in S+T,\exists s\in S,t\in T:v=s+t\)

      Then \(s=a_{1}(2,-i)+a_{2}(1,i)+a_{3}(1+i,-i),t=b_{1}(i,0)+b_{2}(i,2i)+b_{3}(0,1)\)

      Thus \(v=a_{1}(2,-i)+a_{2}(1,i)+a_{3}(1+i,-i)+b_{1}(i,0)+b_{2}(i,2i)+b_{3}(0,1)\)​

      Thus the spanning list of \(S+T\) can be \(\{(2,-i),(1,i),(1+i,-i),(i,0),(i,2i),(0,1)\}\)​

2. (30 pts) Consider the following subspace of \(F^5\): \(U = \{(x, y, x+y, x-y, 2x) \in F^5 : x, y \in F\}.\)

   Find three subspaces \(W_1, W_2, W_3\) of \(F^5\) (none of them equal to \(\{0\}\)) such that \(F^5 = U \oplus W_1 \oplus W_2 \oplus W_3.\)

   Since \(U = \{(x, y, x+y, x-y, 2x) \in F^{5} : x, y \in F\}\), then \(U=\langle\left(1,0,1,1,2\right),\left(0,1,1,-1,0\right)\rangle\)​

   Let \(W_1=\lang(0,0,1,0,0)\rang,W_2=\lang(0,0,0,1,0)\rang,W_3=\lang(0,0,0,0,1)\rang\)​

   Let's check \(F^5 = U \oplus W_1 \oplus W_2 \oplus W_3.\)

   Assume \((0,0,0,0,0)=a_1(1,0,1,1,2)+a_2(0,1,1,-1,0)+a_3(0,0,1,0,0)+a_4(0,0,0,1,0)+a_5(0,0,0,0,1)\)​

   Then \(\begin{cases} a_1=0\\a_2=0\\a_1+a_2+a_3=0\\a_1-a_2+a_4=0\\2a_1+a_5=0 \end{cases}\), thus we have \(a_1=a_2=a_3=a_4=a_5=0\)​

   Thus this is direct sum, then \(W_1=\lang(0,0,1,0,0)\rang,W_2=\lang(0,0,0,1,0)\rang,W_3=\lang(0,0,0,0,1)\rang\) is valid

3. (25 pts) Let \(A = (a_{ij}) \in M_n(F)\) be an upper triangular matrix, that is \(A =\begin{pmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ 0 & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & a_{nn}\end{pmatrix}.\)​

   Suppose that the diagonal entries of \(A\) are not equal to 0 (\(a_{ii} \neq 0\) for every \(i = 1, \dots, n\)) and denote by \(v_1, \dots, v_n\) the columns of \(A\). Prove that \(\{v_{1}, \dots, v_{n}\}\) is linearly independent and spans every vector in \(F^n\).

   Here \(v_{1}=(a_{11},0,...,0),v_{2}=(a_{12},a_{22},0,...,0),...,v_{n}=(a_{1n},...,a_{nn})\)

   Then consider the list \(\{v_{1}, \dots, v_{n}\}\). NTP: it's linearly independent and spanning list in \(F^n\)

   Let \(0=b_1v_1+...+b_nv_n\), then \(\begin{cases} b_{n}a_{nn}=0 \\ b_{n}a_{(n-1)n}+b_{n-1}a_{(n-1)(n-1)}=0 \\ ... \\ b_{n}a_{1n}+b_{n-1}a_{1\left(n-1\right)}+b_{n-2}a_{1\left(n-2\right)}+\cdots+b_{1}a_{11}=0 \end{cases}\)

   Then since \(a_{ii} \neq 0\) for every \(i = 1, \dots, n\), we have \(b_n=0,b_{n-1}=0,...,b_1=0\)

   Thus it is linearly independent

   ---

   Let \(\forall (x_{1},...,x_{n})\in F^{n}\), then consider \((x_{1},...,x_{n})=b_{1}v_{1}+...+b_{n}v_{n}\)​

   We have \(\begin{cases} x_{n}=b_{n}a_{nn}\Rightarrow b_{n}=\frac{x_{n}}{a_{nn}} \\ x_{n-1}=b_{n}a_{\left(n-1\right)n}+b_{n-1}a_{(n-1)\left(n-1\right)}\Rightarrow b_{n-1}=\frac{x_{n-1}-b_{n}a_{\left(n-1\right)n}}{a_{\left(n-1\right)\left(n-1\right)}} \\ .... \\ x_{1}=b_{n}a_{1n}+b_{n-1}a_{1\left(n-1\right)}+b_{n-2}a_{1\left(n-2\right)}+\cdots+b_{1}a_{11}\Rightarrow b_{1}=\frac{x_1-b_{n}a_{1n}-b_{n-1}a_{1\left(n-1\right)}-\cdots-b_2a_{12}}{a_{11}} \end{cases}\)

   Thus for all \(\forall (x_{1},...,x_{n})\in F^{n}\), exists \(b_{1}=\frac{x_{n}}{a_{nn}},\ldots,b_{n-1}=\frac{x_{n-1}-b_{n}a_{\left(n-1\right)n}}{a_{\left(n-1\right)n-1}} ,b_{n}=\frac{x_{n}}{a_{nn}}\in F:\) \((x_{1},...,x_{n})=b_{1}v_{1}+...+b_{n}v_{n}\).

   Thus it spans every vector in \(F^n\)​

4. (15 pts) Prove or give a counterexample: if \(\{v_1, v_2, \dots, v_n\}\) is a linearly independent list of vectors in \(V\), then
   ​\(\{5v_1 - 4v_2, v_2, v_3, \dots, v_n\}\)is linearly independent.

   Consider \(a_1(5v_1-4v_2)+a_2v_2+...+a_nv_n=0\), then \(5a_{1}v_{1}+(a_{2}-4a_{1})v_{2}+a_{3}v_{3}+...+a_{n}v_{n}=0\)​

   Since \(\{v_1, v_2, \dots, v_n\}\) is a linearly independent list, then if \(b_1v_1+...+b_nv_n=0\), we get \(b_1=....=b_n=0\)​

   Let \(b_1=5a_1,b_2=a_2-4a_1,b_i=a_i,\forall i\geq 3\), thus also \(a_1=0,a_2=4a_1,a_3=0,...,a_n=0\)​

   Then \(a_1=a_2=....=a_n=0\)​

   Thus \(\{5v_1 - 4v_2, v_2, v_3, \dots, v_n\}\)is linearly independent.

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