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Homework 5.pdf

Algebra A - Winter 2024

Homework 5

1. (20 pts) Let \(V\) be a vector space with \(\dim(V) = n\). Prove that:

   1. If \(\{v_1, \ldots, v_m\} \subseteq V\) spans \(V\), then \(m \geq n\).

      Since \(\dim(V) = n\), then the length of basis is \(n\).

      Since basis is also a linearly independent, thus there is a linearly independent list and the length of it is also \(n\)

      We have \(\{v_1, \ldots, v_m\} \subseteq V\) spans \(V\)

      Since we know the theorem: Let \(v_1,...,v_n\) be a linear independent list in \(V\). If \(w_1,...,w_m\) is a list in \(V\) such that \(V=\langle w_1,\ldots,w_{m}\rangle\), then \(n\leq m\)

      Thus by theorem, \(m\geq n\)​ 2. If \(\{v_1, \ldots, v_m\} \subseteq V\) is linearly independent, then \(m \leq n\).

      Since we know a theorem that every linearly independent list can be extended to a basis.

      Since the length of basis is \(n\), thus we can add \(k\) vectors into the set (\(k\in[0,n-1]\)) until \(m+k=n\)

      Thus \(m\leq n\)​

2. (20 pts) Prove that there exists a basis \(\{p_0, p_1, p_2, p_3\}\) of \(V = \{p \in F[x] : \deg(p) \leq 3\}\) such that none of the polynomials \(p_0, p_1, p_2, p_3\) has degree \(2\).

   Let \(\deg(p_0)=3\), consider \(\deg(p_1)=3,\deg(p_2)=1,\deg(p_3)=0\)​

   Then consider \(p_0=x^3+x^2,p_1=x^3,p_2=x+1,p_3=1\)

   We need to prove it is a basis

   First, we need to prove it is linearly independent.

   Consider \(0=ax^3+ax^2+bx^3+cx+c+d=\left(a+b\right)x^3+ax^2+cx+c+d\)​

   Thus we have \(a=c=b=d=0\)​

   Then, we prove it can span the vector space.

   Consider arbitrary \(p=ex^3+fx^2+gx+h=\left(a+b\right)x^3+ax^2+cx+c+d\)

   Then \(e=a+b,f=a,g=c,h=c+d\Rightarrow a=f,b=e-f,c=g,d=h-g\) we can express any \(p\in V\)​

   Thus it is a basis

   Thus there exists a basis \(\{p_0, p_1, p_2, p_3\}\) of \(V = \{p \in F[x] : \deg(p) \leq 3\}\) such that none of the polynomials \(p_0, p_1, p_2, p_3\) has degree \(2\).

3. (20 pts) Suppose \(V\) is finite-dimensional, with \(\dim(V) = n \geq 1\). Prove that there exist 1-dimensional subspaces \(U_1, \ldots, U_n\) of \(V\) such that \(V = U_1 \oplus \cdots \oplus U_n\).

   We need to prove \(V=U_1+\cdots+U_{n}\) firstly

   Consider \(B_i=\{u_i\},\forall i\in[1,n]\) is the basis, then \(B=B_1\cup...\cup B_n\) is the basis of \(V\)

   Thus \(V=\left\lbrace\lambda_1u_1+\cdots+\lambda_{n}u_{n}\right\rbrace=U_1+\ldots+U_{n}\).

   Let \(0=\lambda_1u_1+\cdots+\lambda_{n}u_{n}\), since \(B\) is a basis then is a linearly independent

   Thus \(\lambda_1=...=\lambda_n=0\), thus we only have the trivial solution which is \(0\) vector

   Thus \(V = U_1 \oplus \cdots \oplus U_n\)

4. (40 pts) Let \(W_1\) and \(W_2\) be the following subspaces of \(\mathbb{R}^6\):

   \(W_1 = \{(x_1, x_2, x_3, x_4, x_5, x_6) \in \mathbb{R}^6 : x_1 + x_2 + x_3 = 0, x_4 + x_5 + x_6 = 0\}\),

   \(W_2 =\lang(1, -1, 1, -1, 1, -1), (1, 0, 2, 1, 0, 0), (1, 0, -1, -1, 0, 1), (2, 1, 0, 0, 0, 0)\rang\).

   1. Find a basis and the dimension of \(W_1 \cap W_2\). Describe the space implicitly by equations.

      Consider \(v\in W_1\cap W_2\)​

      Let \(W_2=\langle v_1,v_2,v_3,v_4\rangle\). Then we know \(v\in W_2\) can be expressed as \(v=\lambda_1v_1+...+\lambda_4v_4\)​

      Also \(v\in W_1\), we have \(\left(\lambda_1+\lambda_2+\lambda_3+2\lambda_4\right)-\lambda_1+\lambda_4+\lambda_1+2\lambda_2-\lambda_3=\lambda_1+3\lambda_2+3\lambda_4=0\)​

      And \(\left(-\lambda_1+\lambda_2-\lambda_3\right)+\lambda_1-\lambda_1+\lambda_3=-\lambda_1+\lambda_2=0\)​

      Thus \(\lambda_1=-\frac34\lambda_4,\lambda_2=-\frac34\lambda_4\)​

      Thus \(v=-\frac34\lambda_4v_1-\frac34\lambda_4v_2+\lambda_3v_3+\lambda_4v_4=\lambda_3v_3+\lambda_4\left(-\frac34v_1-\frac34v_2+v_4\right)\)​

      Thus the basis is \((1,0,-1,-1,0,1)\) and \((2,7,-9,0,-3,3)\)​

      The dimension is \(2\)

      Thus \(v=a(1,0,-1,-1,0,1)+b(2,7,-9,0,-3,3)\)

      Thus \(\begin{cases}x_1=a+2b\\ x_2=7b\\ x_3=-a-9b\\ x_4=-a\\ x_5=-3b\\ x_6=a+3b\end{cases}\Rightarrow a=-x_4,b=\frac{x_2}{7}\Rightarrow\begin{cases}x_1=-x_4+\frac27x_2\\ x_3=x_4-\frac97x_2\\ x_5=-\frac37x_2\\ x_6=-x_4+\frac37x_2\end{cases}\)​

      Describe: \(v=\left\lbrace\left(x_1,x_2,x_3,x_4,x_5,x_6\right)\in W_1\cap W_2:x_1=-x_4+\frac27x_2,x_3=x_4-\frac97x_2,x_5=-\frac37x_2,x_6=-x_4+\frac37x_2\right\rbrace\)​ 2. Find a basis and the dimension of \(W_1 + W_2\). Describe the space implicitly by equations.

      For \(W_2 =\lang(1, -1, 1, -1, 1, -1), (1, 0, 2, 1, 0, 0), (1, 0, -1, -1, 0, 1), (2, 1, 0, 0, 0, 0)\rang\)​

      \(W_2=\begin{pmatrix}1 & 1 & 1 & 2\\ -1 & 0 & 0 & 1\\ 1 & 2 & -1 & 0\\ -1 & 1 & -1 & 0\\ 1 & 0 & 0 & 0\\ -1 & 0 & 1 & 0\end{pmatrix}\)​

      After eliminating, we have \(W_2=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}\), thus \(\dim(W_2)=4\), it is linearly independent

      Thus by formular \(\dim(W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim\left(W_1\cap W_2\right)=4+4-2=6\)​

      Thus the basis of \(W_1+W_2\) is \((1,0,0,0,0,0),(0,1,0,0,0,0),(0,0,1,0,0,0),(0,0,0,1,0,0),(0,0,0,0,1,0),(0,0,0,0,0,1)\)​

      Describe: \(\left\lbrace\left(a,b,c,d,e,f\right):a,b,c,d,e,f\in\mathbb{R}\right\rbrace\) 3. Decide which of the following vectors are in \(W_1 \cap W_2\) and which of them are in \(W_1 + W_2\):

      \((1, 1, -2, -2, 1, 1)\), \((0, 0, 0, 1, 0, -1)\), \((1, 1, 1, 0, 0, 0)\), \((3, 0, 0, 1, 1, 3)\), \((-1, 2, 5, 6, 5, 4)\).

      All of them is in \(W_1+W_2\) since \(W_1+W_2\) is the whole space

      Let's check \(x_2,x_4\), since we have \(v=\left\lbrace\left(x_1,x_2,x_3,x_4,x_5,x_6\right)\in W_1\cap W_2:x_1=-x_4+\frac27x_2,x_3=x_4-\frac97x_2,x_5=-\frac37x_2,x_6=-x_4+\frac37x_2\right\rbrace\)​

      1. \((1, 1, -2, -2, 1, 1)\)​

         \(x_2=1,x_4=-2\), then \(x_1=2+\frac27\neq1\), thus it is not in \(W_1\cap W_2\)​ 2. \((0, 0, 0, 1, 0, -1)\)​

         \(x_2=0,x_4=1\), then \(x_1=-1\neq0\), thus it is not in \(W_1\cap W_2\) 3. \((1, 1, 1, 0, 0, 0)\)​

         \(x_2=1,x_4=0\), then \(x_1=\frac27\neq1\), thus it is not in \(W_1\cap W_2\) 4. \((3, 0, 0, 1, 1, 3)\)​

         \(x_2=0,x_4=1\), then \(x_1=-1\neq3\), thus it is not in \(W_1\cap W_2\) 5. \((-1, 2, 5, 6, 5, 4)\).

         \(x_2=2,x_4=6\), then \(x_1=-6+\frac27\cdot2\neq-1\), thus it is not in \(W_1\cap W_2\)

      Thus the vectors are all in the \(W_1+W_2\) but not in the \(W_1\cap W_2\)

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