11.15 Linear independent and basis

1. Suppose \(\{v_1, \dots, v_n\}\) is linearly independent in \(V\) and let \(w \in V\). Show that \(\{v_1, \dots, v_n, w\}\) is linearly independent if and only if \(w \notin \langle v_1, \dots, v_n \rangle\).

   Proof

   \(\Rightarrow\)) Suppose \(w \in \langle v_1, \dots, v_n \rangle\), then \(w=\lambda_1v_1+...+\lambda_nv_n\)​

   Since \(\{v_1, \dots, v_n, w\}\) is linearly independent, then consider \(0=a_1v_1+...+a_{n}v_{n}+a_{n+1}w\), we have \(a_1=0,...,a_n=0,a_{n+1}=0\)​

   Also \(0=a_1v_1+...+a_{n}v_{n}+a_{n+1}w=\left(a_1+\lambda_1\right)v_1+\cdots+\left(a_{n}+\lambda_{n}\right)v_{n}\)

   Since \(\{v_1, \dots, v_n\}\) is linearly independent, then \(a_1=-\lambda_1=0,...,a_{n}=-\lambda_{n}=0\)

   Which means \(w=0\) that contradicts to \(\{v_1, \dots, v_n, w\}\) is linearly independent since the coefficient of \(w\) can be any

   Thus \(w \notin \langle v_1, \dots, v_n \rangle\).

   \(\Leftarrow\)) Since \(w \notin \langle v_1, \dots, v_n \rangle\), consider \(0=a_1v_1+...+a_{n}v_{n}+a_{n+1}w\).

   If \(a_{n+1}=0\), since \(\{v_1, \dots, v_n\}\) is linearly independent, then we are done.

   If \(a_{n+1}\neq 0\), then \(w=-\frac{a_1}{a_{n+1}}v_1-\cdots-\frac{a_{n}}{a_{n+1}}v_{n}\) contradiction!!

   Thus\(a_{n+1}=0\)​ is the only option

2. Consider the set \(S = \{(1, 3, -3), (2, 3, -4), (1, -3, 1)\}\) of vectors in \(\mathbb{R}^3\) and let \(W\) be the set spanned by the vectors in \(S\).

   1. Is \(S\) linearly independent? If not, find a non-trivial linear relation between the vectors.

      \(0 = a(1, 3, -3) + b(2, 3, -4) + c(1, -3, 1)\)

      \(\begin{pmatrix} \begin{array}{ccc|c} 1 & 3 & -3 & 0 \\ 2 & 3 & -4 & 0 \\ 1 & -3 & 1 & 0 \end{array} \end{pmatrix} \Rightarrow \begin{pmatrix} \begin{array}{ccc|c} 1 & 0 & -1 & 0 \\ 0 & 1 & -\frac{3}{2} & 0 \\ 0 & 0 & 0 & 0 \end{array} \end{pmatrix}\)

      No, \(\langle(1,\frac32,1)\rangle\) 2. Find the dimension of \(W\) and give a basis for \(W\).

      \(\dim(W)=2\)​

      Basis: \(\left\lbrace\left(1,3,-3\right),\left(2,3,-4\right)\right\rbrace\) 3. Describe implicitly the subspace \(W\). That is, find a homogeneous system of equations for which the solution space is \(W\).

      \((x,y,z)=a(1,3,-3)+b(2,3,-4)\)

      \(\begin{pmatrix}\begin{array}{cc|c}1 & 2 & x\\ 3 & 3 & y\\ -3 & -4 & z\end{array}\end{pmatrix}\Rightarrow\begin{pmatrix}\begin{array}{cc|c}1 & 0 & x+2y+2z\\ 0 & 1 & -y-z\\ 0 & 0 & 3z+3x+2y\end{array}\end{pmatrix}\Rightarrow2y+3z+3x=0\)

      \(\Rightarrow\left(x,y,\frac{-3x-2y}{3}\right)\Rightarrow W=\langle(1,0,-1),(0,1,-\frac23)\rangle\) 4. Decide if the vectors \((4, -5, 1)\), \((5, 15, 5)\), and \((-5, 15, -5)\) are in \(W\).

      no, no, no

3. Suppose \(p_0, p_1, \dots, p_m\) are polynomials in \(V = \{p \in \mathbb{F}[x] : \deg(p) \leq m\}\) such that \(p_k(2) = 0\) for each \(k\). Prove that \(\{p_0, p_1, \dots, p_m\}\) is not linearly independent in \(V\).

   Proof

   Suppose it is linearly independent (hypothesis)

   Since we know \(1,x,x^2,...,x^m\) is a basis for \(\mathbb{F}[x]\), thus \(\dim V=m+1\)

   Also, we have \(p_0,...,p_m\), then by theorem it is a basis by hypothesis

   Thus there must exists a polynomial s.t. \(q(2)\neq0\), for example: \(q(z) = 1, z, z^2, \dots, z^m \neq 0 \text{ for } z = 2\)​

4. Let \(V\) be the real vector space spanned by the rows of the matrix \(A = \begin{pmatrix} 3 & 21 & 0 & 9 & 0 \\ 1 & 7 & -1 & -2 & -1 \\ 2 & 14 & 0 & 6 & 1 \\ 6 & 42 & -1 & 13 & 0 \end{pmatrix}\).

   (a) Find a basis for \(V\).

   \(\begin{pmatrix}1 & 7 & 0 & 3 & 0\\ 0 & 0 & 1 & 5 & 0\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0\end{pmatrix}\), thus the basis is \(\{(1,7,0,3,0),(0,0,1,5,0),(0,0,0,0,1)\}\)​

   (b) Describe all vectors \((x_1, x_2, x_3, x_4, x_5)\) which are elements in \(V\).

   \((x_1,x_2,x_3,x_4,x_5)=a(1,7,0,3,0)+b(0,0,1,5,0)+c(0,0,0,0,1)\)

   Then we have \(\begin{cases}a=x_1\\7a=x_2\\b=x_3\\3a+5b=x_4\\c=x_5\end{cases}\)

   Thus we have \((x_1,7x_1,x_3,3x_1+5x_3,x_5)\)​

   (c) If \((x_1, x_2, x_3, x_4, x_5)\) is in \(V\), what are its coordinates in the basis chosen in (a)?

   \([x_1,x_3,x_5]\)

5. Suppose \(U_1, \dots, U_m\) are finite-dimensional subspaces of \(V\) such that \(U_1 + \dots + U_m\) is a direct sum. Prove that \(U_1 \oplus \dots \oplus U_m\) is finite-dimensional and \(\dim(U_1 \oplus \dots \oplus U_m) = \dim(U_1) + \dots + \dim(U_m)\).

   Proof

   Since \(U_1,...,U_m\) are finite-dimensional subspaces of \(V\), then consider \(B_{i}=\left\lbrace u_1^{i},\ldots,u_{k_i}^{i}\right\rbrace\) for some \(k_i\) is the basis of \(U_i\)

   Since \(U_1+...+U_m\) is direct sum, thus \(v\) can be written in a unique way.

   Thus \(v=u_1+...+u_{m}=\left(\lambda_1^1u_1^1+\ldots+\lambda_{k_1}^1u_{k_1}^1\right)+...+\left(\lambda_1^{m}u_1^{m}+\ldots+\lambda_{k_{m}}^{m}u_{k_{m}}^{m}\right)\) is unique

   Then Since \(B_{i}=\left\lbrace u_1^{i},\ldots,u_{k_i}^{i}\right\rbrace\) for some \(k_i\) is the basis of \(U_i\)

   Thus \(B=B_1\cup B_2\cup...\cup B_m\) spans \(U_1 \oplus \dots \oplus U_m\), then it is the basis of \(U_1\oplus\dots\oplus U_{m}\)

   Thus \(U_1 \oplus \dots \oplus U_m\) is finite-dimensional

   Also, Since \(B=B_1\cup B_2\cup...\cup B_m\) is the basis of \(U_1\oplus\dots\oplus U_{m}\), then \(\dim(U_1\oplus\ldots\oplus U_{m})=\left|B\right|=k_1+k_2+\ldots+k_{m}\)​

   And \(\dim(U_1)+\ldots+\dim(U_{m})=\left|B_1\right|+\ldots+\left|B_{m}\right|=k_1+k_2+\ldots+k_{m}\)​

   Thus \(\dim(U_1 \oplus \dots \oplus U_m) = \dim(U_1) + \dots + \dim(U_m)\).

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