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Summary
- A subset $S$ of $F^{n}$ is called a subspace of $F^{n}$ if
- $0\in S$;
- If $u\in S$ and $v\in S$, then $u+v\in S$;
- If $u\in S$ and $t\in F$, then $tu\in S$.
- Let $X_1, \cdots, X_m\in F^{n}$. Then the set consisting of all linear combinations $x_1X_1+\cdots+x_mX_m$, where $x_1, \cdots, x_m\in F$, is a subspace of $F^{n}$. This subspace is called the subspace spanned or generated by $X_1, \cdots, X_m$ and is denoted by $\big < X_1, \cdots, X_m\big > $. We also call $X_1, \cdots, X_m$ a spanning family for $S=\big < X_1, \cdots, X_m\big > $.
- Let $A\in M_{m\times n}(F)$. Then the set of vectors $X\in F^{n}$ satisfying $AX=0$ is a subspace of $F^{n}$ called the null space of $A$ and is denoted by $N(A)$.
- If $A\in M_{m\times n}(F)$, the subspace generated by the columns of $A$ is the subspace of $F^{m}$ and is called the column space of $A$, $C(A)$. Similarly, the subspace generated by the rows of $A$ is the subspace of $F^{n}$ and is called the row space of $A$, $R(A)$.
- Suppose each of $X_1, \cdots, X_r$ is a linear combination of $Y_1, \cdots, Y_s$. Then any linear combination of $X_1, \cdots, X_r$ is a linear combination of $Y_1, \cdots, Y_s$. That is, $\big < X_1, \cdots, X_r\big > \subseteq\big < Y_1, \cdots, Y_s\big > $.
- Subspaces $\big < X_1, \cdots, X_r\big > $ and $\big < Y_1, \cdots, Y_s\big > $ are equal if each of $X_1, \cdots, X_r$ is a linear combination of $Y_1, \cdots, Y_s$ and each of $Y_1, \cdots, Y_s$ is a linear combination of $X_1, \cdots, X_r$.
- $\big < X_1,\cdots, X_r, Z_1, \cdots, Z_t\big > $ and $\big < X_1,\cdots, X_r\big > $ are equal if each of $Z_1, \cdots, Z_t$ is a linear combination of $X_1,\cdots, X_r$.
- If $A$ is row equivalent to $B$, then $R(A)=R(B)$. (Note that it is not always true that $C(A)=C(B)$)
- Linearly dependent and linear independent
Vectors $X_1, \cdots, X_m$ in $F^{n}$ are said to be linearly dependent if there exist scalars $x_1, \cdots, x_m$, not all zero, such that $$x_1X_1+\cdots +x_mX_m=0.$$ $X_1, \cdots, X_m$ are called linearly independent if they are not linearly dependent. Hence $X_1, \cdots, X_m$ are linearly independent if and only if the equation $$x_1X_1+\cdots +x_mX_m=0$$ has only the trivial solution $x_1=0,\cdots, x_m=0$. - A family of $m$ vectors in $F^{n}$ will be linearly dependent if $m > n$. Equivalently, any linearly independent family of $m$ vectors in $F^{n}$ must satisfy $m \leq n$.
- A family of $s$ vectors in $\big < X_1, \cdots, X_r\big > $ will be linearly dependent if $s > r$. Equivalently, a linearly independent family of $s$ vectors in $\big < X_1, \cdots, X_r\big > $ must have $s \leq r$.
- Left-to-right test
Vectors $X_1, \cdots, X_m$ in $F^{n}$ are linearly independent if
- $X_1 \neq 0$;
- For each $k$ with $1 < k \leq m$, $X_k$ is not a linear combination of $X_1, \cdots, X_{k-1}$.
- Every subspace $S$ of $F^{n}$ can be represented in the form $S=\big < X_1\cdots X_m\big > $, where $m \leq n$.
- Suppose that $A$ is row equivalent to $B$ and let $c_1, \cdots, c_r$ be distinct integers satisfying $1 \leq c_i \leq n$. Then
- Columns $A_{* c_1}, \cdots, A_{* c_r}$ of $A$ are linearly dependent if and only if the corresponding columns of $B$ are linearly dependent; indeed more is true: $$x_1A_{*c_1}+\cdots + x_rA_{*c_r}=0\Leftrightarrow x_1B_{*c_1}+\cdots + x_rB_{*c_r}=0$$
- Columns $A_{* c_1}, \cdots, A_{* c_r}$ of $A$ are linearly independent if and only if the corresponding columns of $B$ are linearly independent.
- If $1 \leq c_{r+1} \leq n$ and $c_{r+1}$ is distinct from $c_1, \cdots, c_r$, then $$A_{*c_{r+1}}= z_1A_{*c_{1}}+ \cdots + z_rA_{*c_{r}}\Leftrightarrow B_{*c_{r+1}}= z_1B_{*c_{1}}+ \cdots + z_rB_{*c_{r}}$$
- Basis
Vectors $X_1, \cdots, X_m$ belonging to a subspace $S$ are said to form a basis of $S$ if
- Every vector in $S$ is a linear combination of $X_1, \cdots, X_m$;
- $X_1, \cdots, X_m$ are linearly independent.
- A subspace of the form $\big < X_1, \cdots, X_m\big > $, where at least one of $X_1, \cdots, X_m$ is non-zero, has a basis $X_{c_1}, \cdots, X_{c_r}$, where $1 \leq c_1 < \cdots < c_r \leq m$.
- Left-to-right algorithm
- A subspace of the form $\big < X_1, \cdots, X_m\big > $. Let $c_1$ be the least index $k$ for which $X_k$ is non-zero. If $c_1=m$ or if all the vectors $X_k$ with $k > c_1$ are linear combinations of $X_{c_1}$, terminate the algorithm and let $r = 1$. Otherwise let $c_2$ be the least integer $k > c_1$ such that $X_k$ is not a linear combination of $X_{c_1}$.
- If $c_2=m$ or if all the vectors $X_k$ with $k > c_2$ are linear combinations of $X_{c_1}$ and $X_{c_2}$, terminate the algorithm and let $r = 2$. Eventually the algorithm will terminate at the $r$th stage, either because $c_r=m$, or because all vectors $X_k$ with $k > c_r$ are linear combinations of $X_{c_1}, \cdots, X_{c_r}$.
- $X_{c_1}, \cdots, X_{c_r}$ is called the left-to-right basis for the subspace $\big < X_1, \cdots, X_m\big > $. For example, $2X, -Y$ is the left-to-right basis of $\big < 0, 2X, X, -Y, X+Y\big > $
- Dimension
Any two bases for a subspace $S$ must contain the same number of elements. This number is called the dimension of $S$ and is written $\dim S$. Naturally we define $\dim{0} = 0$. - A linearly independent family of $m$ vectors in a subspace $S$, with $\dim S=m$, must be a basis for $S$.
- Rank and nullity
- column rank $A = \dim C(A)$;
- row rank $A = \dim R(A)$;
- nullity $A= \dim N(A)$.
- Let $A\in M_{m\times n}(F)$. Then
- column rank $A =$ row rank $A$;
- column rank $A +$ nullity $A = n$.
- The common value if column rank $A$ and row rank $A$ is called the rank of $A$.
- Every linearly independent family of vectors in a subspace $S$ can be extended to a basis of $S$.
Problems 3.6
1. Which of the following subsets of $\mathbb{R}^2$ are subspaces? (a) $[x, y]$ satisfying $x=2y$; (b) $[x, y]$ satisfying $x=2y$ and $2x=y$; (c) $[x, y]$ satisfying $x=2y+1$; (d) $[x, y]$ satisfying $xy=0$; (e) $[x, y]$ satisfying $x \geq 0$ and $y \geq 0$.
Solution:
(a) (1) $[0, 0]\in S$; (2) Suppose $[x_1, y_1]\in S$ and $[x_2, y_2]\in S$, so $x_1+x_2 = 2y_1+2y_2=2(y_1+y_2)$, that is $[x_1+x_2, y_1+y_2] = [x_1, y_1] + [x_2, y_2]\in S$; (3) Suppose $t\in R$, so $tx = t\cdot2y=2(ty)$, that is $[tx, ty]=t[x, y]\in S$. Hence this is subspace. (b) We have $x = 2y=4x\Rightarrow x=y=0$. That is, $S=\{[0, 0]\}$ which is consisting of zero vector, and it is always subspace. (c) $[0, 0]\notin S$ hence it is not subspace. (d) $xy=0\Rightarrow x=0$ or $y=0$. But it is not closed under addition, for example, $[0, 1] + [1, 0]=[1, 1]\notin S$. Hence it is not subspace. (e) This is not closed under scalar multiplication. For example, $[1, 0] \in S$ and $t=-1\in R$, but $t[1, 0]=[-1, 0]\notin S$. Hence it is not subspace.
2. If $X$, $Y$, $Z$ are vectors in $\mathbb{R}^n$, prove that $$\big < X, Y, Z\big > = \big < X+Y, X+Z, Y+Z \big > $$
Solution:
First, each of $X+Y$, $X+Z$, $Y+Z$ is a linear combination of $X$, $Y$, $Z$. So $$\big < X+Y, X+Z, Y+Z \big > \subseteq \big < X, Y, Z\big > $$ On the other hand, $$\begin{cases}X={1\over2}(X+Y) + {1\over2}(X+Z) - {1\over2}(Y+Z)\\ Y={1\over2}(X+Y) - {1\over2}(X+Z) + {1\over2}(Y+Z)\\ Z=-{1\over2}(X+Y) + {1\over2}(X+Z) + {1\over2}(Y+Z) \end{cases}$$ That is, each of $X$, $Y$, $Z$ is a linear combination of $X+Y$, $X+Z$, $Y+Z$, so $$\big < X, Y, Z\big > \subseteq \big < X+Y, X+Z, Y+Z \big > $$ Thus, $\big < X, Y, Z\big > = \big < X+Y, X+Z, Y+Z \big > $.
3. Determine if $X_1 = \begin{bmatrix}1\\0\\ 1\\ 2 \end{bmatrix}$, $X_2 = \begin{bmatrix}0\\1\\ 1\\ 2 \end{bmatrix}$, and $X_3 = \begin{bmatrix}1\\ 1\\ 1\\ 3 \end{bmatrix}$ are linearly independent in $\mathbb{R}^4$.
Solution:
Suppose we have $xX_1+yX_2+zX_3=0$, that is $$\begin{cases}x + z=0\\ y+z =0\\ x+y+z = 0\\ 2x+2y+3z = 0 \end{cases} \Rightarrow x=y=z=0$$ Thus they are linearly independent.
4. For which real number $\lambda$ are the following vectors linearly independent in $\mathbb{R}^3$? $$X_1=\begin{bmatrix}\lambda\\ -1\\ -1 \end{bmatrix}, X_2=\begin{bmatrix}-1\\ \lambda\\ -1 \end{bmatrix}, X_3=\begin{bmatrix}-1\\ -1\\ \lambda \end{bmatrix}.$$
Solution:
Suppose there exists $x$, $y$, $z$ satisfying $xX_1+yX_2+zX_3= 0$. That is $$\begin{cases}\lambda x -y -z=0\\ -x + \lambda y-z =0\\ -x -y +\lambda z = 0\end{cases}$$ $$\Rightarrow \begin{bmatrix}\lambda & -1& -1\\ -1& \lambda& -1\\ -1& -1& \lambda \end{bmatrix}$$ $$\Rightarrow\begin{cases}R_1 + \lambda R_3\\ R_2-R_3\\ -R_3 \end{cases}\begin{bmatrix}0 & -\lambda-1& \lambda^2-1\\ 0& \lambda+1& -1-\lambda\\ 1& 1 & -\lambda \end{bmatrix}$$ If $\lambda = -1$, then $$\begin{bmatrix}0 & 0& 0\\ 0& 0& 0\\ 1& 1 & 1 \end{bmatrix}\Rightarrow x = -y-z$$ which means there are non-trivial solutions. Thus $\lambda \neq -1$. We have $$\begin{bmatrix}0 & -\lambda-1& \lambda^2-1\\ 0& \lambda+1& -1-\lambda\\ 1& 1 & -\lambda \end{bmatrix}$$ $$\Rightarrow\begin{cases}{1\over\lambda+1} R_1\\ {1\over\lambda+1} R_2 \end{cases} \begin{bmatrix}0 & -1& \lambda-1\\ 0& 1& -1\\ 1& 1 & -\lambda \end{bmatrix}$$ $$\Rightarrow\begin{cases}R_1 + R_2\\ R_3-R_2 \end{cases} \begin{bmatrix}0 & 0& \lambda-2\\ 0& 1& -1\\ 1& 0 & -\lambda+1 \end{bmatrix}$$ If $\lambda = 2$, then $$\begin{bmatrix}0 & 0& 0\\ 0& 1& -1\\ 1& 0 & -1 \end{bmatrix} \Rightarrow \begin{cases}x = z\\ y = z \end{cases}$$ which means there are non-trivial solutions. Thus $\lambda \neq 2$. Hence we have $$\begin{bmatrix}0 & 0& \lambda-2\\ 0& 1& -1\\ 1& 0 & -\lambda+1 \end{bmatrix} \Rightarrow {1\over\lambda-2}R_1 \begin{bmatrix}0 & 0& 1\\ 0& 1& -1\\ 1& 0 & -\lambda+1 \end{bmatrix}$$ $$\Rightarrow \begin{cases}R_2+R_1\\ R_3+(\lambda-1)R_1 \end{cases} \begin{bmatrix}0 & 0& 1\\ 0& 1& 0\\ 1& 0 & 0 \end{bmatrix} \Rightarrow x= y= z = 0$$ Therefore, it is linearly independent when $\lambda\neq -1, 2$.
5. Find bases for the row, column and null spaces of the following matrix over $\mathbb{Q}$: $$A=\begin{bmatrix}1& 1& 2& 0& 1\\ 2& 2& 5& 0& 3\\ 0& 0& 0& 1& 3\\ 8& 11& 19& 0& 11 \end{bmatrix}$$
Solution:
$$A=\begin{bmatrix}1& 1& 2& 0& 1\\ 2& 2& 5& 0& 3\\ 0& 0& 0& 1& 3\\ 8& 11& 19& 0& 11 \end{bmatrix}$$ $$\Rightarrow\begin{cases}R_2- 2R_1\\ R_4-8R_1\end{cases} \begin{bmatrix}1& 1& 2& 0& 1\\ 0& 0& 1& 0& 1\\ 0& 0& 0& 1& 3\\ 0& 3& 3& 0& 3 \end{bmatrix}$$ $$\Rightarrow{1\over3} R_4 \begin{bmatrix}1& 1& 2& 0& 1\\ 0& 0& 1& 0& 1\\ 0& 0& 0& 1& 3\\ 0& 1& 1& 0& 1 \end{bmatrix}$$ $$\Rightarrow \begin{cases}R_1-R_4\\ R_4-R_2\end{cases} \begin{bmatrix}1& 0& 1& 0& 0\\ 0& 0& 1& 0& 1\\ 0& 0& 0& 1& 3\\ 0& 1& 0& 0& 0 \end{bmatrix}$$ $$\Rightarrow R_1-R_2 \begin{bmatrix}1& 0& 0& 0& -1\\ 0& 0& 1& 0& 1\\ 0& 0& 0& 1& 3\\ 0& 1& 0& 0& 0 \end{bmatrix}$$ $$\Rightarrow \begin{bmatrix}1& 0& 0& 0& -1\\ 0& 1& 0& 0& 0\\ 0& 0& 1& 0& 1\\ 0& 0& 0& 1& 3 \end{bmatrix} = B$$ Thus, the bases of $R(A)$ and $C(A)$ are: $$\begin{cases}R(A): & [1, 0, 0, 0, -1], [0, 1, 0, 0, 0], [0, 0, 1, 0, 1], [0, 0, 0, 1, 3]\\ C(A): & [1, 2, 0, 8]^{t}, [1, 2, 0, 11]^{t}, [2, 5, 0, 19]^{t}, [0, 0, 1, 0]^{t}\\ \end{cases}$$ For $N(A)$, we have $AX=0$ where $A=[x_1, x_2, x_3, x_4, x_5]^{t}$, that is $$\begin{cases}x_1=x_5\\ x_2=0\\x_3=-x_5\\ x_4=-3x_5 \end{cases}\Rightarrow X=\begin{bmatrix}x_5\\ 0\\ -x_5\\ -3x_5\\ x_5 \end{bmatrix} = x_5\begin{bmatrix}1\\ 0\\ -1\\ -3\\ 1 \end{bmatrix}$$ Hence the basis of $N(A)$ is $[1, 0, -1, -3, 1]^{t}$.
6. Find bases for the row, column and null spaces of the following matrix over $\mathbb{Z}_2$: $$A=\begin{bmatrix}1& 0& 1& 0& 1\\ 0& 1& 0& 1& 1\\ 1& 1& 1& 1& 0\\ 0& 0& 1& 1& 0 \end{bmatrix}$$
Solution:
Recall that $1 + 1 = 0 \Rightarrow -1=1$ in $\mathbb{Z}_2$ field.
$$A=\begin{bmatrix}1& 0& 1& 0& 1\\ 0& 1& 0& 1& 1\\ 1& 1& 1& 1& 0\\ 0& 0& 1& 1& 0 \end{bmatrix}$$ $$\Rightarrow R_3-R_1 \begin{bmatrix}1& 0& 1& 0& 1\\ 0& 1& 0& 1& 1\\ 0& 1& 0& 1& 1\\ 0& 0& 1& 1& 0 \end{bmatrix}$$ $$\Rightarrow \begin{cases}R_3-R_2\\ R_1-R_4\end{cases} \begin{bmatrix}1& 0& 0& 1& 1\\ 0& 1& 0& 1& 1\\ 0& 0& 0& 0& 0\\ 0& 0& 1& 1& 0 \end{bmatrix}$$ $$\Rightarrow R_3\leftrightarrow R_4\begin{bmatrix}1& 0& 0& 1& 1\\ 0& 1& 0& 1& 1\\ 0& 0& 1& 1& 0 \\ 0& 0& 0& 0& 0 \end{bmatrix}$$ The bases of $R(A)$ and $C(A)$ are: $$\begin{cases}R(A): & [1, 0, 0, 1, 1], [0, 1, 0, 1, 1], [0, 0, 1, 1, 0]\\ C(A): & [1, 0, 1, 0]^{t}, [0, 1, 1, 0]^{t}, [1, 0, 1, 1]^{t}\end{cases}$$ For $N(A)$ we have $$\begin{cases}x_1=-x_4-x_5=x_4+x_5\\ x_2=-x_4-x_5=x_4+x_5\\ x_3= -x_4=x_4\\ x_4=x_4\\x_5=x_5 \end{cases}\Rightarrow X=x_4\begin{bmatrix}1\\ 1\\ 1\\ 1\\ 0 \end{bmatrix} + x_5\begin{bmatrix}1\\ 1\\ 0\\ 0\\ 1 \end{bmatrix}$$ Hence the basis of $N(A)$ is $[1, 1, 1, 1, 0]^{t}, [1, 1, 0, 0, 1]^{t}$.
7. Find bases for the row, column and null spaces of the following matrix over $\mathbb{Z}_5$: $$A=\begin{bmatrix}1& 1& 2& 0& 1 & 3\\ 2& 1& 4& 0& 3 & 2\\ 0& 0& 0& 1& 3 &0\\ 3& 0& 2& 4& 3 & 2 \end{bmatrix}$$
Solution:
Recall that for $x \notin \{0, 1, 2, 3, 4\}$ we have $\begin{cases}x - 5n & x > 0\\ 5n - x & x < 0 \end{cases}$, for example, $8 = 8 - 5 = 3$, $-8 = 10 - 8 = 2$. $$A=\begin{bmatrix}1& 1& 2& 0& 1 & 3\\ 2& 1& 4& 0& 3 & 2\\ 0& 0& 0& 1& 3 &0\\ 3& 0& 2& 4& 3 & 2 \end{bmatrix}$$ $$\Rightarrow \begin{cases}R_2-2R_1\\ R_4-3R_1 \end{cases} \begin{bmatrix}1& 1& 2& 0& 1 & 3\\ 0& 4& 0& 0& 1 & 1\\ 0& 0& 0& 1& 3 &0\\ 0& 2& 1& 4& 0 & 3 \end{bmatrix}$$ $$\Rightarrow\begin{cases}2R_1-R_4\\ R_2-2R_4 \end{cases} \begin{bmatrix}2& 0& 3& 1& 2 & 3\\ 0& 0& 3& 2& 1 & 0\\ 0& 0& 0& 1& 3 &0\\ 0& 2& 1& 4& 0 & 3 \end{bmatrix}$$ $$\Rightarrow \begin{cases} R_1-R_2\\ 3R_4-R_2\end{cases} \begin{bmatrix}2& 0& 0& 4& 1 & 3\\ 0& 0& 3& 2& 1 & 0\\ 0& 0& 0& 1& 3 &0\\ 0& 1& 0& 0& 4 & 4 \end{bmatrix}$$ $$\Rightarrow \begin{cases} R_1-4R_3\\ R_2-2R_3\end{cases} \begin{bmatrix}2& 0& 0& 0& 4 & 3\\ 0& 0& 3& 0& 0 & 0\\ 0& 0& 0& 1& 3 &0\\ 0& 1& 0& 0& 4 & 4 \end{bmatrix} = \begin{bmatrix}2& 0& 0& 0& 4 & 8\\ 0& 0& 3& 0& 0 & 0\\ 0& 0& 0& 1& 3 &0\\ 0& 1& 0& 0& 4 & 4 \end{bmatrix}$$ $$\Rightarrow \begin{bmatrix}1& 0& 0& 0& 2 & 4\\ 0& 1& 0& 0& 4 & 4\\ 0& 0& 1& 0& 0 & 0\\ 0& 0& 0& 1& 3 &0\end{bmatrix}$$ Thus the bases of $R(A)$ and $C(A)$ are: $$\begin{cases}R(A):& [1, 0, 0, 0, 2, 4], [0, 1, 0, 0, 4, 4], [0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 3, 0] \\ C(A): & [1, 2, 0, 3]^{t}, [1, 1, 0, 0]^{t}, [2, 4, 0, 2]^{t}, [0, 0, 1, 4]^{t}\end{cases}$$ For $N(A)$, we have $$\begin{cases}x_1=-2x_5-4x_6=3x_5+x_6\\ x_2 = -4x_5 - 4x_6= x_5+x_6\\ x_3=0\\ x_4 = -3x_5=2x_5\\ x_5 = x_5\\ x_6 = x_6 \end{cases}\Rightarrow X=x_5\begin{bmatrix}3\\1\\0\\2\\1\\0 \end{bmatrix} + x_6 \begin{bmatrix}1\\1\\0\\0\\0\\1 \end{bmatrix}$$ Hence the basis of $N(A)$ is $[3, 1, 0, 2, 1, 0]^{t}, [1, 1, 0, 0, 0, 1]^{t}$.
8. Find bases for the row, column and null spaces of the matrix $A$ defined in section 1.6, Problem 17.
Solution:
Recall that in section 1.6 Problem 17 we have the following result:
$$A=\begin{bmatrix}1& a& b& a\\ a& b& b& 1\\ 1& 1& 1& a \end{bmatrix}$$ $$\Longleftrightarrow B=\begin{bmatrix}1& 0& 0& 0\\ 0& 1& 0& b\\0& 0& 1& 1 \end{bmatrix}$$ And the addition and multiplication tables are in the following table.
Thus the bases of $R(A)$ and $C(A)$ are $$\begin{cases}R(A): &[1, 0, 0, 0], [0, 1, 0, b], [0, 0, 1, 1] \\ C(A): & [1, a, 1]^{t}, [a, b, 1]^{t}, [b, b, 1]^{t} \end{cases}$$ For $N(A)$ we have $$\begin{cases}x_1=0\\ x_2 = -bx_4 = bx_4\\ x_3 = -x_4=x_4\\ x_4=x_4 \end{cases} \Rightarrow X=x_4\begin{bmatrix}0\\ b\\1\\1 \end{bmatrix}$$ Hence the basis of $N(A)$ is $[0, b, 1, 1]^{t}$.
9. If $X_1, \cdots, X_m$ form a basis for a subspace $S$, prove that $$X_1, X_1+X_2, \cdots, X_1+\cdots+X_m$$ also form a basis of $S$.
Solution:
There are two phases, firstly prove linearly independent and then prove every vector in $S$ is expressible by the basis.\\
Suppose that there exists $x_1, \cdots, x_m$ such that $$x_1X_1+x_2(X_1+X_2)+\cdots+x_m(X_1+\cdots+X_m)=0$$ $$\Rightarrow (x_1+x_2+\cdots+x_m)X_1+(x_2+\cdots+x_m)X_2+\cdots+x_mX_m = 0$$ Since $X_1, \cdots, X_m$ is linearly independent, so $$\begin{cases}x_1+x_2+\cdots+x_m = 0\\ x_2+\cdots+x_m =0\\ \vdots\\ x_m=0 \end{cases} \Rightarrow \begin{cases}x_1=0\\ x_2=0\\ \vdots\\ x_m=0\end{cases}$$ which indicates that $X_1, X_1+X_2, \cdots, X_1+\cdots+X_m$ are linearly independent. \\
Next, since $X_1, \cdots, X_m$ is a basis of $S$, thus suppose a vector $X\in S$ and we have $$X = a_1X_1+\cdots+a_mX_m = x_1X_1+x_2(X_1+X_2)+\cdots+x_m(X_1+\cdots+X_m)$$ $$= (x_1+x_2+\cdots+x_m)X_1+(x_2+\cdots+x_m)X_2+\cdots+x_mX_m$$ $$\Rightarrow \begin{cases}a_1 = x_1+x_2+\cdots+x_m\\ a_2 = x_2+\cdots+x_m\\ \vdots\\ a_m = x_m \end{cases} \Rightarrow \begin{cases}x_1=a_1-a_2\\ x_2=a_2-a_3\\ \vdots\\ x_m=a_m\end{cases}$$ which means an arbitrary vector $X\in S$ can be expressed by $X_1, X_1+X_2, \cdots, X_1+\cdots+X_m$.
10. Let $A=\begin{bmatrix}a&b&c\\1&1&1\end{bmatrix}$. Classify $a$, $b$, $c$ such that (a) rank $A = 1$; (b) rank $A = 2$.
Solution:
(a) $$\begin{bmatrix}a&b&c\\1&1&1\end{bmatrix}\Rightarrow \begin{bmatrix}0&b-a&c-a\\1&1&1\end{bmatrix}$$ Thus when $b-a=c-a=0\Rightarrow a=b=c$ then rank $A=1$.
(b) Similarly to (a), $b-a \neq 0$ or $c-a\neq 0$, that is, at least two of $a$, $b$, $c$ are distinct then rank $A=2$.
11. Let $S$ be a subspace of $F^n$ with $\dim S=m$. If $X_1, \cdots, X_m$ are vectors in $S$ with the property that $S=\big < X_1, \cdots, X_m\big > $, prove that $X_1, \cdots, X_m$ form a basis for $S$.
Solution:
Consider Problem 9, we only need to prove the dependency of $X_1, \cdots, X_m$ since $S=\big < X_1, \cdots, X_m\big > $ has been given. \\
Without loss of generality, suppose $X_1, \cdots, X_m$ are linear dependent and hence one of the vectors can be expressed by others: $$X_m= x_1X_1+\cdots+x_{m-1}X_{m-1}$$ where $x_i$ are not all zero for $i=1, \cdots, m-1$. This equation indicates that $S$ can be spanned by $m-1$ vectors, say $X_1, \cdots, X_{m-1}$. This is contradict to $S=\big < X_1, \cdots, X_m\big > $.
12. Find a basis for the subspace $S$ of $\mathbb{R}^3$ defined by the equation $$x + 2y + 3z =0$$ Verify that $Y_1=[-1, -1, 1]^{t}\in S$ and find a basis for $S$ which includes $Y_1$.
Solution:
First, $(-1)+2(-1)+3=0\Rightarrow Y_1\in S$. Then, we have
$$x + 2y + 3z =0\Rightarrow x=-2y-3z\Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix} = y\begin{bmatrix} -2\\ 1\\ 0 \end{bmatrix} + z\begin{bmatrix}-3\\ 0\\ 1 \end{bmatrix}$$ Hence $[-2, 1, 0]^{t}$ and $[-3, 0, 1]^{t}$ form a basis for $S$.
13. Let $X_1, \cdots, X_m$ be vectors in $F^n$. If $X_i = X_j$, where $i < j$, prove that $X_1, \cdots, X_m$ are linearly dependent.
Solution:
Without loss of generality, suppose $X_1 = X_2$, then we have $$1\cdot X_1+(-1)\cdot X_2+0\cdot X_3+\cdots + 0\cdot X_m=0$$ which means $X_1, \cdots, X_m$ are linearly dependent.
14. Let $X_1, \cdots, X_{m+1}$ be vectors in $F^n$. Prove that $$\dim \big < X_1, \cdots, X_{m+1}\big > = \dim \big < X_1, \cdots, X_{m} \big > $$ if $X_{m+1}$ is a linear combination of $X_1, \cdots, X_{m}$, but $$\dim \big < X_1, \cdots, X_{m+1}\big > = \dim \big < X_1, \cdots, X_{m} \big > +1$$ if $X_{m+1}$ is not a linear combination of $X_1, \cdots, X_{m}$.
Deduce that the system of linear equations $AX=B$ is consistent, if and only if $$\text{rank}\ [A|B] = \text{rank}\ A.$$
Solution:
(1) Since $X_{m+1}$ is a linear combination of $X_1, \cdots, X_{m}$, so we have $$\big < X_1, \cdots, X_{m+1}\big > = \big < X_1, \cdots, X_{m} \big > $$ $$\Rightarrow \dim \big < X_1, \cdots, X_{m+1}\big > = \dim \big < X_1, \cdots, X_{m} \big > $$
(2) Suppose $X_{c_1}, \cdots, X_{c_r}$ is a basis of $\big < X_1, \cdots, X_{m} \big >$. Since $X_{m+1}$ is not a linear combination of $X_1, \cdots, X_{m}$, so we have $$\big < X_1, \cdots, X_{m+1}\big > = \big < X_{c_1}, \cdots, X_{c_r}, X_{m+1} \big >$$ $$\Rightarrow \dim \big < X_1, \cdots, X_{m+1}\big > = \dim \big < X_{c_1}, \cdots, X_{c_r}, X_{m+1} \big >$$ $$= r+1 = \dim \big < X_1, \cdots, X_{m} \big > +1$$
(3) Since $AX=B$ is consistent, we have $$B = x_1A_{*1}+\cdots+x_nA_{*n}$$ where $X=[x_1, \cdots, x_n]^{t}$. So if $AX=B$ is soluble then $B$ is a linear combination of columns of $A$, that is $B\in C(A)$. From part (1) we know that $B\in C(A)$ if and only if $$\dim C(A|B) = \dim C(A)$$ that is, $\text{rank}\ [A|B] = \text{rank}\ A$.
15. Let $a_1, \cdots, a_n$ be elements of $F$, not all zero. Prove that the set of vectors $[x_1, \cdots, x_n]^{t}$ where $x_1, \cdots, x_n$ satisfy $$a_1x_1+\cdots+a_nx_n=0$$ is a subspace of $F^{n}$ with dimension equal to $n-1$.
Solution:
Denote $S$ be the set of vectors $[x_1, \cdots, x_n]^{t}$. Then $S = N(A)$, where $A$ is the row matrix $[a_1, \cdots, a_n]$. And rank $A=1$ since $A\neq0$. Thus we have $$\dim S=\dim N(A)=n-\text{rank}\ A=n-1$$
16. Prove the following theorems:
(1) Suppose each of $X_1, \cdots, X_r$ is a linear combination of $Y_1, \cdots, Y_s$. Then any linear combination of $X_1, \cdots, X_r$ is a linear combination of $Y_1, \cdots, Y_s$. That is, $\big < X_1, \cdots, X_r\big > \subseteq\big < Y_1, \cdots, Y_s\big > $.
(2) Subspaces $\big < X_1, \cdots, X_r\big > $ and $\big < Y_1, \cdots, Y_s\big > $ are equal if each of $X_1, \cdots, X_r$ is a linear combination of $Y_1, \cdots, Y_s$ and each of $Y_1, \cdots, Y_s$ is a linear combination of $X_1, \cdots, X_r$.
(3) $\big < X_1,\cdots, X_r, Z_1, \cdots, Z_t\big > $ and $\big < X_1,\cdots, X_r\big > $ are equal if each of $Z_1, \cdots, Z_t$ is a linear combination of $X_1,\cdots, X_r$.
(4) A family of $s$ vectors in $\big < X_1, \cdots, X_r\big > $ will be linearly dependent if $s > r$. Equivalently, a linearly independent family of $s$ vectors in $\big < X_1, \cdots, X_r\big > $ must have $s \leq r$.
Solution:
(1) Since each of $X_1, \cdots, X_r$ is a linear combination of $Y_1, \cdots, Y_s$, so we have $$\begin{cases}X_1=a_{11}Y_1+\cdots+a_{1s}Y_s\\ \vdots\\ X_r=a_{r1}Y_1+\cdots+a_{rs}Y_s\end{cases}$$ Now let $X$ is a linear combination of $X_1, \cdots, X_r$, that is $$X=x_1X_1+\cdots+x_rX_r$$ $$=x_1(a_{11}Y_1 + \cdots + a_{1s}Y_s) + \cdots +x_r(a_{r1}Y_1+\cdots+a_{rs}Y_s)$$ $$=y_1Y_1 + \cdots + y_sY_s$$ where $y_i = \displaystyle\sum_{j=1}^{r}x_{j}a_{ji}$ for $i = 1, \cdots, s$. This shows that $X$ is also a linear combination of $Y_1, \cdots, Y_s$.
(2) According to part (1), we know that $$\big < X_1, \cdots, X_r\big > \subseteq\big < Y_1, \cdots, Y_s\big > $$ and $$\big < Y_1, \cdots, Y_s\big > \subseteq \big < X_1, \cdots, X_r\big > $$ Thus $\big < X_1, \cdots, X_r\big > = \big < Y_1, \cdots, Y_s\big > $.
(3) Since each of $Z_1, \cdots, Z_t$ is a linear combination of $X_1,\cdots, X_r$, so each of $X_1,\cdots, X_r, Z_1, \cdots, Z_t$ is a linear combination of $X_1,\cdots, X_r$.
On the other hand, each of $X_1,\cdots, X_r$ is a linear combination of $X_1,\cdots, X_r, Z_1, \cdots, Z_t$. According to part (2), we know that $$\big < X_1,\cdots, X_r, Z_1, \cdots, Z_t\big > = \big < X_1,\cdots, X_r\big > $$
(4) Suppose $Y_1, \cdots, Y_s$ are vector in $\big < X_1, \cdots, X_r\big > $. We have $$\begin{cases}Y_1=a_{11}X_1+\cdots+a_{1r}X_r\\ \vdots\\ Y_s=a_{s1}X_1+\cdots+a_{sr}X_r\end{cases}$$ Then assume we have coefficients $y_1, \cdots, y_s$, and $$y_1Y_1+\cdots+y_sY_s = y_1(a_{11}X_1+\cdots+a_{1r}X_r) + \cdots + y_s(a_{s1}X_1 + \cdots + a_{sr}X_r)$$ $$=x_1X_1 + \cdots + x_rX_r$$ where $x_i = \displaystyle\sum_{j=1}^{s}y_{j}a_{ji}$ for $i = 1, \cdots, r$. That is, $$\begin{cases}a_{11}y_1+\cdots+a_{s1}y_s = x_1\\ \vdots \\ a_{1r}y_1+\cdots+a_{sr}y_s = x_r\end{cases}$$ The homogeneous system $x_i=0$ for $i = 1, \cdots, r$, has non-trivial solutions ($y_j$ for $j =1, \cdots, s$) if $s > r$, otherwise $s \leq r$.
That is, $Y_1, \cdots, Y_s$ will be linearly dependent if $s > r$, otherwise it will be linearly independent.
17. Let $R$ and $S$ be subspaces of $F^n$, with $R \subseteq S$. Prove that $$\dim R\leq \dim S$$ and that equality implies $R=S$.
Solution:
Suppose $X_1, \cdots, X_r$ form a basis of $R$, and $Y_1, \cdots, Y_s$ form a basis of $S$. Since $R\subseteq S$, so we have $$\big < Y_1, \cdots, Y_s\big > = \big < X_1, \cdots, X_r, Y_1, \cdots, Y_s\big > $$ Applying left-to-right algorithm, we can form a basis on the right hand of the above equation, which is an extension of $X_1, \cdots, X_r$, say $X_1, \cdots, X_r, \cdots, X_s$. That is, $$\dim R = r \leq s = \dim S$$ Obviously, from the above extension of bases, when $r=s$ it will be $R=S$.
18. Let $R$ and $S$ be subspaces of $F^{n}$. If $R\cup S$ is a subspace of $F^{n}$, prove that $R\subseteq S$ or $S\subseteq R$.
Solution:
Suppose $R\nsubseteq S$ and $S\nsubseteq R$, that is, there exists $u$ and $v$ such that $u\in R$ but $u\notin S$, and $v\in S$ but $v\notin R$. We have $$u, v \in R\cup S \Rightarrow u+v\in R\cup S$$ $$\Rightarrow u+v\in R\ \text{or}\ u+v\in S$$ Assume that $u+v\in R$, since $u\in R$, $-u\in R$ (since $R$ is a subspace and also a field which satisfying 10 properties in Chapter 1). So we have $$v=(u+v)+(-u)\in R\Rightarrow v\in R$$ which is contradiction. Thus, we can conclude that $R\subseteq S$ or $S\subseteq R$.
19. Let $X_1, \cdots, X_r$ be a basis for a subspace $S$. Prove that all bases for $S$ are given by the family $Y_1, \cdots, Y_r$, where $$Y_i=\sum_{j=1}^{r}a_{ij}X_j,$$ and where $A=[a_{ij}]\in M_{r\times r}(F)$ is a non-singular matrix.
Solution:
First, we will prove $Y_i$ is a basis for $S$ if $A$ is non-singular matrix. Then we will prove $A$ is non-singular matrix if $Y_i$ is a basis for $S$.
(1) From the given condition, we have $$\begin{cases}Y_1 = a_{11}X_1 + \cdots + a_{1r}X_r\\ \vdots\\ Y_r=a_{r1}X_1+\cdots + a_{rr}X_r\end{cases}$$ where $A=[a_{ij}]$ is a non-singular matrix, $B=[Y_1, \cdots, Y_r]^{t}$, $X=[X_1, \cdots, X_r]^{t}$. Thus $X=A^{-1}B$, which means $X$ is soluble in terms of $Y$. So by Problem 16 part (1) we have $$S =\big < X_1, \cdots, X_r \big > = \big < Y_1, \cdots, Y_r \big > $$ And (by problem 11) hence $Y_1, \cdots, Y_r$ is a basis for $S$.
(2) On the other hand, if $A$ is non-singular, then the rows of $A$ are linearly independent. That is, assuming $$x_1[a_{11}, \cdots, a_{1r}]+ \cdots+ x_r[a_{r1}, \cdots, a_{rr}] = [0, \cdots, 0]$$ We need to prove that all $x_i$ are zeros. So we have $$\begin{cases}a_{11}x_1+\cdots + a_{r1}x_r = 0 \\ \vdots \\ a_{1r}x_1+\cdots + a_{rr}x_r =0\end{cases}$$ Hence $$x_1Y_1+\cdots+x_rY_r$$ $$ = x_1(a_{11}X_1 + \cdots + a_{1r}X_r) + \cdots + x_r(a_{r1}X_1+\cdots + a_{rr}X_r)$$ $$= (a_{11}x_1+\cdots + a_{r1}x_r)X_1 + \cdots + ( a_{1r}x_1+\cdots + a_{rr}x_r)X_r$$ $$=0\cdot X_1+\cdots+0\cdot X_r = 0$$ Since $Y_i$ are linearly independent (i.e. basis), thus $x_1 = \cdots = x_r =0$. That is, the rows of $A$ are linearly independent which implies $A$ is non-singular matrix.
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