Homework 5

Homework 5.pdf

(25 pts) Prove the following statement only by using dimensions: the only subspaces of $\R^{2}$ are $\R^{2}$, ${\{(0,0)\}}$ and $L={\{\lambda(a,b):\lambda\in \R\}}$ for $(a, b) \in \R^{2}$.

Let $V$ be a subspace of $\mathbb{R}^2$ $\Rightarrow(0,0) \in V$ by definition. We have 2 cases.
1. $V$ has only one element $\Rightarrow V$ must be $V = \{(0,0)\}$
2. $V$ has more than one element $\Rightarrow\exists v=\left(a,b\right)\in V$ with $v\neq(0,0)$
  
  Since $V$ is a subspace, $\lambda v=\lambda\left(a,b\right)\in V$ for all $\lambda \in \mathbb{R}$, then $M\subseteq V$
  
  If $M=V$, we are done
  
  If $M\subset V$, then $\exists w\in V,w\neq \lambda v$
  
  Since we have $v,w\in V$ such that $w\neq\lambda v$
  
  Thus $\forall x,y\in\mathbb{R},(x,y)\in\mathbb{R}^2$, we have $(x,y)=\lambda v+\mu w$ for $\lambda,\mu\in\mathbb{R}$
  
  This is possible since $w\neq\lambda v$ for $\lambda \in \mathbb{R}$
  
  Therefore $V=\mathbb{R}^2$.
Thus the only subspaces of $\mathbb{R}^2$ are $\mathbb{R}^2$, $\{(0,0)\}$ and $M=\{\lambda(a,b):a,b\in\mathbb{R},\lambda\in\mathbb{R}\}$
(20 pts) Decide whether the following statements are true or false. Prove or give a counterexample.
1. Let $V$ be a vector space with $\dim(V) = 42$. If $V_1$ and $V_2$ are subspaces of $V$ with $\dim(V_1) = 33$ and $\dim(V_2) = 9$ such that $V = V_1 + V_2$, then $V = V_1 \oplus V_2$.
  
  True, proof
  
  Since we know $\dim(V_{1}+V_{2})=\dim(V_{1})+\dim(V_{2})-\dim(V_{1}\cap V_{2})$
  
  Then $42=33+9-\dim(V_1\cap V_2)$, then $\dim(V_1\cap V_2)=0$, thus $V_{1}\cap V_{2}=\{0\}$
  
  Thus $V = V_1 \oplus V_2$. 2. Let $V$ be a vector space with $\dim(V) = 42$. If $V_1$ and $V_2$ are subspaces of $V$ with $\dim(V_1) = 33$ and $\dim(V_2) = 9$ such that $V_1 \cap V_2 \neq {\{0\}}$, then $V = V_1 \oplus V_2$.
  
  False, counterexample
  
  $V=\left\lbrace(a_{1},...,a_{42}):a_{1},\ldots,a_{42}\in\mathbb{R}\right\rbrace$ and $V_{1}=\left\lbrace(a_{1},...,a_{33},\underbrace{0,\ldots,0}_{9\text{ zeros}}):a_{1} ,\ldots,a_{33}\in\mathbb{R}\right\rbrace$ and $V_{2}=\left\lbrace\left(\underbrace{0,\ldots,0}_{32\text{ zeros}},a_{33},\ldots, a_{41},0\right):a_{33},\ldots,a_{41}\in\mathbb{R}\right\rbrace$
  
  Then clearly $V_1$ and $V_2$ are subspaces of $V$ with $\dim(V_1) = 33$ and $\dim(V_2) = 9$ such that $V_1 \cap V_2 \neq {\{0\}}$
  
  But $V\neq V_{1}\oplus V_{2}$ since $a_{42}=0$ in $V_1+V_2$, then $V_1\oplus V_2\subsetneq V$ 3. Let $V_1$ and $V_2$ be subspaces of $F^8$ such that $\dim(V_1) = \dim(V_2) = 4$, then $F^8 = V_1 \oplus V_2$.
  
  False, counterexample
  
  Let $V=\left\lbrace(a_{1},...,a_{8}):a_{1},\ldots,a_{8}\in F\right\rbrace$ and $V_{1}=\left\lbrace(a_{1},...,a_{4},\underbrace{0,\ldots,0}_{4\text{ zeros}}):a_{1} ,\ldots,a_{4}\in F\right\rbrace$ and $V_{2}=\left\lbrace\left(\underbrace{0,\ldots,0}_{3\text{ zeros}},a_{3},\ldots,a_{7} ,0\right):a_{3},\ldots,a_{7}\in F\right\rbrace$
  
  Here $\dim(V_1) = \dim(V_2) = 4$, but $V\neq V_{1}\oplus V_{2}$ since $a_{8}=0$ in $V_1+V_2$, then $V_1\oplus V_2\subsetneq V$ 4. Let $V_1$ and $V_2$ be subspaces of $F^8$ such that $\dim(V_1) = \dim(V_2) = 5$, then $V_1 \cap V_2 = {\{0\}}$.
  
  False, counterexample
  
  Let $V=\left\lbrace(a_{1},...,a_{8}):a_{1},\ldots,a_{8}\in F\right\rbrace$ and $V_{1}=\left\lbrace(a_{1},...,a_{5},\underbrace{0,\ldots,0}_{3\text{ zeros}}):a_{1} ,\ldots,a_{5}\in F\right\rbrace$ and $V_{2}=\left\lbrace\left(\underbrace{0,\ldots,0}_{3\text{ zeros}},a_{3},\ldots,a_{7} ,a_{8}\right):a_{3},\ldots,a_{8}\in F\right\rbrace$
  
  Here $\dim(V_1) = \dim(V_2) = 5$, but $V_{1}\cap V_{2}=\left\lbrace\left(0,0,0,a_{3},a_{4},a_{5},0,0,0\right):a_{3},a_{4} ,a_{5}\in F\right\rbrace\neq{\{0\}}$.
(25 pts) Prove that the elementary row operation of interchanging two rows of a matrix can be accomplished by a finite sequence of elementary row operations of the other two types (that is, multiplying a row by a constant and/or adding a multiple of a row to another row).

Assume we have $\begin{cases} a_{11}x_{1}+\dots+a_{1n}x_{n}=b_{1} \\ \vdots \\ a_{i1}x_{1}+\dots+a_{in}x_{n}=b_{i},\quad\text{equation i} \\ a_{j1}x_{1}+\dots+a_{jn}x_{n}=b_{j},\quad\text{equation j} \\ \vdots \\ a_{m1}x_{1}+\dots+a_{mn}x_{n}=b_{m} \end{cases}$

Then $R_{i}+R_{j}\to R_{j}$

Then we get $\begin{cases} a_{11}x_{1}+\dots+a_{1n}x_{n}=b_{1} \\ \vdots \\ a_{i1}x_{1}+\dots+a_{in}x_{n}=b_{i},\quad\text{equation i} \\ a_{i1}x_{1}+\dots+a_{in}x_{n}+\left(a_{j1}x_{1}+\dots+a_{jn}x_{n}\right)=b_{i}+b_{j},\quad\text{equation j} \\ \vdots \\ a_{m1}x_{1}+\dots+a_{mn}x_{n}=b_{m} \end{cases}$

Then $R_{i}-R_{j}\to R_{i}$

Then we get $\begin{cases} a_{11}x_{1}+\dots+a_{1n}x_{n}=b_{1} \\ \vdots \\ -\left(a_{j1}x_{1}+\dots+a_{jn}x_{n}\right)=-b_{j},\quad\text{equation i} \\ a_{i1}x_{1}+\dots+a_{in}x_{n}+\left(a_{j1}x_{1}+\dots+a_{jn}x_{n}\right)=b_{i}+b_{j},\quad\text{equation j} \\ \vdots \\ a_{m1}x_{1}+\dots+a_{mn}x_{n}=b_{m} \end{cases}$

Then $R_{i}+R_{j}\to R_{j}$

Then we get $\begin{cases} a_{11}x_{1}+\dots+a_{1n}x_{n}=b_{1} \\ \vdots \\ -\left(a_{j1}x_{1}+\dots+a_{jn}x_{n}\right)=-b_{j},\quad\text{equation i} \\ a_{i1}x_{1}+\dots+a_{in}x_{n}=b_{i},\quad\text{equation j} \\ \vdots \\ a_{m1}x_{1}+\dots+a_{mn}x_{n}=b_{m} \end{cases}$

Then $-R_{i}\to R_{i}$

Then we get $\begin{cases} a_{11}x_{1}+\dots+a_{1n}x_{n}=b_{1} \\ \vdots \\ a_{j1}x_{1}+\dots+a_{jn}x_{n}=b_{j},\quad\text{equation i} \\ a_{i1}x_{1}+\dots+a_{in}x_{n}=b_{i},\quad\text{equation j} \\ \vdots \\ a_{m1}x_{1}+\dots+a_{mn}x_{n}=b_{m} \end{cases}$
(30 pts) Let $A = \begin{pmatrix} a & 1 & a^2 \\ 1 & a & 1 \\ 1 & a^2 & 2a \end{pmatrix}$
1. By applying elementary row operations, determine for which values of $a \in R$ the matrix $A$ is invertible.
  
  Consider augmented matrix: $A=\left( \begin{array}{ccc|ccc} a & 1 & a^2 &1 &0 & 0\\ 1 & a & 1 & 0&1&0\\ 1 & a^2 & 2a &0 &0&1 \end{array} \right)$
  
  Let $R_2-R_3\to R_2,R_1-aR_3\to R_1$, we get $\left( \begin{array}{ccc|ccc} 0 & 1-a^3 & -a^2 & 1 & 0 & -a \\ 0 & a-a^2 & 1-2a & 0 & 1 & -1 \\ 1 & a^2 & 2a & 0 & 0 & 1 \end{array}\right)$
  
  Let $R_1 \leftrightarrow R_3$, we get $\left( \begin{array}{ccc|ccc} 1 & a^2 & 2a & 0 & 0 & 1 \\ 0 & a\left(1-a\right) & 1-2a & 0 & 1 & -1 \\ 0 & 1-a^3 & -a^2 & 1 & 0 & -a \end{array}\right)$
  
  Case 1: $a(1-a) = 0$ iff $a=0$ or $a=1$.
  - If $a=0$, $\left( \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 & -1 \\ 0 & 1 & 0 & 1 & 0 & 0 \end{array}\right)$
    $R_2 \leftrightarrow R_3$: $\left( \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & -1 \end{array}\right)$
    Thus, A is invertible when $a=0$.
  - If $a=1$, $\left( \begin{array}{ccc|ccc} 1 & 1 & 2 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 & 1 & -1 \\ 0 & 0 & -1 & 1 & 0 & -1 \end{array}\right)$
    $R_{3}-R_{2}\to R_{3}$, $\left( \begin{array}{ccc|ccc} 1 & 1 & 2 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 & -1 & 0 \end{array}\right)$
    Thus, A is not invertible when $a=1$.
  Case 2: $a(1-a) \neq 0$ that is $a \neq 0$ and $a \neq 1$
  $R_{3}-\frac{1-a^{3}}{a(1-a)}R_{2}\to R_{3}$. Since $a \neq 1$, $1-a \neq 0$, then $R_{3}-\frac{1+a+a^{2}}{a}R_{2}\to R_{3}$.
  $\left( \begin{array}{ccc|ccc} 1 & a^2 & 2a & 0 & 0 & 1 \\ 0 & a(1-a) & 1-2a & 0 & 1 & -1 \\ 0 & 0 & \frac{a^{3} + a^{2} + a - 1}{a} & 1 & -\frac{1+a+a^{2}}{a} & \frac{a+1}{a} \end{array}\right)$
  
  Since we have know this matrix has two pivots at row 1 column 1 and row 2 column 2 since $a(1-a)\neq 0$
  
  Then the matrix is invertible iff $a^{3} + a^{2} + a - 1 \neq 0$.
  
  And we know A is invertible for $a=0$ and $a\neq1$. If $a=0$, then $a^3+a^2+a-1\neq 0$
  
  Thus A is invertible if and only if $a^3 + a^2 + a - 1 \neq 0$ and $a\neq 1$. 2. Give the inverse of $A$ for the values of $a$ found in 1.
  
  If $a=0$, then we get $\left( \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & -1 \end{array}\right)$, thus the inverse of $A$ is $ \begin{pmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & -1 \end{pmatrix}$
  
  If $a \neq 0$, $a \neq 1$, and $p(a)=a^{3}+a^{2}+a-1\neq0$
  
  We have $\left( \begin{array}{ccc|ccc} 1 & a^2 & 2a & 0 & 0 & 1 \\ 0 & a(1-a) & 1-2a & 0 & 1 & -1 \\ 0 & 0 & \frac{p(a)}{a} & 1 & -\frac{1+a+a^2}{a} & \frac{a+1}{a} \end{array}\right)$
  
  Let $\frac{a}{p(a)}R_{3}\to R_{3}$, then $\left( \begin{array}{ccc|ccc} 1 & a^2 & 2a & 0 & 0 & 1 \\ 0 & a(1-a) & 1-2a & 0 & 1 & -1 \\ 0 & 0 & 1 & \frac{a}{p(a)} & -\frac{1+a+a^{2}}{p(a)} & \frac{a+1}{p(a)} \end{array}\right)$
  
  Let $R_{1}-2aR_{3}\to R_{1}$ and $R_{2}-(1-2a)R_{3}\to R_{2}$
  $\left( \begin{array}{ccc|ccc} 1 & a^2 & 0 & -\frac{2a^2}{p(a)} & \frac{2a(1+a+a^2)}{p(a)} & 1-\frac{2a(a+1)}{p(a)} \\ 0 & a(1-a) & 0 & -\frac{a(1-2a)}{p(a)} & 1+\frac{(1-2a)(1+a+a^2)}{p(a)} & -1-\frac{(1-2a)(a+1)}{p(a)} \\ 0 & 0 & 1 & \frac{a}{p(a)} & -\frac{1+a+a^2}{p(a)} & \frac{a+1}{p(a)} \end{array}\right)$
  
  Then we have $\left( \begin{array}{ccc|ccc} 1 & a^2 & 0 & -\frac{2a^2}{p(a)} & \frac{2a(1+a+a^2)}{p(a)} & \frac{a^3-a^2-a-1}{p(a)} \\ 0 & a(1-a) & 0 & -\frac{a(1-2a)}{p(a)} & \frac{-a^3}{p(a)} & \frac{-a^2(a-1)}{p(a)} \\ 0 & 0 & 1 & \frac{a}{p(a)} & -\frac{1+a+a^2}{p(a)} & \frac{a+1}{p(a)} \end{array}\right)$
  
  Let $\frac{1}{a(1-a)}R_{2}\to R_{2}$, then$\left( \begin{array}{ccc|ccc} 1 & a^2 & 0 & -\frac{2a^2}{p(a)} & \frac{2a(1+a+a^2)}{p(a)} & \frac{a^3-a^2-a-1}{p(a)} \\ 0 & 1 & 0 & -\frac{1-2a}{(1-a)p(a)} & \frac{-a^2}{(1-a)p(a)} & \frac{-a(a-1)}{(1-a)p(a)} \\ 0 & 0 & 1 & \frac{a}{p(a)} & -\frac{1+a+a^2}{p(a)} & \frac{a+1}{p(a)} \end{array}\right)$
  
  Then $\left( \begin{array}{ccc|ccc} 1 & a^2 & 0 & -\frac{2a^2}{p(a)} & \frac{2a(1+a+a^2)}{p(a)} & \frac{a^3-a^2-a-1}{p(a)} \\ 0 & 1 & 0 & \frac{2a-1}{(1-a)p(a)} & \frac{-a^2}{(1-a)p(a)} & \frac{a}{p(a)} \\ 0 & 0 & 1 & \frac{a}{p(a)} & -\frac{1+a+a^2}{p(a)} & \frac{a+1}{p(a)} \end{array}\right)$
  
  Let $R_{1}-a^{2}R_{2}\to R_{1}$, then $\left( \begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{-a^{2}}{(1-a)p(a)} & \frac{a(2-a^{3})}{(1-a)p(a)} & -\frac{a^{2}+a+1}{p(a)} \\ 0 & 1 & 0 & \frac{2a-1}{(1-a)p(a)} & \frac{-a^{2}}{(1-a)p(a)} & \frac{a}{p(a)} \\ 0 & 0 & 1 & \frac{a}{p(a)} & -\frac{1+a+a^{2}}{p(a)} & \frac{a+1}{p(a)} \end{array}\right)$
  
  Thus $A^{-1} = \frac{1}{p(a)} \begin{pmatrix} \frac{-a^2}{1-a} & \frac{a(2-a^3)}{1-a} & -(a^2+a+1) \\ \frac{2a-1}{1-a} & \frac{-a^2}{1-a} & a \\ a & -(1+a+a^2) & a+1 \end{pmatrix}$ where $p(a) = a^3 + a^2 + a - 1$.

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