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\ovingnr{06}
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\fag{MA1201 Linear Algebra and Geometry }
\institutt{Department of Mathematical Sciences }
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\begin{document}

\section*{Compulsory exercises}
Hand in your solutions to these exercises. All answers must be justified.

\subsection*{Chapter 3.5 - Cross product}
{\bf Exercise \exnum} Do exercise 13 in chapter 3.5 of Elementary Linear Algebra.

\ifLF{
  The triangle is spanned by $A-C = (3,-2)$ and $B-C = (4,2)$. The area of a triangle is half that of the associated parallellogram thus the area is 
  \begin{align*}
    \left|\det\left(\begin{bmatrix}
      3 & -2\\
      4 & 2
    \end{bmatrix}\right)\right| = 14
  \end{align*}
}\fi

{\bf Exercise \exnum} Do exercise 21 in chapter 3.5 of Elementary Linear Algebra.
\ifLF{
  \begin{align*}
    \mathbf u \cdot (\mathbf v \times \mathbf w) =
    \det\left(\begin{bmatrix}
      -2&0&6\\
      1& -3& 1\\
      -5& -1& 1
    \end{bmatrix}\right) = -92
  \end{align*}
}\fi

\subsection*{Chapter 4.1 - Real vector spaces}

Here are the vector space axioms as listed on page 203.
\begin{enumerate}
  \item If $\mathbf u$ and $\mathbf v$ are in $V$, then $\mathbf u + \mathbf v \in V$.
  \item $\mathbf u + \mathbf v = \mathbf v + \mathbf u$
  \item $(\mathbf u + \mathbf v) + \mathbf w = \mathbf u + (\mathbf v + \mathbf w)$
  \item There exists an object in $V$, called the zero vector, that is denoted by $\mathbf 0$ and has the property that $\mathbf 0 + \mathbf u = \mathbf u + \mathbf 0 = \mathbf u$ for all $\mathbf u$ in $V$.
  \item For each $\mathbf u$ in $V$, there is an object $-\mathbf u$ in $V$, called a negative of $\mathbf u$, such that $\mathbf u+(-\mathbf u) = (-\mathbf u)+\mathbf u = \mathbf 0$.
  \item  If $k$ is any scalar and $\mathbf u \in V$, then $k\mathbf u \in V$.
  \item  $k(\mathbf u+\mathbf v)=k\mathbf u+k\mathbf v$
  \item $(k+m)\mathbf u=k\mathbf u+m\mathbf u$
  \item $k(m\mathbf u) = (km)(\mathbf u)$
  \item $1\mathbf u = \mathbf u$ 
\end{enumerate}

{\bf Exercise \exnum} Do exercise 3-9 in chapter 4.1 of Elementary Linear Algebra.

\ifLF{
  3, 4, 6, and 9 are vector spaces. In exercise 5 axiom 5 and 6 fails. In exercise 7 axiom 8 fails. In exercise 8 axiom 1, 4, and 6 fails, and axiom 5 is illdefined without axiom 4.
}\fi

\subsection*{Chapter 4.2 - Subspaces}

{\bf Exercise \exnum} Do exercise 5 in chapter 4.2 of Elementary Linear Algebra.
\ifLF{
  \begin{enumerate}[label=(\alph*)]
    \item Let $p(x) = a_1x + a_2x^2 + a_3x^3$ and $q(x) = b_1x+b_2x^2+b_3x^3$ be polynomials in $P_3$ with constant term $0$, and let $k\in\mathbb R$ be a scalar. Then $p(x)+q(x) = (a_1+b_1)x+(a_2+b_2)x^2 + (a_3+b_3)x^3$ and $kp(x)=ka_1x + ka_2x^2 + ka_3x^3$ also have constant term 0. Thus the set of such polynomials is a subspace.
    \item Let $p(x) = a_0 + a_1x + a_2x^2 + a_3x^3$ and $q(x) = b_0 + b_1x+b_2x^2+b_3x^3$ be such that $$\sum_{i=0}^3 a_i = 0 = \sum_{i=0}^3 b_i,$$ and let $k\in\mathbb R$ be a scalar. Then $p(x)+q(x) = (a_0 + b_0) + (a_1+b_1)x + (a_2+b_2)x^2 + (a_3+b_3)x^3$ satisfies $$\sum_{i=0}^3 (a_i+b_i) = \sum_{i=0}^3 a_i+\sum_{i=0}^3 b_i = 0,$$ and $kp(x)$ satisfies $$\sum_{i=0}^3 ka_i = k\sum_{i=0}^3 a_i = 0.$$ So the set of such polynomials is a subspace.
  \end{enumerate}
}\fi

\end{document}