Exercise 1.20

Question

\begin{enumerate}[label=(\alph*)]
\item Let $S=\left\{\mathbf{A x} \mid \mathbf{x} \in \Re^{n}\right\}$, where $\mathbf{A}$ is a given matrix. Show that $S$ is a subspace of $\Re^{n}$.
\item Assume that $S$ is a proper subspace of $\Re^{n}$. Show that there exists a matrix $\mathbf{B}$ such that $S=\left\{\mathbf{y} \in \Re^{n} \mid \mathbf{B y}=\mathbf{0}\right\} .$ Hint: Use vectors that are orthogonal to $S$ to form the matrix $\mathbf{B}$.
\item Suppose that $V$ is an $m$-dimensional affine subspace of $\Re^{n}$, with $m<$ $n$. Show that there exist linearly independent vectors $\mathbf{a}_{1}, \ldots, \mathbf{a}_{n-m}$, and scalars $b_{1}, \ldots, b_{n-m}$, such that
$$
V=\left\{\mathbf{y} \mid \mathbf{a}_{i}^{\prime} \mathbf{y}=b_{i}, i=1, \ldots, n-m\right\}
$$
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(\alph*)]
\item \begin{proof} Following the definition on p.29, fix arbitrary $a,b\in\mathfrak{R}$ and $\mathbf{y}_1,\mathbf{y}_2 \in S$, or equivalently, $\mathbf{x}_1,\mathbf{x}_2\in\mathfrak{R}^n$ such that $\mathbf{A}\mathbf{x}_1
eq\mathbf{A}\mathbf{x}_2$. We then have
$$a\mathbf{y}_1 + b\mathbf{y}_2
= a\mathbf{A}\mathbf{x}_1 + b\mathbf{A}\mathbf{x}_2
= \mathbf{A}\left(a\mathbf{x}_1 + b\mathbf{x}_2ight).
$$
Since $\mathbf{x}_1,\mathbf{x}_2\in\mathfrak{R}^n$, $a\mathbf{x}_1 + b\mathbf{x}_2 \in \mathfrak{R}^n$ as well. Thus, the above vector is an element of $S$ by construction. \end{proof}

\item \begin{proof} Let $m := \dim(S) < n$. Recall that we can find $n-m$ linearly independent vectors $\mathbf{b}_1,\ldots,\mathbf{b}_{n-m}$ in the orthogonal complement $S^{\perp}$ of $S$. Define
$$B:= \begin{bmatrix} - & \mathbf{b}_{1}' & - \ & \vdots & \ - & \mathbf{b}_{n-m}' & - \end{bmatrix}.$$
We argue that $S=\left\{\mathbf{y} \in \Re^{n} \mid \mathbf{B y}=\mathbf{0}ight\}$.
\begin{itemize}
\item[$\implies$] Suppose that $\mathbf{x} \in S$. Then $\mathbf{Bx} = \sum_{i=1}^{n-m}\mathbf{b}_i'\mathbf{x} = \sum_{i=1}^{n-m} \mathbf{0} = \mathbf{0}$, as desired.

\item[$\impliedby$] Let $\mathbf{x} \in \mathfrak{R}^n$ such that $\mathbf{Bx} = \mathbf{0}$. By orthogonal projection theorem, we write $\mathbf{x} = \mathbf{v} + \mathbf{w}$ for some $\mathbf{v} \in S$ and $\mathbf{w} \in S^{\perp}$. Since $\mathbf{x}$ is perpendicular to every vector in $\mathbf{S}^\perp$, we conclude that $\mathbf{w} = \mathbf{0}$ and thus $\mathbf{x} \in S$.
\end{itemize}
\end{proof}

\item \begin{proof}
We can, 	extit{per definitionem}, find a proper linear subspace $V \subset \mathfrak{R}^n$ and the translation vector $\mathbf{x}_0 \in \mathfrak{R}^n$ such that $V = V_0 + \mathbf{x}_0$. By the previous part of this exercise, we can find a $n-m$ full row rank matrix $\mathbf{B}$ such that this vector subspace can be represented as the null space $V_0 = \{\mathbf{y} \mid \mathbf{By} = \mathbf{0}\}$. Change of variables and some trivial matrix arithmetic then gives
$$
V = \mathbf{x}_0 + \left\{ \mathbf{y} \mid \mathbf{B}\mathbf{y} = \mathbf{0}ight\}
= \left\{ \mathbf{x}_0 + \mathbf{y} \mid \mathbf{B}\mathbf{y} = \mathbf{0}ight\}
= \left\{ \mathbf{x} \mid \mathbf{B}\mathbf{x} = \mathbf{B}\mathbf{x}_0ight\}.
$$
Setting $\mathbf{b} := \mathbf{B}\mathbf{x}_0$ yields the desired result.
\end{proof}
\end{enumerate}

Exercise 1.20

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Exercise 1.20

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