Suppose that $V$ is an $m$-dimensional affine subspace of $\mathbb R^n$, with $m < n$. show that there exist linearly independent vectors $a_1, \dots, a_{n-m}$, and scalars $b_1, \dots , b_{n-m}$ such that $$ V = \{y : a_i'y = b_i, \quad i = 1,...,n-m\} $$

Where I am: So I know that we are working with a subspace that if represented in matrix form has less rows than columns, i.e. one or more dimension of the subspace could be unbounded. I also know that an affine subspace would be like a subspace that doesn't include the origin (it has been shifted away from another parallel subspace). So using all of this, to my best knowledge: I need to show that there is an Ax = b equation where all the vectors a$_i$ that make up A are linearly independent but not zero!! (Affine subspace)

Where to go with this though?? I was thinking about playing around with the vectors (2,1,1) and (1,2,1) which (I think) satisfy the the linear independent and affine requirements, but I am not sure.

Any help here would be much appreciated. I am not a wise man, and sorry for the horibly phrased question! :C


Your intuition is good: if $V \subset \mathbb R^n$ is an $m$-dimensional affine space, then there exists some vector $t \in \mathbb R^n$ such that the translation $V - t =: U$ is an $m$-dimensional linear subspace of $\mathbb R^n$. We may pick an orthonormal basis $u_1, \dots, u_m$, of $U$, and this extends to an orthonormal basis $u_1, \dots, u_m, a_1, \dots, a_{n-m}$ of $V$. As described, we have $u \perp a_i$ for every $u \in U$, i.e., $$ U = \{ u \in \mathbb R^n : a_i' u = 0, \quad i = 1, \dots, n-m\}, $$ where $a_i'$ denotes the transpose of the column vector $a_i$. Then \begin{align*} V &= U + t \\ &= \{u + t \in \mathbb R^n : a_i' u = 0 \quad\forall i \in \{1, \dots, n-m\}\} \\ &= \{v \in \mathbb R^n : a_i' (v - t) = 0 \quad\forall i \in \{1, \dots, n-m\}\} \\ &= \{v \in \mathbb R^n : a_i' v = a_i' t \quad\forall i \in \{1, \dots, n-m\} \} && \text{substitute } v = u+t \end{align*} and the desired scalars are $b_i = a_i ' t$.

  • $\begingroup$ awesome! Thank you so much, this is very clear to me, way more so than my textbook. $\endgroup$ – wbrugato Apr 14 '16 at 14:56

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