Reduction of an affine hyperplane

Question

For a $0 \neq v \in \mathbb{R}^{n}$ and a number $b \in \mathbb{R}$, the set $W^{n} \left( v,b \right) = \left\{ x \in \mathbb{R}^{n} | \langle x| v\rangle =b\right\}$ is called an affine hyperplane in $\mathbb{R}^{n}$.
For any affine hyperplane $W \subseteq \mathbb{R}^{n}$ show that there exists $i \in [n]$ such that for any number $t \in \mathbb{R}$ the set $A = \left\{ y \in \mathbb{R}^{n-1} | \left( y_{1}, \ldots, y_{i-1},t,y_{i+1},\ldots,y_{n} \right) \in W\right\}$ is an affine hyperplane in $\mathbb{R}^{n-1}$.

My try was:
Let $W = W^{n} \left( v,b \right)$ be an affine hyperplane for some $0 \neq v \in \mathbb{R}^{n}$ and a number $b \in \mathbb{R}$. Denote $v = \left( v_{1}, \ldots ,v_{n}\right)$. Because of the fact that $v \neq 0$, there exists some $i \in [n]$ such that $v_{i} \neq 0$, so let's choose this $i$. Now let $t \in \mathbb{R}$ be a number and $y \in A$, such that $y \in \mathbb{R}^{n-1}$, $y = \left( y_{1}, \ldots, y_{i-1},y_{i+1},\ldots,y_{n} \right)$ satisfies $\left( y_{1}, \ldots, y_{i-1},t,y_{i+1},\ldots,y_{n} \right) \in W$, meaning:
$\langle \left( y_{1}, \ldots, y_{i-1},t,y_{i+1},\ldots,y_{n} \right) | v \rangle = y_{1}v_{1} + \ldots + y_{i-1}v_{i-1} + tv_{i} + y_{i+1}v_{i+1} + \ldots + y_{n}v_{n} = b$, therefore: $y_{1}v_{1} + \ldots + y_{i-1}v_{i-1} + y_{i+1}v_{i+1} + \ldots + y_{n}v_{n} = b - tv_{i}$.
So let's denote $b' = b - tv_{i}$ and $v' = \left( v_{1}, \ldots, v_{i-1},v_{i+1},\ldots,v_{n} \right)$ and we get that $A = W^{n-1} \left( v', b'\right)$ is an affine hyperplane in $\mathbb{R}^{n-1}$.
But then I thought to myself what if $v_{i}$ is the only cooardinate that satisfies $v_{i} \neq 0$ and all the other cooardinates are zero? so of course $v' = 0$ and $W^{n-1} \left( v', b'\right)$ is not an affine hyperplane.
I would be glad if someone can help me with it.

Answer 1

Hint

You were quite on the spot...

For $t \in \mathbb R$ and $i \in [n]$, let $A_t^i = \left\{ y \in \mathbb{R}^{n-1} | \left( y_{1}, \ldots, y_{i-1},t,y_{i+1},\ldots,y_{n} \right) \in W\right\}$.

As you said, $v \neq 0$. If $v_j \neq 0$ for all $j \in [n] = \{1, \dots, n\}$, let $i=1$ and otherwise chose $i$ such that $v_i = 0$.

Then prove that $A_t^i$ is a hyperplane of $\mathbb R^{n-1}$ for all $t \in \mathbb R$. For that use the fact that one of the $v_j$ with $j \neq i$ is not equal to zero.