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Show or give counter example: Let $\vec{b}\in \mathbb{R}^n$ and define $S=\{\vec{y}\in\mathbb{R}^n|\vec{y}^{\,T} \vec{b} = 1\}$. S is a subspace of $\mathbb{R^n}$

I tried using the subspace test

1.Let $\vec{y}$, $\vec{x} \in \mathbb{R}^n$ such that $\vec{y}^{\,T} \vec{b}= 1$ and $\vec{x}^{\,T} \vec{b}= 1$

$\begin{bmatrix} y_1&\\ .&\\ y_n \end{bmatrix}^{T}$ $\vec{b}$ + $\begin{bmatrix} x_1&\\ .&\\ x_n \end{bmatrix}^{T}$ $\vec{b} =$ $\begin{bmatrix} y_1&.&.&y_n \end{bmatrix} \vec{b} + \begin{bmatrix} x_1&.&.&x_n \end{bmatrix} \vec{b}$

$= \begin{bmatrix} y_1 + x_1&.&.&y_n + x_n \end{bmatrix} \vec{b}$

since $\vec{z} = \vec{y} + \vec{x} \in \mathbb{R}^n$, $ \exists \vec{b}$ such that $\vec{z}^{\,T}\vec{b}=1$

closed under addition.

  1. show closed under scalar multiplication...

I am supposed to give a brief explanation. There must be a counter-example, but I am not able to come up with one. I am a bit confused by the definition of set $S$.

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    Is the zero vector in $S$ ? – 2017-02-05
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    I should note that $S$ is *not* closed under addition. If $x,y$ are such that $\langle x,b\rangle=\langle y,b\rangle=1$, then $\langle x+y,b\rangle = 2$ and hence $x+y$ is not in $S$. – 2017-02-05
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    It is not closed under scalar multiplication either. For $\langle y,b\rangle=1$, we have $\langle cy,b\rangle=c\langle y,b\rangle=c$. The set $S$ actually looks like a hyperplane that does not pass through the origin. – 2017-02-05
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    It is not closed under multiplication or addition and does not contain the zero vector. Affine perhaps, but definitely not a linear subspace. – 2017-02-05

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