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[SOLVED] Decide if the sets are subspaces or affine subspaces

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
Hey!! :eek:

We have the subsets \begin{equation*}V:=\left \{\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}\mid x_1=0\right \}, \ \ \ W:=\left \{\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}\mid x_2=2\right \}, \ \ \ S:=\left \{\lambda \begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}\mid \lambda \in \mathbb{R}\right \}, \\ T:=\left \{\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}+\lambda\begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}\mid \lambda \in \mathbb{R}\right \}\end{equation*}

I want to check which are subspaces and which are affine subspaces.

I have done the following:

  • We consider the subset $V$.
    1. It holds that $V\neq \emptyset$, since for example $\begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}$ is in $V$.
    2. We consider two elements of $V$, $v_1=\begin{pmatrix}0 \\ x_2 \\ x_3\end{pmatrix}$, $v_2=\begin{pmatrix}0 \\ \tilde{x}_2 \\ \tilde{x}_3\end{pmatrix}$. Then the sum is $v_1+v_2=\begin{pmatrix}0 \\ x_2+\tilde{x}_2 \\ x_3+\tilde{x}_3\end{pmatrix}\in V$.
    3. Let $v=\begin{pmatrix}0 \\ x_2 \\ x_3\end{pmatrix}\in V$ and $\alpha\in \mathbb{R}$. Then $\alpha\cdot v=\begin{pmatrix}0 \\ \alpha x_2 \\ \alpha x_3\end{pmatrix}\in V$.
    That means that $V$ is a subspace.

    $$$$
  • We consider the subset $W$.
    1. It holds that $W\neq \emptyset$, since for example $\begin{pmatrix}0 \\ 2 \\ 0\end{pmatrix}\in W$.
    2. Let $w_1=\begin{pmatrix}x_1 \\ 2 \\ x_3\end{pmatrix}, w_2=\begin{pmatrix}\tilde{x}_1 \\ 2 \\ \tilde{x}_3\end{pmatrix}\in W$. Then $w_1+w_2=\begin{pmatrix}x_1+\tilde{x}_1 \\ 4 \\ x_3+\tilde{x}_3\end{pmatrix}\notin W$.
    So $W$ is not a subspace.

    We can write this subset in the form:
    \begin{equation*}W:=\left \{\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}\mid x_2=2\right \}=\left \{\begin{pmatrix}x_1 \\ 2 \\ x_3\end{pmatrix}\right \}=\left \{\begin{pmatrix}0 \\ 2 \\ 0\end{pmatrix}+\begin{pmatrix}x_1 \\ 0 \\ x_3\end{pmatrix}\right \}\end{equation*}

    We show that the set $\tilde{W}=\left \{\begin{pmatrix}x_1 \\ 0 \\ x_3\end{pmatrix}\right \}$ is s subspace, and then $W=\left \{\begin{pmatrix}0 \\ 2 \\ 0\end{pmatrix}+\tilde{w}\mid \tilde{w}\in \tilde{W}\right \} $ is an affine subspace.
    1. It holds that $\tilde{W}\neq \emptyset$, since for example the vector $\begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}$ is contained.
    2. Let $\tilde{w}_1=\begin{pmatrix}x_1 \\ 0 \\ x_3\end{pmatrix}, \tilde{w}_2=\begin{pmatrix}\tilde{x}_1 \\ 0 \\ \tilde{x}_3\end{pmatrix}\in \tilde{W}$. Then $\tilde{w}_1+\tilde{w}_2=\begin{pmatrix}x_1+\tilde{x}_1 \\ 0 \\ x_3+\tilde{x}_3\end{pmatrix}\in \tilde{W}$.
    3. Let $\tilde{w}=\begin{pmatrix}x_1 \\ 0 \\ x_3\end{pmatrix}\in \tilde{W}$ and $\alpha\in \mathbb{R}$. Then $\alpha\cdot \tilde{w}=\begin{pmatrix}\alpha x_1 \\ 0 \\ \alpha x_3\end{pmatrix}\in \tilde{W}$.
    Therefore $\tilde{W}$ is a subspace and so $W$ is an affine subspace.

    $$$$
  • We consider the subset $S$.
    1. It holds that $S\neq \emptyset$, since $\begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}$ is in $S$.
    2. Let $s_1=\begin{pmatrix}\lambda_1 \\ 0 \\ -\lambda_1 \end{pmatrix}, s_2=\begin{pmatrix}\lambda_2 \\ 0 \\ -\lambda_2\end{pmatrix}\in S$. Then $s_1+s_2=\begin{pmatrix}\lambda_1 \\ 0 \\ -\lambda_1 \end{pmatrix}+\begin{pmatrix}\lambda_2 \\ 0 \\ -\lambda_2\end{pmatrix}=\left (\lambda_1+\lambda_2\right )\cdot \begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}\in S$.
    3. Let $s=\lambda\begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}\in S$ and $\alpha\in \mathbb{R}$. Then $\alpha\cdot s=\alpha\cdot \lambda\begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}=\left (\alpha\cdot \lambda\right )\begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}\in S$.
    Therefore $S$ is a subspace.



    $$$$

  • We consider the subset $T$.

    It holds that \begin{equation*}T:=\left \{\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}+\lambda\begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}\mid \lambda \in \mathbb{R}\right \}=\left \{\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}+s\mid s\in S\right \}\end{equation*}

    Since $S$ is a subspace it follows that $T$ is an affine subspace.


Is everything correct and complete? (Wondering)

Does it hold in general that a subset is either a subspace or an affine subspace? (Wondering)
 
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Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
Is everything correct and complete?
Yep. All in order. (Nod)

Does it hold in general that a subset is either a subspace or an affine subspace?
Nope. (Shake)

Suppose we add a single non-zero vector to any of these spaces that is not already in it.
Let's say we add $(1,0,0)$ to them.
Are they still subspaces or affine subspaces then? (Wondering)
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Yep. All in order. (Nod)



Nope. (Shake)

Suppose we add a single non-zero vector to any of these spaces that is not already in it.
Let's say we add $(1,0,0)$ to them.
Are they still subspaces or affine subspaces then? (Wondering)
??? Yes, it is an affine subset. (I would not use the term "affine subspace".)

An example of a subset of a vector space that is neither a subspace nor an affine subset is [tex]\{\begin{pmatrix}x \\ y \\ z \end{pmatrix}| x^2+ y^2+ z^2|= 1\}[/tex].
 
Last edited:

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
Ahh ok!! Thank you!! (Malthe)