# Decide if the sets are subspaces or affine subspaces

• MHB
• mathmari
In summary, the conversation discussed subsets of a vector space and their properties as subspaces or affine subspaces. It was shown that $V$ and $S$ are subspaces, while $W$ is not a subspace but an affine subspace. $T$ was also shown to be an affine subspace. It was mentioned that in general, a subset can be either a subspace or an affine subspace, but not both. The addition of a non-zero vector to these spaces would result in an affine subset. However, an example of a subset that is neither a subspace nor an affine subset was given.
mathmari
Gold Member
MHB
Hey!

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)

Last edited by a moderator:
mathmari said:
Is everything correct and complete?

Yep. All in order. (Nod)

mathmari said:
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)

Klaas van Aarsen said:
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 $$\{\begin{pmatrix}x \\ y \\ z \end{pmatrix}| x^2+ y^2+ z^2|= 1\}$$.

Last edited by a moderator:
Ahh ok! Thank you! (Malthe)

## 1. What is a subspace?

A subspace is a subset of a vector space that is closed under addition and scalar multiplication. This means that if you add two vectors or multiply a vector by a scalar within the subspace, the result will still be within the subspace.

## 2. How do you determine if a set is a subspace?

To determine if a set is a subspace, you need to check if it satisfies the two conditions of closure under addition and scalar multiplication. This can be done by checking if the set contains the zero vector, and if any two vectors in the set can be added or multiplied together to result in a vector that is also in the set.

## 3. What is an affine subspace?

An affine subspace is a subset of a vector space that is not necessarily closed under addition and scalar multiplication, but still maintains the structure of a vector space. This means that the set contains a point that serves as an origin, and all other points in the set can be reached by adding a vector to the origin.

## 4. How do you determine if a set is an affine subspace?

To determine if a set is an affine subspace, you need to check if it contains an origin point and if all other points in the set can be reached by adding a vector to the origin. If these conditions are met, then the set is an affine subspace.

## 5. Can a set be both a subspace and an affine subspace?

Yes, a set can be both a subspace and an affine subspace. This means that the set is closed under addition and scalar multiplication, but also contains an origin point and can be reached by adding a vector to the origin. In this case, the set is both a subset and a subset with an origin point.

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