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My (introductory) linear algebra textbook states the following:

The equation of the plane through three noncollinear points $A$, $B$, and $C$ in space is $x = A + su + tv$, where $u$ and $v$ denote the vectors beginning at $A$ and ending at $B$ and $C$, respectively, and $s$ and $t$ denote arbitrary real numbers. An important special case occurs when $A$ is the origin. In this case, the equation of the plane simplifies to $x = su + tv$, and the set of all points in this plane is a subspace of $\mathbb{R}^3$.

I am specifically confused about how the set of all points in the plane is a subspace of $\mathbb{R}^3$.

I understand what subspace are, but I fail to see how the set of all points in the plane $x = su + tv$ is a subspace of $\mathbb{R}^3$.

I would greatly appreciate it if someone could please take the time to elaborate on and demonstrate this concept. Please use elementary concepts and mathematical language in any explanation.

2 Answers 2

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First of all note that the fact that $A=0$ is very important, as this implies that the identity of $\mathbb{R}^3$ is in the subspace, which is a must. Anyway to understand why the plane is a subspace notice that any plane is generated by two non-parallel vectors. In other words every point in a plane is a linear combination of the two vectors. So hence we have that the plane $\pi$ is equal to the $\text{Span}(\vec{u},\vec{v})$, so hence it's a subspace.

In fact you can easily veryfy that the set contains the identity element is closed under both addition and scalar multiplication using the properties of vector algebra.

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    I understand. Thank you for the response. Would you mind please also answering my query in the comments to Arun's answer? – 2017-02-11
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    @ThePointer Yes, $x = su + tv$ is the equation of a plane, but $x$ is a vector (the RHS is summation of vectors) hence a point in the plane. In fact the plane $\pi$ is defined as $\pi = \{\vec{x} | \vec{x} = s\vec{u} + t\vec{v}; s,t \in \mathbb{R} \}$ – 2017-02-11
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    ahh, I understand. So the equation of a plane is just also just the equation of a vector (in this case, all possible linear combinations of two vectors)? – 2017-02-11
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    @ThePointer You're right. – 2017-02-11
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Since it is a Subset of $\mathbb{R}^3$ we only need to show it is closed under addition and scalar multiplication. So let $x,y$ be two points in the given plane, then $$x=s_1u+t_1v, y=s_2u+t_2v$$ So $$x+y=(s_1+s_2)u+(t_1+t_2)v$$ is in the plane and $$\alpha x=(\alpha s_1) u+(\alpha t_1)v$$ is also in the plane. Hence the given plane is a subspace of $\mathbb{R}^3$.

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    Do you mean $x$ and $y$ are planes -- not points? – 2017-02-11
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    $x,y$ are points in the plane. thanks!! stefan4024 for the edit – 2017-02-11
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    But it says that $x = su + tv$ is the equation of a plane starting at the origin. Doesn't that mean $x$ and $y$ are both planes? – 2017-02-11
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    $x=su+tv$ is the equation of a plane i.e the set of all points $x$ of the form $x=su+tv$. you have written it correctly in the Title of the question. – 2017-02-11