Every Plane Through the Origin in the Three Dimensional Space is a Subspace

Problems and solutions in Linear Algebra

Problem 294

Prove that every plane in the $3$-dimensional space $\R^3$ that passes through the origin is a subspace of $\R^3$.

 
FavoriteLoadingAdd to solve later

Sponsored Links

Proof.

Each plane $P$ in $\R^3$ through the origin is given by the equation
\[ax+by+cz=0\] for some real numbers $a, b, c$.
That is, the plane $P$ is a set of vectors $\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}$ satisfying the equation $ax+by+cz=0$:
\[P=\left\{\, \begin{bmatrix}
x \\
y \\
z
\end{bmatrix}\in \R^3 \quad \middle| \quad ax+by+cz=0 \,\right \}.\]


Now the equation can be written as the matrix equation
\[A\mathbf{x}=\mathbf{0},\] where $A$ is the $1\times 3$ matrix $A$, $\mathbf{x}\in \R^3$, and $\mathbf{0}$ is the $1$-dimensional zero vector given by
\[A=\begin{bmatrix}
a & b & c \\
\end{bmatrix}, \mathbf{x}=\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}, \mathbf{0}=[0].\]


Thus, the plane can be written as
\begin{align*}
P&=\{\mathbf{x}\in \R^3 \mid A\mathbf{x}=\mathbf{0}\}\\
\end{align*}
and this is the definition of the nullspace $\calN(A)$ of $A$. That is
\[P=\calN(A).\]


Therefore, the plane $P$ is the nullspace of the $1\times 3$ matrix $A$.
Since the nullspace of a matrix is always a subspace, we conclude that the plane $P$ is a subspace of $\R^3$.
Therefore, every plane in $\R^3$ through the origin is a subspace of $\R^3$.


FavoriteLoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Linear Algebra
Linear algebra problems and solutions
The Subset Consisting of the Zero Vector is a Subspace and its Dimension is Zero

Let $V$ be a subset of the vector space $\R^n$ consisting only of the zero vector of $\R^n$. Namely $V=\{\mathbf{0}\}$....

Close