# Hyperplane in $n$-Dimensional Space Through Origin is a Subspace

## Problem 352

A hyperplane in $n$-dimensional vector space $\R^n$ is defined to be the set of vectors
$\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}\in \R^n$ satisfying the linear equation of the form
$a_1x_1+a_2x_2+\cdots+a_nx_n=b,$ where $a_1, a_2, \dots, a_n$ (at least one of $a_1, a_2, \dots, a_n$ is nonzero) and $b$ are real numbers.
Here at least one of $a_1, a_2, \dots, a_n$ is nonzero.

Consider the hyperplane $P$ in $\R^n$ described by the linear equation
$a_1x_1+a_2x_2+\cdots+a_nx_n=0,$ where $a_1, a_2, \dots, a_n$ are some fixed real numbers and not all of these are zero.
(The constant term $b$ is zero.)

Then prove that the hyperplane $P$ is a subspace of $R^{n}$ of dimension $n-1$.

## Proof.

### The hyperplane is a subspace.

In a set theoretical notation, the hyperplane is give by
$P=\left\{\, \mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}\in \R^n \quad \middle | \quad a_1x_1+a_2x_2+\cdots+a_nx_n=0 \,\right\}$

The defining relation
$a_1x_1+a_2x_2+\cdots+a_nx_n=0$ can be written as
$\begin{bmatrix} a_1 & a_2 & \dots & a_n \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} =0.$

Let $A$ be the $1\times n$ matrix $A=\begin{bmatrix} a_1 & a_2 & \dots & a_n \end{bmatrix}$.
Then the above equation is simply $A\mathbf{x}=0$, and the hyperplane $P$ is described by
$P=\{\mathbf{x}\in \R^n \mid A\mathbf{x}=0\},$ and this is exactly the definition of the null space of $A$. Namely, we have
$P=\calN(A).$ Since every null space of a matrix is a subspace, it follows that the hyperplane $P$ is a subspace of $\R^n$.

### The dimension of the hyperplane is $n-1$.

Because $P=\calN(A)$, the dimension of $P$ is the nullity of the matrix $A$.
Since not all of $a_i$’s are zero, the rank of the matrix $A$ is $1$. Then by the rank-nullity theorem, we have
$\text{rank of A } + \text{ nullity of A }=n.$ Hence the nullity of $A$ is $n-1$.

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