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 $n-1$.

In a set theoretical notation, the hyperplane is given 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 $n-1$.

Rank and Nullity of a Matrix, Nullity of Transpose
Let $A$ be an $m\times n$ matrix. The nullspace of $A$ is denoted by $\calN(A)$.
The dimension of the nullspace of $A$ is called the nullity of $A$.
Prove the followings.
(a) $\calN(A)=\calN(A^{\trans}A)$.
(b) $\rk(A)=\rk(A^{\trans}A)$.
Hint.
For part (b), […]

Given a Spanning Set of the Null Space of a Matrix, Find the Rank
Let $A$ be a real $7\times 3$ matrix such that its null space is spanned by the vectors
\[\begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}, \begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}, \text{ and } \begin{bmatrix}
1 \\
-1 \\
0
[…]

Prove a Given Subset is a Subspace and Find a Basis and Dimension
Let
\[A=\begin{bmatrix}
4 & 1\\
3& 2
\end{bmatrix}\]
and consider the following subset $V$ of the 2-dimensional vector space $\R^2$.
\[V=\{\mathbf{x}\in \R^2 \mid A\mathbf{x}=5\mathbf{x}\}.\]
(a) Prove that the subset $V$ is a subspace of $\R^2$.
(b) Find a basis for […]

Linear Transformation to 1-Dimensional Vector Space and Its Kernel
Let $n$ be a positive integer. Let $T:\R^n \to \R$ be a non-zero linear transformation.
Prove the followings.
(a) The nullity of $T$ is $n-1$. That is, the dimension of the nullspace of $T$ is $n-1$.
(b) Let $B=\{\mathbf{v}_1, \cdots, \mathbf{v}_{n-1}\}$ be a basis of the […]

Linear Dependent/Independent Vectors of Polynomials
Let $p_1(x), p_2(x), p_3(x), p_4(x)$ be (real) polynomials of degree at most $3$. Which (if any) of the following two conditions is sufficient for the conclusion that these polynomials are linearly dependent?
(a) At $1$ each of the polynomials has the value $0$. Namely $p_i(1)=0$ […]

Idempotent Matrices are Diagonalizable
Let $A$ be an $n\times n$ idempotent matrix, that is, $A^2=A$. Then prove that $A$ is diagonalizable.
We give three proofs of this problem. The first one proves that $\R^n$ is a direct sum of eigenspaces of $A$, hence $A$ is diagonalizable.
The second proof proves […]