# Find the Dimension of the Subspace of Vectors Perpendicular to Given Vectors

## Problem 578

Let $V$ be a subset of $\R^4$ consisting of vectors that are perpendicular to vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$, where
$\mathbf{a}=\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \quad \mathbf{b}=\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \quad \mathbf{c}=\begin{bmatrix} 0 \\ 1 \\ -1 \\ 0 \end{bmatrix}.$

Namely,
$V=\{\mathbf{x}\in \R^4 \mid \mathbf{a}^{\trans}\mathbf{x}=0, \mathbf{b}^{\trans}\mathbf{x}=0, \text{ and } \mathbf{c}^{\trans}\mathbf{x}=0\}.$

(a) Prove that $V$ is a subspace of $\R^4$.

(b) Find a basis of $V$.

(c) Determine the dimension of $V$.

## Proof.

### (a) Prove that $V$ is a subspace of $\R^4$.

Observe that the conditions
$\mathbf{a}^{\trans}\mathbf{x}=0, \mathbf{b}^{\trans}\mathbf{x}=0, \text{ and } \mathbf{c}^{\trans}\mathbf{x}=0$ can be combined into the following matrix equation
$A\mathbf{x}=\mathbf{0},$ where
$A=\begin{bmatrix} 1 & 0 & 1 & 0 \\ 1 &1 & 0 & 0 \\ 0 & 1 & -1 & 0 \end{bmatrix}$ and $\mathbf{0}$ is the three dimensional zero vector.
Note that the rows of the matrix $A$ are $\mathbf{a}^{\trans}$, $\mathbf{b}^{\trans}$, and $\mathbf{c}^{\trans}$.
It follows that the subset $V$ is the null space $\calN(A)$ of the matrix $A$.
Being the null space, $V=\calN(A)$ is a subspace of $\R^4$.
(See the post “The Null Space (the Kernel) of a Matrix is a Subspace of $\R^n$“.)

### (b) Find a basis of $V$.

In the proof of Part (a), we saw that $V=\calN(A)$.
To find a basis, we determine the solutions of $A\mathbf{x}=\mathbf{0}$.
Applying elementary row operations to the augmented matrix, we see that
\begin{align*}
\left[\begin{array}{rrrr|r}
1 & 0 & 1 & 0 &0 \\
1 & 1 & 0 & 0 & 0 \\
0 & 1 & -1 & 0 & 0
\end{array}\right] \xrightarrow{R_2-R_1}
\left[\begin{array}{rrrr|r}
1 & 0 & 1 & 0 &0 \\
0 & 1 & -1 & 0 & 0 \\
0 & 1 & -1 & 0 & 0
\end{array}\right]\6pt] \xrightarrow{R_3-R_2} \left[\begin{array}{rrrr|r} 1 & 0 & 1 & 0 &0 \\ 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]. \end{align*} It follows that the general solution is given by \[x_1=-x_3, x_2=x_3. The vector form solution is
$\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}=\begin{bmatrix} -x_3 \\ x_3 \\ x_3 \\ x_4 \end{bmatrix}=x_3\begin{bmatrix} -1 \\ 1 \\ 1 \\ 0 \end{bmatrix}+x_4\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}.$

Hence we have
\begin{align*}
V&=\calN(A)\\
-1 \\
1 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
0 \\
0 \\
0 \\
1
\end{bmatrix}, \text{ where $x_3, x_4\in \R$} \,\right\}\\[6pt] &=\Span \left\{\, \begin{bmatrix}
-1 \\
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
0 \\
0 \\
1
\end{bmatrix} \,\right\}.
\end{align*}
Let $B:=\left\{\, \begin{bmatrix} -1 \\ 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \,\right\}$.
Then we just showed that $B$ is a spanning set for $V$.
It is straightforward to see that $B$ is linearly independent.
Hence $B$ is a basis for $V$.

### (c) Determine the dimension of $V$.

As the basis $B$ for $V$ that we obtained in Part (b) consists of two vectors, the dimension of the subspace $V$ is $\dim(V)=2$.

Let $V$ be a subspace of $\R^n$. Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\}$ is a basis of the subspace $V$....