# 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 $V$.

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## 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)\\

&=\left\{\, \mathbf{x}\in \R^4 \quad \middle|\quad \mathbf{x}=x_3\begin{bmatrix}

-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 $\dim(V)=2$.

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