Linear Algebra

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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