Linear Algebra

The Set of Vectors Perpendicular to a Given Vector is a Subspace

Problem 659

Fix the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$, and let $\R^3$ be the vector space of $3 \times 1$ column vectors. Define
\[W = \{ \mathbf{v} \in \R^3 \mid \mathbf{b} \mathbf{v} = 0 \}.\] Prove that $W$ is a vector subspace of $\R^3$.

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Proof.

We verify the subspace criteria: the zero vector $\mathbf{0}$ of $\R^3$ is in $W$, and $W$ is closed under addition and scalar multiplication.

First, the zero element in $\R^3$ is $\mathbf{0}$, the $3 \times 1$ column vector whose entries are all $0$. Then clearly $\mathbf{b} \mathbf{0} = 0$, and so $\mathbf{0} \in W$.

Next, suppose $\mathbf{v} , \mathbf{w} \in W$, and $c \in \mathbb{R}$. Then $\mathbf{b} \mathbf{v} = \mathbf{b} \mathbf{w} = 0$, and so
\[\mathbf{b} ( \mathbf{v} + \mathbf{w} ) = \mathbf{b} \mathbf{v} + \mathbf{b} \mathbf{w} = 0.\] Thus, $\mathbf{v} + \mathbf{w} \in W$.

Because, again, $\mathbf{b} \mathbf{v} = \mathbf{0}$, we have
\[\mathbf{b} ( c \mathbf{v} ) = c \mathbf{b} \mathbf{v} = c \mathbf{0} = \mathbf{0}.\] Thus $c \mathbf{v} \in W$. These three criteria show that $W$ is a vector subspace of $\R^3$.

Comment.

We can generalize the problem with an arbitrary $1\times 3$ row vector $\mathbf{b}$.

The proof is almost identical.
(Look at the proof. We didn’t use components of the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$.)

Note that vectors $\mathbf{u}, \mathbf{v}\in \R^3$ is said to be perpendicular if
\[\mathbf{u}\cdot \mathbf{v}=\mathbf{u}^{\trans}\mathbf{v}=0.\]

Thus, the result of the problem says that for a fixed vector $\mathbf{u}\in \R^3$, the set of vectors $\mathbf{v}$ that are perpendicular to $\mathbf{u}$ is a subspace in $\R^3$.
(Note that we appy the problem to $\mathbf{b}=\mathbf{u}^{\trans}$.)

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