Subset of Vectors Perpendicular to Two Vectors is a Subspace

Problems and solutions in Linear Algebra

Problem 119

Let $\mathbf{a}$ and $\mathbf{b}$ be fixed vectors in $\R^3$, and let $W$ be the subset of $\R^3$ defined by
\[W=\{\mathbf{x}\in \R^3 \mid \mathbf{a}^{\trans} \mathbf{x}=0 \text{ and } \mathbf{b}^{\trans} \mathbf{x}=0\}.\]

Prove that the subset $W$ is a subspace of $\R^3$.
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We prove the following criteria for the subset $W$ to be a subspace of $\R^3$.

(a) The zero vector $\mathbf{0} \in \R^3$ is in $W$.

(b) If $\mathbf{x}, \mathbf{y} \in W$, then $\mathbf{x}+\mathbf{y}\in W$.

(c) If $\mathbf{x} \in W$ and $c\in \R$, then $c\mathbf{x} \in W$.

For (a), note that $\mathbf{a}^{\trans} \mathbf{0}=0$ and $\mathbf{b}^{\trans} \mathbf{0}=0$. Thus the zero vector $\mathbf{0}\in \R^3$ is in $W$.

To check (b), let $\mathbf{x}, \mathbf{y} \in W$. Then we have the following relations.
\[\mathbf{a}^{\trans} \mathbf{x}=0 \text{ and } \mathbf{b}^{\trans} \mathbf{x}=0,
\text{ and }
\mathbf{a}^{\trans} \mathbf{y}=0 \text{ and } \mathbf{b}^{\trans} \mathbf{y}=0.\tag{*}\] To show that $\mathbf{x}+\mathbf{y} \in W$, we need to show that
\[\mathbf{a}^{\trans}(\mathbf{x}+\mathbf{y})=0 \text{ and } \mathbf{b}^{\trans}(\mathbf{x}+\mathbf{y})=0.\]

We first compute
\mathbf{a}^{\trans}(\mathbf{x}+\mathbf{y}) &=\mathbf{a}^{\trans}\mathbf{x}+\mathbf{a}^{\trans}\mathbf{y}\\
&= 0+0=0
by the relations (*).
Similarly, we have
\mathbf{b}^{\trans}(\mathbf{x}+\mathbf{y}) &=\mathbf{b}^{\trans}\mathbf{x}+\mathbf{b}^{\trans}\mathbf{y}\\
&= 0+0=0
by the relations (*).

Thus the vector $\mathbf{x}+\mathbf{y}$ satisfies the defining relations for $W$, hence $\mathbf{x}+\mathbf{y} \in W$.

Finally, to prove (c), let $\mathbf{x} \in W$ and let $c\in \R$.
Since $\mathbf{x} \in W$, we have
\[\mathbf{a}^{\trans} \mathbf{x}=0 \text{ and } \mathbf{b}^{\trans} \mathbf{x}=0.\] Multiplying by the scalar $c$ from the left, we obtain
\[\mathbf{a}^{\trans} (\mathbf{cx})=0 \text{ and } \mathbf{b}^{\trans} (\mathbf{cx})=0.\] (Note that since $c$ is a scalar, we can switch the order of the product of $c$ and $\mathbf{a}^{\trans}$. Same for $c$ and $\mathbf{b}^{\trans}$.)

These equalities proves that the vector $c\mathbf{x}$ satisfies the defining relation for $W$. Thus $c\mathbf{x} \in W$.

Therefore the criteria (a)-(c) are all met, and we conclude that $W$ is a subspace of $\R^3$.

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