What is the dimension of $\{x\}^{\perp}\cap \{y\}^{\perp}$

Question

Let $V$ be a real inner product space of dimension $10$. Let $x,y\in V$ be non-zero vectors such that $\langle x,y\rangle=0$. Then the dimension of $\{x\}^{\perp}\cap \{y\}^{\perp}$_______?

So by $\{x\}^{\perp}=\{s\in V: \langle s,x \rangle=0\}$, then how to calculate the dimension of $\{x\}^{\perp}\cap \{y\}^{\perp}$? I can think of only 3 elements $0$, $x$ and $y$ belong to that intersection. How can I do this. Please help me to solve this. Thanks.

Answer 1

Suppose $B=\{e_1,\ldots,e_{10}\}$ is an orthonormal basis of $V$ and $$x=x_1e_1+\cdots+x_{10}e_{10},\quad y=y_1e_1+\cdots+y_{10}e_{10}.$$ Then, $$v=a_1e_1+\cdots+a_{10}e_{10}\in\{x\}^{\perp}\cap \{y\}^{\perp}\Leftrightarrow \begin{cases} x_1a_1+\cdots+x_{10}a_{10}=0 \\ y_1a_1+\cdots+y_{10}a_{10}=0\end{cases}.$$ As $x\ne 0,y\ne 0$ and $\langle x,y \rangle=0$, $\{x,y\}$ is linearly independent so,
$$\dim \left(\{x\}^{\perp}\cap \{y\}^{\perp}\right)=10-\text{rank }\begin{bmatrix}{x_1}&{\ldots}&x_{10}\\{y_1}&{\ldots}&y_{10}\end{bmatrix}\underbrace{=}_{\{x,y\}\text{ lin. indep.}}10-2=8.$$

Answer 2

It is a general fact that, for subspaces of $V$, $$ U_1^{\perp}\cap U_2^{\perp}=(U_1+U_2)^\perp $$ In your case $U_1$ is the span of $x$ and $U_2$ is the span of $y$; since these are nonzero and orthogonal, $U_1+U_2$ has $\{x,y\}$ as a basis.

Therefore $$\DeclareMathOperator{\SP}{span} \dim(\{x\}^\perp\cap\{y\}^\perp)= \dim((\SP\{x\})^\perp\cap(\SP\{y\})^\perp)= \dim(\SP\{x,y\})^\perp)=10-2 $$ because if $U$ is a $k$-dimensional subspace of the $n$-dimensional subspace $V$, $$ \dim U^\perp=n-k $$