Consider $\mathbb{R}^{4}$ with the inner product $\langle x,y \rangle=\sum _{i=0}^{4} x_iy_i$, for $x=(x_1,x_2,x_3,x_4)$ and $y=(y_1,y_2,y_3,y_4).$ Let $M=\{(x_1,x_2,x_3,x_4)\in \mathbb{R}^{4}:x_1=x_3\}$ and $M^{\perp}$ denote the orthogonal complement of $M$. What is the dimention of $M^{\perp}$?
It is well clear that dimension of $M$ is $3$, so as a matrix it has rank $3$ and nullity $1$. I think the sum of dimension of $M$ and $M^{\perp}$ should be equal to dimension of $\mathbb{R}^{4}$ i.e. $4$. As $M^{\perp}$ is orthogonal complement so inner product of any vector from $M^{\perp}$ and any vector from $M$ should be zero.
Am I thinking in a right direction?
