Dimension of a subspace of finite-dimensional product space $V$ equals $\dim V - 1$ Suppose $w$ is a nonzero vector in a finite dimensional inner product space $V$. Let $P = \{ v \in V | \langle v,w\rangle = 0\}$. Show that $\dim P = \dim V - 1$ where $P$ is a subspace of $V$.
 A: Let $v=(x_1,x_2,\ldots,x_n), w=(a_1,a_2, \ldots, a_n)$ Then the condition $<v,w>=0$ is equivalent to the equation $a_1 x_1+a_2 x_2+\cdots a_n x_n =0.$ The solutions of the equation forms a vector space. It is easy to see that  $n-1$ vectors $(a_1,0,0,-\frac{a_i}{a_1},0,\ldots)$  are linearly independed. Thus   $\dim P=n-1.$
A: We have a short exact sequence
$$0\to P\to V\to k\to 0$$
where on the left is inclusion and on the right $v\mapsto\langle v,w\rangle$ (surjectivity follows from $\frac1{\langle w,w\rangle}w\mapsto 1$), hence $\dim V=\dim P+\dim k$.
A: Hint: If $e_1, \ldots, e_n$ is any orthonormal basis of $P$ then $\dfrac{w}{\|w\|}, e_1, \ldots, e_n$ is an orthonormal basis of $V$.
A: Since $P$ is the kernel of the linear mapping $v\mapsto \langle v,w\rangle$, it can be described (in any coordinates) as a nullspace of the matrix $w^T$. Now use what you know about matrix row reduction to finish the problem.
A: Let $w=(a_1,\ldots, a_n), v=(x_1,\ldots, x_n)$. Then $P$ is defined by one equation $a_1x_1+\ldots +a_nx_n=0$, hence $\dim P=\dim V-1$.
A: just imagine is as a n dimensional space. w is one dimension. except it there remains n dimension which are orthogonal/perpendicular to w, which is exactly what P is , set of vectors orthogonal to w.
