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I am poor in mathematics and want to learn few fundamental ethics to understand some of advanced things;

For plane $i$, denote $n_i\in\mathbb{R}^3$ and $o_i\in\mathbb{R}^3$ respectively as its normal vector and a point on the plane. Then

(1). How the distance between a point $p$ and the plane can be expressed as

$$\left\|\left(\frac{n_i n_i^T}{n_i^Tn_i}\right)(p-o_i)\right\|$$


(2) Let $P_i=\frac{n_i n_i^T}{n_i^Tn_i}$ and $P_i$ is considered as an orthogonal projection matrix then,

what is the meaning of being $P_i=P_i^T$ and $P_i^2=P_i$. Is there any special conditions should meet to be like that?

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up vote 2 down vote accepted

First, vectors are considered a column vectors by default, so ${n_i}^T$ is a row vector. Then the denominator is the matrix product ${n_i}^Tn_i$, which is the same as the inner product $n_i\cdot n_i$, that is, it gives the square of the length of $n_i$: $${n_i}^Tn_i=\|n_i\|^2\,.$$ Now introduce $m_i:=n_i/\|n_i\|$, this is already a unit vector ($\|m_i\|=1$). We have $$m_i\,m_i^T=\frac{n_i\,{n_i}^T}{\|n_i\|^2}=P_i\,.$$ We also have $P^2=m_im_i^T\cdot m_im_i^T=m_i\,\|m_i\|^2\, m_i^T=m_i\cdot1\cdot m_i^T=P_i$, and ${P_i}^T=(m_im_i^T)^T=((m_i)^T)^Tm_i^T=m_im_i^T=P_i$, so $P_i$ is indeed an orthogonal projection matrix.

Say, $v$ is parallel to the plane $i$, so that $v\perp n_i$, or, $v\perp m_i$. Then we have $m_i^Tv= m_i\cdot v=0$, so also $P_iv=0$. On the other hand $P_i(m_i)=m_i$. So, $P_i$ projects orthogonally to the line of $m_i$.

If you want the projection to the plane, then you have to take $I-P_i$ where $I$ is the identity matrix.

For (1), by the avove, we get that $P_i(p-o_i)$ is the orthogonal projection of the vector $p-o_i$ (this one goes from point $o_i$ to point $p$) to the line of $m_i$, so its length is indeed the distance from $p$ to the plane.

Alternatively, we can calculate it as $$|(p-o_i)\cdot m_i|\,,$$ maybe it's a bit more simple..

Update: For a linear transformation $T:V\to V$ to be idempotent (i.e. $T^2=T$), is equivalent to that $T$ is a projection, but not necessarily orthogonal, i.e. there are subspaces $U,W$ such that $T$ is the projection onto $U$ along $W$. Here $U$ and $W$ are disjoint: $U\cap W=\{0\}$, and together span the whole $V$, i.e. every vector $v$ can be uniquely written in the form $v=u+w$ for $u\in U,\,w\in W$. [This fact is denoted by $V=U\oplus W$.]

Now, if $T^2=T$, then let $U:={\rm ran\,}T=\{Tv\mid v\in V\}$, and let $W:=\ker T=\{v\mid Tv=0\}$. Try to prove the statements above w.r.t $U$ and $W$, and that $$T(u+w)=u$$ for all $u\in U$ and $w\in W$.

The additional condition that $T^*=T$ ensures in addition that $W\perp U$, so it is an orthogonal projection.

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really appreciate your explanation. if you can say something on Q2, it is also nice. (i amended the way i asked the question 2). thanks again – gnp Jun 12 '13 at 13:52
Can you answer the question? – Orient Jun 9 '14 at 8:17

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