Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
    
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 at 8:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.