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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

2 vectors are given: $$ \vec{a_1}=\begin{bmatrix}1 && 2 && -2\end{bmatrix}^T \\ \vec{a_2}=\begin{bmatrix}-2 && 2 && 1\end{bmatrix}^T $$

To calculate their projection matrices, I used the formula: $$ P = \frac{\vec{a}{\vec{a^T}}}{\vec{a^T}{\vec{a}}} $$

and got the following matrices:

$$ \vec{P_1}=\begin{bmatrix}1 && 2 && -2 \\ 2 && 4 && -4 \\ -2 && -4 && 4 \end{bmatrix} \\ \vec{P_2}=\begin{bmatrix}4 && -4 && -2 \\ -4 && 4 && 2 \\ -2 && 2 && 1\end{bmatrix} $$

And their product is a zero matrix, but I can't explain why.

Is it because the vectors are orthogonal?

share|cite|improve this question
1  
yes. $P_1P_2= a_1a_1^Ta_2a_2^T= a_10a_2^T=0$.Here by $0$ I mean zero matrix of the corresponding dimension. – Alexander Vigodner Jun 20 '14 at 15:33
up vote 1 down vote accepted

Notice that the two vectors are orthogonal so the kernel of the projection $P_1$ is the image of the projection $P_2$ and then their product is the zero matrix

share|cite|improve this answer
    
I do not undestand how the kernel is an image of the other projection, can you please explain please? – Mark Jun 20 '14 at 15:51
1  
$P_1$ is the orthogonal projection onto $\operatorname{span}(a_1)$ so it's the image of $P_1$ and $\operatorname{span}(a_2)$ is the kernel since $a_1\perp a_2$. – user63181 Jun 20 '14 at 15:56
    
thank you, a lot clearer now – Mark Jun 20 '14 at 16:28
    
You're welcome. – user63181 Jun 20 '14 at 16:28

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.