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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?

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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 at 15:33
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1 Answer 1

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

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I do not undestand how the kernel is an image of the other projection, can you please explain please? –  Mark Jun 20 at 15:51
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$P_1$ 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 $a_1\perp a_2$. –  Sami Ben Romdhane Jun 20 at 15:56
    
thank you, a lot clearer now –  Mark Jun 20 at 16:28
    
You're welcome. –  Sami Ben Romdhane Jun 20 at 16:28
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