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This is my HW (Linear algebra 2) and I need to find projection matrix, kernel and image of the projection.

V=$R^2$

So I have: subspace of V

$sp{(2,-3)}$

Than I found the projection matrix is:

$$ \begin{matrix} 4/13 & -6/13\\ -6/13 & 9/13 \\ \end{matrix} $$

but now I need to find kernel and image... I don't remember how to do that and I searched and google and I know I saw I need to find Ax=0 So is my kernel is 0?! I don't totally understand this

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What does "$sp(2,-3)$" mean? –  wj32 Dec 5 '12 at 11:58
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I would guess the space is $\mathbb{R}^2$ and $sp(2,-3)$ is the span of the vector $(2,-3)$. But the OP should answer that. –  Julian Kuelshammer Dec 5 '12 at 12:01
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so is it always depend on the determinant?? if it was non zero? and what about the image? –  Mary Dec 5 '12 at 12:19
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You can find the kernel by solving the linear equation $Ax=0$. What the image is, is obvious if you recall what the projection map does, i.e. that it projects onto the given subspace. You should include some more information about what you already know. –  Julian Kuelshammer Dec 5 '12 at 12:21
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What is your definition of projection matrix? How did you come up with this projection matrix? –  Julian Kuelshammer Dec 5 '12 at 12:23

1 Answer 1

up vote 0 down vote accepted

The projection you constructed is precisely the orthogonal projection onto the span of $(2,-3)$. So the span of $(2,-3)$ is the image.

Being an orthogonal projection, its kernel is the orthogonal of its image, so the kernel is $V^\perp$, which you can write as the span of $(3,2)$.

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So this will be the same for every projection matrix? like if I have to find for y=1/2x the image is also R^2 and the kernel is sp(-1,2)??? –  Mary Dec 5 '12 at 14:44
    
My bad, I didn't read the question carefully when typing. The image can never be all of $\mathbb R^2$, unless your projection is the identity matrix. I've edited the answer. –  Martin Argerami Dec 5 '12 at 14:48
    
Thanks, so following my previous comment the image will be sp (2,1) and kernel as I said? –  Mary Dec 5 '12 at 15:05
    
Yes.$\ \ \ \ \ \ \ $ –  Martin Argerami Dec 5 '12 at 15:30
    
Thank you very much!! –  Mary Dec 5 '12 at 15:31

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