Kernel and Image of projection Matrix

Question

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

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 05 '12 at 12:01
The determinant of your matrix is zero, so the kernel will not be trivial. — Julian Kuelshammer, Dec 05 '12 at 12:18
so is it always depend on the determinant?? if it was non zero? and what about the image? — Mary, Dec 05 '12 at 12:19
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 05 '12 at 12:21
@JulianKuelshammer don't know nothing more than I mentioned.. — Mary, Dec 05 '12 at 12:22
What is your definition of projection matrix? How did you come up with this projection matrix? — Julian Kuelshammer, Dec 05 '12 at 12:23
as we learned at class there is a pattern for projection matrix that is -> 1/(a^2+b^2)*{{a^2,ab},{ab,b^2}} — Mary, Dec 05 '12 at 12:27

Martin Argerami · Accepted Answer · 2012-12-05T14:47:58.483

0

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 $(3,2)$.

edited Dec 05 '12 at 14:47

answered Dec 05 '12 at 14:26

Martin Argerami

205,756

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 05 '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 05 '12 at 14:48
Thanks, so following my previous comment the image will be sp (2,1) and kernel as I said? – Mary Dec 05 '12 at 15:05
Yes.$\ \ \ \ \ \ \ $ – Martin Argerami Dec 05 '12 at 15:30

Kernel and Image of projection Matrix

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