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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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    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}}2012-12-05

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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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    Yes.$\ \ \ \ \ \ \ $2012-12-05