2

I am trying to find the image and kernel of the orthogonal projection onto the plane x+2y+3z=0. The following is how I solved the problem, but it appears to be incorrect. What is wrong with my solution?

I began by finding the matrix for this transformation, which I believe is\begin{bmatrix}1/196&2/196&3/196\\2/196&4/196&6/196\\3/196&6/196&9/196\end{bmatrix} The rref form of this is \begin{bmatrix}1&2&3\\0&0&0\\0&0&0\end{bmatrix} My undestanding was that the kernel would be the set of inputs that the transform would make into the 0 vector, which I found to be the span of \begin{bmatrix}-2\\1\\0\end{bmatrix} and \begin{bmatrix}-3\\0\\1\end{bmatrix} And the image to be simply the span of the first column in the original matrix, since the first column of the rref form is the only one with a leading 1. So, span of \begin{bmatrix}1/196\\2/196\\3/196\end{bmatrix} Is this reasoning incorrect?

Emory Attikson
  • 61

1 Answers1

1

You seem to have gone wrong right off the bat in computing the projection matrix. The kernel and image that you’ve come up with for it are correct, but that matrix doesn’t represent orthogonal projection onto the given plane, nor, for that matter, any projection whatsoever. The matrix obviously has rank 1, but since the image of the projection must be the given plane, it should have rank 2. As well, one of the defining properties of a projection $P$ is that $P^2=P$, but that doesn’t hold for the matrix in your question.

It looks like you might’ve computed the projection onto the plane’s normal and then scaled incorrectly: the common denominator $196$ is equal to $14^2$, i.e., the square of the square of the length of the normal that can be extracted from the plane’s equation. Working out the projection onto the normal (correctly) is a good start to finding the matrix that projects onto the plane, but you then have to subtract this projection from the original vector, i.e., subtract the resulting matrix from the identity.

That said, there’s no need to construct any matrix whatsoever to solve this problem. By definition, the image of a projection onto a plane is the plane itself. The kernel or any orthogonal projection is the orthogonal complement of the image, which in this case is the set of vectors normal to the plane.

amd
  • 53,693