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So, let's say we want the projection of vector v [2, 1] on to L (line L) L = C[2, 1]), where vector x is a vector that lies along the Line.

I know that one way to figure out the projection would be C*vectorX (C representing some scaled version of vector x).

Then, a second way to figure out the projection would be to normalize vector x. then C = (vector v * vector x) * vector x (or vector x as vector u (normalized).

But the final way to figure it out would be to see the projection as a linear transformation. Projection is Matrix A * vector v. For some reason when expressed as a Matrix / Vector product, it is less clear/intuitive for me.

The key point of confusion: I'm not sure for example how we get from a normalized version of vector x to expressing it as a Matrix (let's say Matrix A). Where then we would simply multiply it by vector v.

Hopefully this is clear.

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A fact about projection matrices i hope you would find interesting is that they are always positive semi-definite. All eigenvalues are all ones (for a vector that is in column space) or zeros (for a vector that is not in column space). I am very new here. Sorry for my answer being improper.

I got your question better after reading it again. For $$P = A(A^T A)^{-1}A$$. $$ (A^TA)$$ does the weighting job here. If A were an orthogonal matrix it would be the identity. You know that for a vector v,$$v^Tv$$ is the square of the 2-norm of the vector. This could be the relation you are looking for.

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  • $\begingroup$ I believe you understood my question but it is slightly still unclear. I have not learned eigenvalues yet. Can you show me the math (not just the form/equation) of how we get from a normalized version of vector x to expressing it as a Matrix (let's say Matrix A) and then multiplying it by vector v. $\endgroup$
    – ed8484
    Mar 24 at 14:33
  • $\begingroup$ Specifically, I'm not sure how the identity matrix fits here. $\endgroup$
    – ed8484
    Mar 24 at 16:04
  • $\begingroup$ You have a good question. But I afraid my math skills don't allow me to give the answer you would like. For identity case it is easier. Take a matrix with perpendicular columns. Find the projection matrix for each column separately. Add the matrices and you have your projection matrix. This is because you are projecting onto fully distinct 1-d spaces. You see i am not good at teaching. Don't worry if you don't understand it. Just improve your linear algebra skills. I am sure will get better with further knowledge. $\endgroup$
    – watermelon
    Mar 24 at 16:29
  • $\begingroup$ The off-diagonals entries of matrix A transpose times A, corresponds to multiplication of different column vectors of matrix A. $\endgroup$
    – watermelon
    Mar 24 at 16:33

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