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How does this work for matrices which don't provide an X and a Y term in both elements, such as a multiple of the identity matrix (kI)? You are left with kX equal to X' (and kY=Y') which means that when X' and Y' are substituted into y=mx + c, you get Y'=m(X') + c and hence kY=m(kx) + c and therefore kmx + kc = kmx +c which suggests m can be anything when c=0.
If both y= 5/2x + c and y= -x are both invariant lines and the translated coordinate will remain on those lines, will the intersections between these invariant lines be invariant points, as that seems to me the only way the translated point can satisfy both invariant lines? Or am I missing something?
Good question I hadn't thought about before. I would assume the answer to this is yes, as otherwise there would be points on an invariant line that do not remain on that line. So wherever two invariant lines intersect, that point must be an invariant point.
how does this work when there is an identity matrix or an identity 'like' matrix e.g. swap out the 1 in the identity matrix for -3 as surely it will just get cancelled each time suggesting it has no invarient lines despite this not being the case?
Why are we assume the coefficient of both x and c are 0 to solve for m. How do you know that it’s not 3x-4c=0 for example. Because that could still equal 0
I was watching a video about eigenvectors, and they seemed similar to invariant lines. Are they the same idea, and if so could you use eigenvectors to find the equation of these lines?
Eigenvectors point in the direction of invariant lines. Eigenvalues are effectively the scale factor that tells you how far points have moved along those lines. So if the eigenvalue for a particular eigenvector is 1, then it is a line of invariant points.
We assumed that the invariant lines are of the form y = mx + c, so we can replace y with mx + c. Once we have y' and x', we know that y' = mx' + c as the line that is outputted by the transformation must be the same line as what we started with (so same gradient m and y-intercept c).
wouldnt it be easier as this is under the linear transformations topic to just have y= mx and neglect the c as it should map to the origin if treated as a transformation. I can see this being useful if it it is not but for this topic ??
