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I am interested in finding the t2' with the help of t1, t1' and t2. Actually I am using some projection matrix T (this will be used to project point x and y which is clear in the Image attached) on x and y to get the points t1 and t2. Again I have to use the matrix T on x' and y' to find the t1' and t2'. But I don't want to exactly use the matrix T to do this just to avoid the lots of multiplication.

Is there any way to find t2' with the help of t1' and the result followed in the case of t1 and t1'. 

T*x = t1
T*y = t2
t*x' = t1'

I am attaching picture for better visibility of the question.

enter image description here

Olivier Moindrot
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hawkeye
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1 Answers1

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If there is a linear relationship between the variables, you can use that same relationship after applying T because T, being a linear transformation, will preserve those relationships.

So, let's assume that the relationship you have verifies y - x = y' - x'. This assumption is consistent with the one you wrote in the comment because there y - x = (0, 1) and y'- x' = (0, 1) too.

So, applying T to both sides of the equation we have: T(y - x) = T(y' - x'). Now, using that T preserves linear combinations (i.e., T(u + sv) = T(u) + sT(v) for all vectors u,v and scalars s), we get: Ty - Tx = Ty' - Tx'. With your variable names we get: t2 - t1 = t2' - t1'. Solving for t2' leads to t2' = t2 - t1 + t1'.

If the actual relationship between x, y, x' and y' happens to be different but still linear you can apply the same reasoning to it.

Leandro Caniglia
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  • T(x+1,y+2,z) = T(x,y,x) + T(1,2,0) Is it true ? If not what is the true linear combination of T(x+1,y+2,z). – hawkeye Jan 06 '15 at 12:22
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    It is true. Since `(x+1, y+2, z) = (x, y, z) + (1, 2, 0)` you have `T(x+1, y+2, z) = T(x, y, z) + T(1, 2, 0)`. Of course, provided `T` is a *linear transformation* (e.g., T performs a matrix multiplication). I assume this is your case because you said that T was a projection and projections are linear. – Leandro Caniglia Jan 06 '15 at 12:56