Orthogonal Controls

Here's the picture of xA x for another 2 × 2 matrix A. 

The picture shows you what A does to the standard coordinate grid.  You see that A sends the nice orthogonal (right angled) grid on the left to a non-orthogonal grid on the right.

Move the vector u to see what A does to other grids. Find a location for the vector u that makes θ = π/2.

(Q5) Estimate the angle that u makes with the x-axis when θ = π/2.

Let θ_1 be the angle θ when u = (1) 0 and let θ_2 be the angle θ when u = (0) 1.  

(Q6) What's the exact sum of these two angles?  θ_1 + θ_2 = ___.

(Q7) It turns out that for any matrix A, θ_1 + θ_2 = π.  So for any A, if θ_1<π/2, then θ_2>___ and if θ_1>π/2, then θ_2< ___.  So either way there must be a u between (1) 0and (0) 1 where θ = ___.

Here's the picture of xA x for another 2 × 2 matrix A.

The graphic also shows an orthogonal grid which remains orthogonal under the transformation.  Having an orthogonal grid in the plot on the right makes it easier to solve A x = y .

Control the amounts of u and v to find an x = c_1u + c_2v that A sends to y .

(Q8) The solution to A x = y is x = ___u + ___v .

You can also solve A x = y by eye without moving x .
    On the right you see y = 2 (A u) + 3 (A v).
    This tells you that the x you need is x = 2 u + 3v .

Here's the picture of xA x for another 2 × 2 matrix A.

Try to solve A x = y by eye.  

(Q9) How much A u and A v add up to y ?  Answer: y = __ (A u) + __ (A v).

(Q10) The solution to A x = y is x = __ u + __ v .

Verify your answers to the problems above by moving x to x = -3u + 2v .FormBox[Cell[], TraditionalForm]

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Copyright © 2007 Todd Will
Created by Mathematica  (April 15, 2007)