Use the function to find the image of v and the preimage of w. T(v1, v2) = 2 2 v1 − 2 2 v2, v1 + v2, 2v1 − v2 , v = (1, 1), w = (−7 2 , −2, −22)

Question

Use the function to fine the image of v and the preimage of w

T(v1, v2) = (((sqrt(2)/2)v1) - ((sqrt(2)/2)v2), v1+v2, 2v1-v2) , v = (1,1), w = (-7 sqrt(2), -2, -22)

Answer 1

By the definitions of T(v₁,v₂) and v, we find that the image of v is T(1,1) = (sqrt(2)/2*1 - sqrt(2)/2*1,1 + 1, 2*1 - 1) = (0,2,1). By the definitions of T(v₁,v₂) and v, we find that the preimage of w must satisfy (sqrt(2)/2*v₁ - sqrt(2)/2*v₂,v₁ + v₂, 2v₁ - v₂) = (-7sqrt(2), -2, -22). This further implies that sqrt(2)/2*v₁ - sqrt(2)/2*v₂ = -7sqrt(2), v₁ + v₂ = -2, and 2v₁ - v₂ = -22. The first of these three equations further implies that v₁ - v₂ = -7sqrt(2)*(2/sqrt(2)) = -14. Adding the equations v₁ - v₂ = -14 and v₁ + v₂ = -2 together gives us 2v₁ = -14 + -1 = -16, which implies that v₁ = -16/2 = -8. Subtracting the equation v₁ - v₂ = -14 from the equation v₁ + v₂ = -2 gives us 2v₂ = -2 - (-14) = -2 + 14 = 12, which implies that v₂ = 12/2 = 6. Since 2*(-8) - 6 = -16 - 6 = -22, we can conclude that the preimage of w is (-8,6).