Homework # 2 Solutions
... Solution: For any vectors u, v ∈ Rn , T (u + v) = A(u + v) + b T (u) + T (v) = Au + b + Av + b = A(u + v) + 2b 6= T (u + v) unless b = 0. Therefore, T is not a linear transformation when b 6= 0. Alternate solution: T (0) = A(0) + b = b 6= 0, so T is not a linear transformation when b 6= 0. 36. Let T ...
... Solution: For any vectors u, v ∈ Rn , T (u + v) = A(u + v) + b T (u) + T (v) = Au + b + Av + b = A(u + v) + 2b 6= T (u + v) unless b = 0. Therefore, T is not a linear transformation when b 6= 0. Alternate solution: T (0) = A(0) + b = b 6= 0, so T is not a linear transformation when b 6= 0. 36. Let T ...
1 Quantization of the Electromagnetic Field
... In either case taking the commutator with the number operator is like taking the time derivative and dividing by the energy of the mode. If a commutator vanishes it vanishes identically for each mode separately. Clearly this cannot happen by just taking a time derivative so neither the electric or t ...
... In either case taking the commutator with the number operator is like taking the time derivative and dividing by the energy of the mode. If a commutator vanishes it vanishes identically for each mode separately. Clearly this cannot happen by just taking a time derivative so neither the electric or t ...
Solution
... Thus if λ 6= ρ we must have hu, vi = 0. In other words, u and v are orthogonal. Problem 2: Show that (F ⊕∞ )∗ ∼ = F ×∞ . Conclude that it is not the case that V and V ∗ are always isomorphic. Solution: Suppose that f ∈ (F ⊕∞ )∗ . Then f is uniquely determined by its value on a basis of F ⊕∞ . Taking ...
... Thus if λ 6= ρ we must have hu, vi = 0. In other words, u and v are orthogonal. Problem 2: Show that (F ⊕∞ )∗ ∼ = F ×∞ . Conclude that it is not the case that V and V ∗ are always isomorphic. Solution: Suppose that f ∈ (F ⊕∞ )∗ . Then f is uniquely determined by its value on a basis of F ⊕∞ . Taking ...
Matrices and Pictures
... There are a large number of pixels at the lower half of the integers, so these values are doubled. The upper half the integers can’t be doubled because the range would be larger than the domain of the colors, so they are set equal to one. Show Slide 9. (Slope) Edge Sharpening: Convolution Convolutio ...
... There are a large number of pixels at the lower half of the integers, so these values are doubled. The upper half the integers can’t be doubled because the range would be larger than the domain of the colors, so they are set equal to one. Show Slide 9. (Slope) Edge Sharpening: Convolution Convolutio ...
Revision 08/01/06
... Four is the product of the two ways from C to E and the two ways from E to G. Similarly, there are 11 ways from C to H: one from C to D to H, four from C to E to H (the product of two from C to E and two ways from E to H), and six from C to F to H (two ways from C to F and three ways from F to H). ...
... Four is the product of the two ways from C to E and the two ways from E to G. Similarly, there are 11 ways from C to H: one from C to D to H, four from C to E to H (the product of two from C to E and two ways from E to H), and six from C to F to H (two ways from C to F and three ways from F to H). ...
