7_Kronecker

# 7_Kronecker - EE 211A Fall Quarter, 2011 Instructor: John...

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Unformatted text preview: EE 211A Fall Quarter, 2011 Instructor: John Villasenor Digital Image Processing I Handout 7 Separability and Kronecker Products: An Example with N = 2 Let v and u be 2 × 2 matrices, where ⎡ྎv(0,0) v(0,1)⎤ྏ ⎡ྎu (0,0) u (0,1)⎤ྏ v = ⎢ྎ ⎥ྏ, u = ⎢ྎ u (1,0) u (1,1) ⎥ྏ ⎣ྏ v(1,0) v(1,1) ⎦ྏ ⎣ྏ ⎦ྏ 1 1 and v(k , l ) = ∑ m=0 ∑ n=0 u(m, n)akl (m, n). If v ʹȃ and u ʹȃ are row-ordered versions of v and u , then we can write ⎡ྎv(0,0)⎤ྏ ⎡ྎa00 (0,0) ⎢ྎ v(0,1) ⎥ྏ ⎢ྎ a (0,0) ⎥ྏ = ⎢ྎ 01 v ʹȃ = ⎢ྎ ⎢ྎ v(1,0) ⎥ྏ ⎢ྎ a10 (0,0) ⎢ྎ ⎥ྏ ⎢ྎ ⎣ྏ v(1,1) ⎦ྏ ⎣ྏ a11 (0,0) a00 (0,1) a00 (1,0) a00 (1,1)⎤ྏ ⎡ྎu (0,0)⎤ྏ a01 (0,1) a01 (1,0) a01 (1,1) ⎥ྏ ⎢ྎ u (0,1) ⎥ྏ ⎥ྏ ⋅ ⎢ྎ ⎥ྏ = au ʹȃ. a10 (0,1) a10 (1,0) a10 (1,1) ⎥ྏ ⎢ྎ u (1,0) ⎥ྏ ⎥ྏ ⎢ྎ ⎥ྏ a11 (0,1) a11 (1,0) a11 (1,1) ⎦ྏ ⎣ྏ u (1,1) ⎦ྏ In the general case, all elements akl (m, n) are independent, and no further simplification is possible. Calculation of v ʹȃ costs N 4 = 16 multiplies. Now suppose that akl (m, n) = b(k , m)c(l , n).The above matrix product is: ⎡ྎv(0, 0) ⎤ྏ ⎡ྎb(0, 0)c(0, 0) b(0, 0)c(0,1) b(0,1)c(0, 0) b(0,1)c(0,1) ⎤ྏ ⎡ྎu (0, 0) ⎤ྏ ⎢ྎ v(0,1) ⎥ྏ ⎢ྎ b(0, 0)c(1, 0) b(0, 0)c(1,1) b(0,1)c(1, 0) b(0,1)c(1,1) ⎥ྏ ⎢ྎ u (0,1) ⎥ྏ ⎥ྏ = ⎢ྎ ⎥ྏ ⋅ ⎢ྎ ⎥ྏ = auʹȃ. vʹȃ = ⎢ྎ ⎢ྎ v(1, 0) ⎥ྏ ⎢ྎ b(1, 0)c(0, 0) b(1, 0)c(0,1) b(1,1)c(0, 0) b(1,1)c (0,1) ⎥ྏ ⎢ྎ u (1, 0) ⎥ྏ ⎢ྎ ⎥ྏ ⎢ྎ ⎥ྏ ⎢ྎ ⎥ྏ ⎣ྏ v(1,1) ⎦ྏ ⎣ྏ b(1, 0)c(1, 0) b(1, 0)c(1,1) b(1,1)c(1, 0) b(1,1)c(1,1) ⎦ྏ ⎣ྏ u (1,1) ⎦ྏ The matrix a can be expressed as a Kronecker product a = b ⊗ c. ⎡ྎb(0,0)c b(0,1)c ⎤ྏ ⇒ a = ⎢ྎ ⎥ྏ. ⎣ྏ b(1,0)c b(1,1)c ⎦ྏ In addition, v can be expressed: ⎡ྎb(0,0) b(0,1)⎤ྏ ⎡ྎu (0,0) u (0,1)⎤ྏ ⎡ྎc(0,0) c(1,0)⎤ྏ T v = bu c = ⎢ྎ ⎥ྏ ⋅ ⎢ྎ ⎥ྏ ⋅ ⎢ྎ ⎥ྏ ⎣ྏ b(1,0) b(1,1) ⎦ྏ ⎣ྏ u (1,0) u (1,1) ⎦ྏ ⎣ྏ c(0,1) c(1,1) ⎦ྏ 1 Verify: ⎡ྎb(0,0) b(0,1)⎤ྏ ⎡ྎc(0,0)u (0,0) + c(0,1)u (0,1) c(1,0)u (0,0) + c(1,1)u (0,1)⎤ྏ v = ⎢ྎ ⎥ྏ ⋅ ⎢ྎ ⎥ྏ ⎣ྏ b(1,0) b(1,1) ⎦ྏ ⎣ྏ c(0,0)u (1,0) + c(0,1)u (1,1) c(1,0)u (1,0) + c(1,1)u (1,1) ⎦ྏ ⇒ v(0,0) = b(0,0)[c(0,0)u (0,0) + c(0,1)u (0,1)] + b(0,1)[c(0,0)u (1,0) + c(0,1)u (1,1)], and so on. On an element by element basis, we can see that if akl (m, n) is separable, where a = b ⊗ c, T then v ʹȃ = a u ʹȃ can be expressed as = c . v b u N 2 ×1 N 2 ×N 2 N 2 ×1 N ×N The total multiplies for the separable case is 2 N 3 . 2 N×N N ×N N ×N ...
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## This note was uploaded on 12/27/2011 for the course EE211A 211A taught by Professor Villasenor during the Spring '11 term at UCLA.

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