Research Space for Linear Algebra & Discrete Mathematics
和平常一樣, 就對 A 做列運算可得 R =1 exp((2/3)πi) exp((4/3)πi)0 1-exp(2πi) exp((2/3)πi)-exp((8/3)πi)0 0 1-exp(2πi)因為 exp(2πi) = cos(2π)+i*sin(2π) = 1且 exp((8/3)*πi) = exp((2/3)πi), 所以 R = 1 exp((2/3)πi) exp((4/3)πi)0 0 00 0 0=> rank(A) = 1
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1 則留言:
和平常一樣, 就對 A 做列運算可得 R =
1 exp((2/3)πi) exp((4/3)πi)
0 1-exp(2πi) exp((2/3)πi)-exp((8/3)πi)
0 0 1-exp(2πi)
因為 exp(2πi) = cos(2π)+i*sin(2π) = 1
且 exp((8/3)*πi) = exp((2/3)πi),
所以 R =
1 exp((2/3)πi) exp((4/3)πi)
0 0 0
0 0 0
=> rank(A) = 1
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