Consider the matrix Y = ⎝ ⎜ ⎛ i 0 2 i 0 1 0 0 0 0 0 0 − 2 i 0 − i ⎠ ⎟ ⎞
If the sum of the eigenvalues can be represented as a + b i and the product of the eigenvalues can be represented as c + d i .
Find the value of a b + d − c
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a 400 points problem!
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Yea! Just find trace of matrix and its determinant and you get 400 point! This must be probably easiest 400 points problem!
sum of eigen values is Tr(Y) and product of eigen values is determinant of Y. Enjoy!!
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Sum of Eigen values is the trace of matrix, thus, a + i b = i + 1 0 0 0 0 0 − i = 1 0 0 0 0 0 + 0 i
Product of eigen values is determinant. d e t ( Y ) = ∣ ∣ ∣ ∣ ∣ ∣ i 0 2 i 0 1 0 0 0 0 0 0 − 2 i 0 − i ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ − 3 i 0 0 0 1 0 0 0 0 0 0 − 2 i 0 − i ∣ ∣ ∣ ∣ ∣ ∣
Now determinant of diagonal matrix is just the product of diagonal elements, thus, d e t ( Y ) = − 3 0 0 0 0 0 + 0 i
Substituting values, 1 0 0 0 0 0 − ( − 3 0 0 0 0 0 ) = 3