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Let M be an n × n matrix with real entries such that M3 = I. Suppose that Mv ≠ v for any nonzero vector v. Then which of the following statements is/are TRUE ?
- a)M has real eigenvalues
- b)M + M–1 has real eigenvalues
- c)n is divisible by 2
- d)n is divisible by 3
Correct answer is option 'B,C'. Can you explain this answer?
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Let M be an n × n matrix with real entries such that M3 = I. Supp...
Statement: Let M be an n x n matrix with real entries such that M^3 = I. Suppose that Mv = v for any nonzero vector v.
To prove: The following statements are true:
a) M has real eigenvalues.
b) M - M^(-1) has real eigenvalues.
c) n is divisible by 2.
d) n is divisible by 3.
Proof:
Statement a: M has real eigenvalues.
Let λ be an eigenvalue of M, and let v be the corresponding eigenvector. We have Mv = λv.
Taking the cube of both sides, we get:
M^3v = λ^3v.
Since M^3 = I, we have I v = λ^3v.
This implies λ^3v - v = 0.
Factoring out v, we get (λ^3 - 1)v = 0.
Since v is nonzero, we must have λ^3 - 1 = 0.
This equation has real solutions for λ, which means M has real eigenvalues.
Statement b: M - M^(-1) has real eigenvalues.
Let λ be an eigenvalue of M - M^(-1), and let v be the corresponding eigenvector. We have (M - M^(-1))v = λv.
Rearranging the equation, we get Mv - M^(-1)v = λv.
Multiplying both sides by M, we get M^2v - v = λMv.
Using the given condition Mv = v, we have M^2v - v = λv.
This can be rearranged as M^2v - (1 + λ)v = 0.
Factoring out v, we get (M^2 - (1 + λ)I)v = 0.
Since v is nonzero, we must have M^2 - (1 + λ)I = 0.
This equation has real solutions for λ, which means M - M^(-1) has real eigenvalues.
Statement c: n is divisible by 2.
Let's assume n is not divisible by 2. Then n can be written as n = 2k + 1 for some positive integer k.
Since M^3 = I, we have (M^2)(M) = I.
Taking the determinant of both sides, we get det(M^2)det(M) = det(I).
Since det(I) = 1, we have (det(M))^3 = 1.
This implies det(M) = 1.
But the determinant of a matrix with real eigenvalues is the product of its eigenvalues, and the product of real numbers is always real.
Therefore, if M has real eigenvalues, we must have det(M) = 1, which contradicts our assumption that n is odd.
Hence, n must be divisible by 2.
Statement d: n is divisible by 3.
Since M^3 = I, the characteristic polynomial of M is p(λ) = λ^3 - 1.
By the fundamental theorem of algebra, this polynomial has exactly 3 complex roots.
If n is not divisible by 3, then the matrix M cannot have exactly 3 eigenvalues
To prove: The following statements are true:
a) M has real eigenvalues.
b) M - M^(-1) has real eigenvalues.
c) n is divisible by 2.
d) n is divisible by 3.
Proof:
Statement a: M has real eigenvalues.
Let λ be an eigenvalue of M, and let v be the corresponding eigenvector. We have Mv = λv.
Taking the cube of both sides, we get:
M^3v = λ^3v.
Since M^3 = I, we have I v = λ^3v.
This implies λ^3v - v = 0.
Factoring out v, we get (λ^3 - 1)v = 0.
Since v is nonzero, we must have λ^3 - 1 = 0.
This equation has real solutions for λ, which means M has real eigenvalues.
Statement b: M - M^(-1) has real eigenvalues.
Let λ be an eigenvalue of M - M^(-1), and let v be the corresponding eigenvector. We have (M - M^(-1))v = λv.
Rearranging the equation, we get Mv - M^(-1)v = λv.
Multiplying both sides by M, we get M^2v - v = λMv.
Using the given condition Mv = v, we have M^2v - v = λv.
This can be rearranged as M^2v - (1 + λ)v = 0.
Factoring out v, we get (M^2 - (1 + λ)I)v = 0.
Since v is nonzero, we must have M^2 - (1 + λ)I = 0.
This equation has real solutions for λ, which means M - M^(-1) has real eigenvalues.
Statement c: n is divisible by 2.
Let's assume n is not divisible by 2. Then n can be written as n = 2k + 1 for some positive integer k.
Since M^3 = I, we have (M^2)(M) = I.
Taking the determinant of both sides, we get det(M^2)det(M) = det(I).
Since det(I) = 1, we have (det(M))^3 = 1.
This implies det(M) = 1.
But the determinant of a matrix with real eigenvalues is the product of its eigenvalues, and the product of real numbers is always real.
Therefore, if M has real eigenvalues, we must have det(M) = 1, which contradicts our assumption that n is odd.
Hence, n must be divisible by 2.
Statement d: n is divisible by 3.
Since M^3 = I, the characteristic polynomial of M is p(λ) = λ^3 - 1.
By the fundamental theorem of algebra, this polynomial has exactly 3 complex roots.
If n is not divisible by 3, then the matrix M cannot have exactly 3 eigenvalues
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Let M be an n × n matrix with real entries such that M3 = I. Suppose that Mv ≠ v for any nonzero vector v. Then which of the following statements is/are TRUE ?a)M has real eigenvaluesb)M + M–1 has real eigenvaluesc)n is divisible by 2d)n is divisible by 3Correct answer is option 'B,C'. Can you explain this answer?
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