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For a 3 x 3 real matrix. Let C(A) denotes the set of the real characteristic roots of A. Suppose C(B) = C(B-1) fora non singular matrix B with no repeated eigenvalues then
- a)either 1 or - 1 must be an eigenvalue of B
- b)I = B2
- c)1 and - 1 are the only possible real eigenvalues of B.
- d)- 1 must be an eigenvalues of B
Correct answer is option 'A'. Can you explain this answer?
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For a 3 x 3 real matrix. Let C(A) denotes the set of the real characte...
Explanation:
To prove that option 'A' is correct, we need to show that either 1 or -1 must be an eigenvalue of matrix B.
Let's assume that B has no repeated eigenvalues. We know that the characteristic polynomial of a matrix is given by det(B - λI), where λ represents the eigenvalue and I is the identity matrix.
Proof:
1. If B is non-singular, then 0 cannot be an eigenvalue of B. This is because if 0 were an eigenvalue, then det(B - 0I) = det(B) = 0, which contradicts the assumption that B is non-singular.
2. Since B has no repeated eigenvalues, the characteristic polynomial of B will have degree 3 and will be of the form (λ - λ1)(λ - λ2)(λ - λ3), where λ1, λ2, and λ3 are the eigenvalues of B.
3. From the given information, we have C(B) = C(B-1), which means that the eigenvalues of B and B-1 are the same.
4. Let's assume that 1 is not an eigenvalue of B. Then 1 cannot be an eigenvalue of B-1 either, as eigenvalues are preserved under inverse operation. Therefore, the characteristic polynomial of B-1 will also be of the form (λ - λ1)(λ - λ2)(λ - λ3), where none of the eigenvalues is 1.
5. Now, consider the equation B(B-1)x = 0, where x is a non-zero vector. Since B is non-singular, it does not have 0 as an eigenvalue. Therefore, the equation Bx = B-1x = 0 has a non-zero solution. This implies that Bx and B-1x are linearly dependent, which means that Bx = k(B-1)x for some scalar k.
6. Multiplying both sides of the equation by B-1, we get B(B-1)x = kB-1x. Since k is a scalar and x is a non-zero vector, we can divide both sides of the equation by x to obtain B(B-1) = kB-1.
7. Now, let's consider the characteristic polynomial of B(B-1). Since B(B-1) is a 3x3 matrix, its characteristic polynomial will also have degree 3. Let's denote this polynomial as p(λ) = (λ - μ1)(λ - μ2)(λ - μ3), where μ1, μ2, and μ3 are the eigenvalues of B(B-1).
8. From the equation B(B-1) = kB-1, we can conclude that p(λ) = k(λ - λ1)(λ - λ2)(λ - λ3). Comparing this with the characteristic polynomial of B(B-1), we can see that μ1 = kλ1, μ2 = kλ2, and μ3 = kλ3.
9. Since the eigenvalues of B(B-1) are the same as those of B, we can equate the coefficients of the corresponding terms in the characteristic polynomials of B(B-1) and B. This gives us μ1 + μ2 + μ3 = λ1 + λ2 +
To prove that option 'A' is correct, we need to show that either 1 or -1 must be an eigenvalue of matrix B.
Let's assume that B has no repeated eigenvalues. We know that the characteristic polynomial of a matrix is given by det(B - λI), where λ represents the eigenvalue and I is the identity matrix.
Proof:
1. If B is non-singular, then 0 cannot be an eigenvalue of B. This is because if 0 were an eigenvalue, then det(B - 0I) = det(B) = 0, which contradicts the assumption that B is non-singular.
2. Since B has no repeated eigenvalues, the characteristic polynomial of B will have degree 3 and will be of the form (λ - λ1)(λ - λ2)(λ - λ3), where λ1, λ2, and λ3 are the eigenvalues of B.
3. From the given information, we have C(B) = C(B-1), which means that the eigenvalues of B and B-1 are the same.
4. Let's assume that 1 is not an eigenvalue of B. Then 1 cannot be an eigenvalue of B-1 either, as eigenvalues are preserved under inverse operation. Therefore, the characteristic polynomial of B-1 will also be of the form (λ - λ1)(λ - λ2)(λ - λ3), where none of the eigenvalues is 1.
5. Now, consider the equation B(B-1)x = 0, where x is a non-zero vector. Since B is non-singular, it does not have 0 as an eigenvalue. Therefore, the equation Bx = B-1x = 0 has a non-zero solution. This implies that Bx and B-1x are linearly dependent, which means that Bx = k(B-1)x for some scalar k.
6. Multiplying both sides of the equation by B-1, we get B(B-1)x = kB-1x. Since k is a scalar and x is a non-zero vector, we can divide both sides of the equation by x to obtain B(B-1) = kB-1.
7. Now, let's consider the characteristic polynomial of B(B-1). Since B(B-1) is a 3x3 matrix, its characteristic polynomial will also have degree 3. Let's denote this polynomial as p(λ) = (λ - μ1)(λ - μ2)(λ - μ3), where μ1, μ2, and μ3 are the eigenvalues of B(B-1).
8. From the equation B(B-1) = kB-1, we can conclude that p(λ) = k(λ - λ1)(λ - λ2)(λ - λ3). Comparing this with the characteristic polynomial of B(B-1), we can see that μ1 = kλ1, μ2 = kλ2, and μ3 = kλ3.
9. Since the eigenvalues of B(B-1) are the same as those of B, we can equate the coefficients of the corresponding terms in the characteristic polynomials of B(B-1) and B. This gives us μ1 + μ2 + μ3 = λ1 + λ2 +
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For a 3 x 3 real matrix. Let C(A) denotes the set of the real characteristic roots of A. Suppose C(B) = C(B-1) fora non singular matrix B with no repeated eigenvalues thena)either 1 or - 1 must be an eigenvalue of Bb)I = B2c)1 and - 1 are the only possible real eigenvalues of B.d)- 1 must be an eigenvalues of BCorrect answer is option 'A'. Can you explain this answer?
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