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Let T be a linear operator on a finite dimensional space V and c is any scalar, then c is characteristic value of T if
- a)the operator (T - cl) is singular
- b)the operator (T - cl) is non-singular
- c)the operator (T - cl) is identity
- d)the operator (T - cl) is zero
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Let T be a linear operator on a finite dimensional space V and c is an...
Let T be a linear operator on a finite dimensional space V and c is any scalar. Then
we need to find the valid condition for which c is characteristic value of T.
Let c be a characteristic value of 7, then there exist a non-zero vector x ∈ V such that
T(x) = cx.
or equivalently (T - cl)x = 0.
Since x ≠ 0, Therefore, T - cl = 0
Taking the determinant on both sides, we get det(T- cl) = 0.
which shows that T - cl is singular.
Thus, if c is a characteristic value of T. Then T- cl is singular.
we need to find the valid condition for which c is characteristic value of T.
Let c be a characteristic value of 7, then there exist a non-zero vector x ∈ V such that
T(x) = cx.
or equivalently (T - cl)x = 0.
Since x ≠ 0, Therefore, T - cl = 0
Taking the determinant on both sides, we get det(T- cl) = 0.
which shows that T - cl is singular.
Thus, if c is a characteristic value of T. Then T- cl is singular.
Most Upvoted Answer
Let T be a linear operator on a finite dimensional space V and c is an...
Explanation:
To determine whether a scalar value c is a characteristic value (or eigenvalue) of a linear operator T on a finite-dimensional space V, we need to consider the operator (T - cl), where l is the identity operator.
Definition:
A scalar value c is a characteristic value of T if there exists a nonzero vector v in V such that (T - cl)v = 0.
Explanation of options:
a) If (T - cl) is singular, it means that there exists a nonzero vector v such that (T - cl)v = 0. This satisfies the definition of a characteristic value, so option 'a' is correct.
b) If (T - cl) is non-singular, it means that for all vectors v, (T - cl)v ≠ 0. In other words, there is no nonzero vector v that satisfies the definition of a characteristic value. Therefore, option 'b' is incorrect.
c) If (T - cl) is the identity operator, it means that (T - cl)v = v for all vectors v. This implies that c = 0, as the identity operator does not change the vector. Therefore, option 'c' is incorrect.
d) If (T - cl) is the zero operator, it means that (T - cl)v = 0 for all vectors v. This implies that c = 0, as the zero operator maps all vectors to the zero vector. Therefore, option 'd' is incorrect.
Conclusion:
Among the given options, only option 'a' satisfies the definition of a characteristic value. Therefore, the correct answer is option 'a'.
To determine whether a scalar value c is a characteristic value (or eigenvalue) of a linear operator T on a finite-dimensional space V, we need to consider the operator (T - cl), where l is the identity operator.
Definition:
A scalar value c is a characteristic value of T if there exists a nonzero vector v in V such that (T - cl)v = 0.
Explanation of options:
a) If (T - cl) is singular, it means that there exists a nonzero vector v such that (T - cl)v = 0. This satisfies the definition of a characteristic value, so option 'a' is correct.
b) If (T - cl) is non-singular, it means that for all vectors v, (T - cl)v ≠ 0. In other words, there is no nonzero vector v that satisfies the definition of a characteristic value. Therefore, option 'b' is incorrect.
c) If (T - cl) is the identity operator, it means that (T - cl)v = v for all vectors v. This implies that c = 0, as the identity operator does not change the vector. Therefore, option 'c' is incorrect.
d) If (T - cl) is the zero operator, it means that (T - cl)v = 0 for all vectors v. This implies that c = 0, as the zero operator maps all vectors to the zero vector. Therefore, option 'd' is incorrect.
Conclusion:
Among the given options, only option 'a' satisfies the definition of a characteristic value. Therefore, the correct answer is option 'a'.
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Let T be a linear operator on a finite dimensional space V and c is any scalar, then c is characteristic value of T ifa)the operator (T - cl) is singularb)the operator (T - cl) is non-singularc)the operator (T - cl) is identityd)the operator (T- cl) is zeroCorrect answer is option 'A'. Can you explain this answer?
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Let T be a linear operator on a finite dimensional space V and c is any scalar, then c is characteristic value of T ifa)the operator (T - cl) is singularb)the operator (T - cl) is non-singularc)the operator (T - cl) is identityd)the operator (T- cl) is zeroCorrect answer is option 'A'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let T be a linear operator on a finite dimensional space V and c is any scalar, then c is characteristic value of T ifa)the operator (T - cl) is singularb)the operator (T - cl) is non-singularc)the operator (T - cl) is identityd)the operator (T- cl) is zeroCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let T be a linear operator on a finite dimensional space V and c is any scalar, then c is characteristic value of T ifa)the operator (T - cl) is singularb)the operator (T - cl) is non-singularc)the operator (T - cl) is identityd)the operator (T- cl) is zeroCorrect answer is option 'A'. Can you explain this answer?.
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