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Let T:V-V be the linear transformation such as Tf(t)=tft), where f(t) is a polynomial. Then, Tis singular Tis non-singular T is onto None of these?
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Let T:V-V be the linear transformation such as Tf(t)=tft), where f(t) ...
Linear transformation T:V-V:
A linear transformation is a function between two vector spaces that preserves vector addition and scalar multiplication. In this case, the linear transformation T maps a vector space V to itself, so T:V-V.
Transformation Tf(t) = tft:
The given transformation Tf(t) = tft takes a polynomial f(t) and multiplies it by the variable t. This means that for any polynomial f(t), the transformed polynomial Tf(t) will have an additional factor of t.
Singular or non-singular:
To determine whether the linear transformation T is singular or non-singular, we need to consider the null space of T. The null space of a linear transformation consists of all vectors in the domain that are mapped to the zero vector in the codomain.
Let's consider the null space of T. Suppose there exists a polynomial f(t) in V such that Tf(t) = tft = 0 for all t. This means that the polynomial f(t) must be identically zero, i.e., all its coefficients are zero. Therefore, the null space of T consists only of the zero polynomial.
Since the null space of T is trivial (i.e., contains only the zero vector), the linear transformation T is non-singular.
Onto or not onto:
To determine whether the linear transformation T is onto, we need to consider the range of T. The range of a linear transformation consists of all vectors in the codomain that are mapped to by at least one vector in the domain.
Let's consider an arbitrary polynomial g(t) in the codomain. Is there a polynomial f(t) in the domain such that Tf(t) = tft = g(t) for all t?
Since the transformation Tf(t) multiplies the polynomial f(t) by t, it is not possible for T to map a polynomial g(t) to any polynomial that has a different degree. Therefore, the range of T consists only of polynomials of the same degree as g(t).
Since the range of T is restricted to polynomials of a specific degree, the linear transformation T is not onto.
Summary:
- The linear transformation T:V-V with Tf(t) = tft is non-singular because its null space consists only of the zero polynomial.
- The linear transformation T is not onto because its range is restricted to polynomials of a specific degree.
A linear transformation is a function between two vector spaces that preserves vector addition and scalar multiplication. In this case, the linear transformation T maps a vector space V to itself, so T:V-V.
Transformation Tf(t) = tft:
The given transformation Tf(t) = tft takes a polynomial f(t) and multiplies it by the variable t. This means that for any polynomial f(t), the transformed polynomial Tf(t) will have an additional factor of t.
Singular or non-singular:
To determine whether the linear transformation T is singular or non-singular, we need to consider the null space of T. The null space of a linear transformation consists of all vectors in the domain that are mapped to the zero vector in the codomain.
Let's consider the null space of T. Suppose there exists a polynomial f(t) in V such that Tf(t) = tft = 0 for all t. This means that the polynomial f(t) must be identically zero, i.e., all its coefficients are zero. Therefore, the null space of T consists only of the zero polynomial.
Since the null space of T is trivial (i.e., contains only the zero vector), the linear transformation T is non-singular.
Onto or not onto:
To determine whether the linear transformation T is onto, we need to consider the range of T. The range of a linear transformation consists of all vectors in the codomain that are mapped to by at least one vector in the domain.
Let's consider an arbitrary polynomial g(t) in the codomain. Is there a polynomial f(t) in the domain such that Tf(t) = tft = g(t) for all t?
Since the transformation Tf(t) multiplies the polynomial f(t) by t, it is not possible for T to map a polynomial g(t) to any polynomial that has a different degree. Therefore, the range of T consists only of polynomials of the same degree as g(t).
Since the range of T is restricted to polynomials of a specific degree, the linear transformation T is not onto.
Summary:
- The linear transformation T:V-V with Tf(t) = tft is non-singular because its null space consists only of the zero polynomial.
- The linear transformation T is not onto because its range is restricted to polynomials of a specific degree.
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Let T:V-V be the linear transformation such as Tf(t)=tft), where f(t) is a polynomial. Then, Tis singular Tis non-singular T is onto None of these?
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Let T:V-V be the linear transformation such as Tf(t)=tft), where f(t) is a polynomial. Then, Tis singular Tis non-singular T is onto None of these? 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:V-V be the linear transformation such as Tf(t)=tft), where f(t) is a polynomial. Then, Tis singular Tis non-singular T is onto None of these? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let T:V-V be the linear transformation such as Tf(t)=tft), where f(t) is a polynomial. Then, Tis singular Tis non-singular T is onto None of these?.
Let T:V-V be the linear transformation such as Tf(t)=tft), where f(t) is a polynomial. Then, Tis singular Tis non-singular T is onto None of these? 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:V-V be the linear transformation such as Tf(t)=tft), where f(t) is a polynomial. Then, Tis singular Tis non-singular T is onto None of these? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let T:V-V be the linear transformation such as Tf(t)=tft), where f(t) is a polynomial. Then, Tis singular Tis non-singular T is onto None of these?.
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