Kreyszig 2.6 Linear Operators

Problem 1. Show that the operators in sections 2.6-2, 2.6-3, and 2.6-4 are linear.

Solution:

To show that the operators in sections 2.6-2, 2.6-3, and 2.6-4 are linear, we need to verify that each operator satisfies the two linearity conditions for all vectors \(x, y\) in the domain and all scalars \(a, b\) in the field over which the vector space is defined:

  1. \(T(x + y) = T(x) + T(y)\) (additivity)

  2. \(T(ax) = aT(x)\) (homogeneity)

Let's consider each operator in turn:

2.6-2 Identity Operator \(I_x\): The identity operator \(I_x\) on a vector space \(X\) is defined by \(I_x(x) = x\) for all \(x \in X\).

  • For additivity, consider two vectors \(x, y \in X\). We need to show that \(I_x(x + y) = I_x(x) + I_x(y)\). Indeed, \(I_x(x + y) = x + y = I_x(x) + I_x(y)\).

  • For homogeneity, consider a scalar \(a\) and a vector \(x \in X\). We need to show that \(I_x(ax) = aI_x(x)\). Indeed, \(I_x(ax) = ax = aI_x(x)\).

2.6-3 Zero Operator \(0_x\): The zero operator \(0_x\) on a vector space \(X\) to another vector space \(Y\) is defined by \(0_x(x) = 0\) for all \(x \in X\), where \(0\) is the zero vector in \(Y\).

  • For additivity, consider two vectors \(x, y \in X\). We have \(0_x(x + y) = 0 = 0 + 0 = 0_x(x) + 0_x(y)\).

  • For homogeneity, for any scalar \(a\) and vector \(x in X\), \(0_x(ax) = 0 = a \cdot 0 = a0_x(x)\).

2.6-4 Differentiation Operator \(D\): Let \(X\) be the vector space of all polynomials on \([a, b]\). The differentiation operator \(D\) is defined by \(D(T(x)) = T'(x)\), where \(T'\) denotes differentiation with respect to \(x\).

  • For additivity, let \(x(t)\) and \(y(t)\) be polynomials in \(X\). Then \(D(x(t) + y(t)) = (x + y)'(t) = x'(t) + y'((t) = D(x(t)) + D(y(t))\).

  • For homogeneity, let \(a\) be a scalar and \(x(t)\) be a polynomial in \(X\). Then \(D(a \cdot x(t)) = (a \cdot x)'(t) = a \cdot x'(t) = a \cdot D(x(t))\).

In all cases, the operators satisfy the linearity conditions, hence they are indeed linear operators.

\(\blacksquare\)

Problem 2. Show that the operators \(T_1, T_2, T_3\), and \(T_4\) from \(\mathbb{R}^2\) into \(\mathbb{R}^2\) defined by

  • \(T_1(\xi_1, \xi_2) = (\xi_1, 0)\)

  • \(T_2(\xi_1, \xi_2) = (0, \xi_2)\)

  • \(T_3(\xi_1, \xi_2) = (\xi_2, \xi_1)\)

  • \(T_4(\xi_1, \xi_2) = (\gamma\xi_1, \gamma\xi_2)\)

respectively, are linear, and interpret these operators geometrically.

Solution: To demonstrate the linearity of operators \(T_1, T_2, T_3\), and \(T_4\), we must verify that each operator satisfies the following properties for all vectors \(\xi, \eta \in \mathbb{R}^2\) and all scalars \(a \in \mathbb{R}\):

  1. Additivity: \(T(\xi + \eta) = T(\xi) + T(\eta)\)

  2. Homogeneity: \(T(a\xi) = aT(\xi)\)

For \(T_1\):

  • Additivity: \(T_1((\xi_1 + \eta_1, \xi_2 + \eta_2)) = (\xi_1 + \eta_1, 0) = (\xi_1, 0) + (\eta_1, 0) = T_1(\xi_1, \xi_2) + T_1(\eta_1, \eta_2)\)

  • Homogeneity: \(T_1(a(\xi_1, \xi_2)) = (a\xi_1, 0) = a(\xi_1, 0) = aT_1(\xi_1, \xi_2)\)

For \(T_2\), additivity and homogeneity can be shown similarly, with \(T_2\) projecting any vector onto the y-axis.

For \(T_3\):

  • Additivity: \(T_3((\xi_1 + \eta_1, \xi_2 + \eta_2)) = (\xi_2 + \eta_2, \xi_1 + \eta_1) = (\xi_2, \xi_1) + (\eta_2, \eta_1) = T_3(\xi_1, \xi_2) + T_3(\eta_1, \eta_2)\)

  • Homogeneity: \(T_3(a(\xi_1, \xi_2)) = (a\xi_2, a\xi_1) = a(\xi_2, \xi_1) = aT_3(\xi_1, \xi_2)\)

For \(T_4\):

  • Additivity: \(T_4((\xi_1 + \eta_1, \xi_2 + \eta_2)) = (\gamma(\xi_1 + \eta_1), \gamma(\xi_2 + \eta_2)) = (\gamma\xi_1, \gamma\xi_2) + (\gamma\eta_1, \gamma\eta_2) = T_4(\xi_1, \xi_2) + T_4(\eta_1, \eta_2)\)

  • Homogeneity: \(T_4(a(\xi_1, \xi_2)) = (a\gamma\xi_1, a\gamma\xi_2) = a(\gamma\xi_1, \gamma\xi_2) = aT_4(\xi_1, \xi_2)\)

Geometric Interpretation:

  • \(T_1\) and \(T_2\) are projection operators onto the x-axis and y-axis respectively.

  • \(T_3\) is a reflection operator across the line \(\xi_1 = \xi_2\).

\(\blacksquare\)

Problem 3. What are the domain, range, and null space of \(T_1, T_2, T_3\) in Problem 2?

Solution:

To determine the domain, range, and null space of the linear operators \(T_1, T_2,\) and \(T_3\), we consider their definitions from Problem 2.

For Operator \(T_1\): \(T_1(\xi_1, \xi_2) = (\xi_1, 0)\)

  • Domain: The domain of \(T_1\) is the entire \(\mathbb{R}^2\).

  • Range: The range of \(T_1\) is the x-axis, given by \(\{(\xi_1, 0) \mid \xi_1 \in \mathbb{R}\}\).

  • Null Space: The null space of \(T_1\) is the set of all vectors that map to the zero vector under \(T_1\), which is \(\{(0, \xi_2) \mid \xi_2 \in \mathbb{R}\}\).

For Operator \(T_2\): \(T_2(\xi_1, \xi_2) = (0, \xi_2)\)

  • Domain: The domain of \(T_2\) is the entire \(\mathbb{R}^2\).

  • Range: The range of \(T_2\) is the y-axis, described by \(\{(0, \xi_2) \mid \xi_2 \in \mathbb{R}\}\).

  • Null Space: The null space of \(T_2\) includes all vectors that \(T_2\) maps to the zero vector, which is \(\{(\xi_1, 0) \mid \xi_1 \in \mathbb{R}\}\).

For Operator \(T_3\): \(T_3(\xi_1, \xi_2) = (\xi_2, \xi_1)\)

  • Domain: The domain of \(T_3\) is the entire \(\mathbb{R}^2\).

  • Range: The range of \(T_3\) is also \(\mathbb{R}^2\) since any vector in \(\mathbb{R}^2\) can be obtained by applying \(T_3\) to some vector in \(\mathbb{R}^2\).

  • Null Space: The null space of \(T_3\) is the set of vectors that are mapped to the zero vector, which is only the zero vector itself \(\{(0, 0)\}\).

These operators' geometric interpretations relate to their ranges and null spaces, with \(T_1\) and \(T_2\) acting as projection operators onto the x-axis and y-axis, respectively, and \(T_3\) mapping vectors onto the line \(\xi_1 = \xi_2\).

\(\blacksquare\)

Problem 4. What is the null space of \(T_4\) in Problem 2? Of \(T_1\) and \(T_2\) in 2.6-7? Of \(T\) in 2.6-4?

Solution:

Given the definitions of the operators \(T_4\), \(T_1\), and \(T_2\) from the provided images, we can find their null spaces.

For Operator \(T_4\) from 2.6-4 (Differentiation):

  • Definition: \(T_4\) is defined on the vector space \(X\) of all polynomials on \([a, b]\) by \(T_4(x(t)) = x'(t)\), where the prime denotes differentiation with respect to \(t\).

  • Null Space: The null space of the differentiation operator consists of all polynomials \(x(t)\) such that \(x'(t) = 0\). Thus, the null space of \(T_4\) is the set of all constant polynomials on \([a, b]\).

For Operator \(T_1\) from 2.6-7 (Cross product with a fixed vector):

  • Definition: \(T_1\) is defined on \(\mathbb{R}^3\) by \(T_1(\vec{x}) = \vec{x} \times \vec{a}\), where \(\vec{a}\) is a fixed vector in \(\mathbb{R}^3\).

  • Null Space: The null space of \(T_1\) includes all vectors \(\vec{x}\) such that \(\vec{x} \times \vec{a} = \vec{0}\), which are the scalar multiples of \(\vec{a}\) including the zero vector.

For Operator \(T_2\) from 2.6-7 (Dot product with a fixed vector):

  • Definition: \(T_2\) is defined on \(\mathbb{R}^3\) by \(T_2(\vec{x}) = \vec{x} \cdot \vec{a}\), where \(\vec{a} = (a_i)\) is a fixed vector in \(\mathbb{R}^3\).

  • Null Space: The null space of \(T_2\) consists of all vectors \(\vec{x}\) that are orthogonal to \(\vec{a}\), which is the orthogonal complement of the vector \(\vec{a}\) in \(\mathbb{R}^3\).

The null spaces reflect the specific transformations these operators perform on their respective vector spaces.

\(\blacksquare\)

Problem 7. Determine if the operators \(T_1\) and \(T_3\) from Problem 2 commute.

Given:

  • \(T_1(\xi_1, \xi_2) = (\xi_1, 0)\)

  • \(T_3(\xi_1, \xi_2) = (\xi_2, \xi_1)\)

Solution:

To check for commutativity, we calculate \((T_1T_3)(\xi_1, \xi_2)\) and \((T_3T_1)(\xi_1, \xi_2)\).

Applying \(T_1\) followed by \(T_3\):

  1. Apply \(T_1\) to \((\xi_1, \xi_2)\):

    \(T_1(\xi_1, \xi_2) = (\xi_1, 0)\)

  2. Then apply \(T_3\) to the result:

    \(T_3(\xi_1, 0) = (0, \xi_1)\)

Applying \(T_3\) followed by \(T_1\):

  1. Apply \(T_3\) to \((\xi_1, \xi_2)\):

    \(T_3(\xi_1, \xi_2) = (\xi_2, \xi_1)\)

  2. Then apply \(T_1\) to the result:

    \(T_1(\xi_2, \xi_1) = (\xi_2, 0)\)

Comparing Results:

  • \(T_1T_3\) yields \((0, \xi_1)\).

  • \(T_3T_1\) yields \((\xi_2, 0)\).

Since \((0, \xi_1) \neq (\xi_2, 0)\) for arbitrary \(\xi_1, \xi_2\), we conclude that \(T_1\) and \(T_3\) do not commute.

Conclusion:

The operators \(T_1\) and \(T_3\) do not satisfy the commutativity property \(T_1T_3 = T_3T_1\) for all vectors in \(\mathbb{R}^2\). Therefore, they are non-commutative.

\(\blacksquare\)

Problem 8. Represent the operators \(T_1, T_2, T_3\), and \(T_4\) from Problem 2 using \(2 \times 2\) matrices.

Given Operators:

  • \(T_1(\xi_1, \xi_2) = (\xi_1, 0)\)

  • \(T_2(\xi_1, \xi_2) = (0, \xi_2)\)

  • \(T_3(\xi_1, \xi_2) = (\xi_2, \xi_1)\)

  • \(T_4(\xi_1, \xi_2) = (\gamma\xi_1, \gamma\xi_2)\)

Matrix Representations:

  • For \(T_1\):

    The matrix representation is: \(T_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\)

  • For \(T_2\):

    The matrix representation is: \(T_2 = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\)

  • For \(T_3\):

    The matrix representation is: \(T_3 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\)

  • For \(T_4\):

    The matrix representation is: \(T_4 = \begin{bmatrix} \gamma & 0 \\ 0 & \gamma \end{bmatrix}\)

Conclusion:

Each operator from Problem 2 can be expressed as a \(2 \times 2\) matrix. These matrices transform vectors in \(\mathbb{R}^2\) by linearly scaling and/or permuting their components as specified by the operators.

\(\blacksquare\)

Problem 9. Elaborate the condition in 2.6-10(a) regarding the existence of an inverse operator, \(T^{-1}\), in the context of the null space of \(T\).

Theorem Interpretation: The theorem from section 2.6-10(a) can be restated in the context of the null space of \(T\) as follows:

  • The inverse operator \(T^{-1}\) from \(\mathcal{R}(T)\) to \(\mathcal{D}(T)\) exists if and only if the only solution to \(Tx = 0\) is the trivial solution \(x = 0\). This is equivalent to saying that the null space of \(T\), denoted \(N(T)\) or \(\text{ker}(T)\), consists solely of the zero vector.

Definitions:

  • Linear Operator: A mapping \(T: \mathcal{D}(T) \rightarrow Y\) between vector spaces \(X\) and \(Y\), adhering to additivity (\(T(x + z) = T(x) + T(z)\)) and homogeneity (\(T(\alpha x) = \alpha T(x)\)), for all \(x, z \in \mathcal{D}(T)\) and scalars \(\alpha\).

  • Inverse Operator: \(T^{-1}: \mathcal{R}(T) \rightarrow \mathcal{D}(T)\) is the reverse mapping such that \(T^{-1}(Tx) = x\) for all \(x \in \mathcal{D}(T)\) and \(T(T^{-1}y) = y\) for all \(y \in \mathcal{R}(T)\).

  • Null Space: Denoted by \(N(T)\) or \(\text{ker}(T)\), it is the set of vectors \(x \in \mathcal{D}(T)\) where \(T(x) = 0\).

In-Depth Analysis of Theorem 2.6-10(a):

This theorem posits that \(T^{-1}\) can only exist if \(Tx = 0\) strictly leads to \(x = 0\). Essentially, \(N(T)\) must be trivial—comprised solely of the zero vector. If \(N(T)\) included any non-zero vectors, \(T\) could not be injective, as it would map distinct vectors to the same point (the zero vector in \(Y\)), contravening the bijective requirement for an inverse function.

Formulating the Condition for Inverse Existence:

The existence condition for \(T^{-1}\) relative to the null space of \(T\) is that \(N(T) = \{0\}\). This reflects the injectivity of \(T\).

Examples:

  • For an Injective Operator: A matrix representation of \(T\) as \(A\) with no linearly dependent rows or columns ensures \(N(T) = \{0\}\), affirming the existence of \(T^{-1}\).

  • For a Non-Injective Operator: Should \(T\) be depicted by a matrix \(A\) containing a zero row, \(N(T)\) would be non-trivial, housing non-zero vectors, thus negating the presence of \(T^{-1}\).

Conclusion:

The theorem outlined in 2.6-10(a) underscores a pivotal tenet in linear algebra: the invertibility of a linear operator is inherently dependent on the exclusivity of the zero vector in its null space. An operator \(T\) is invertible if and only if \(N(T)\) is trivial, serving as a vital criterion for \(T\)'s injectivity.

\(\blacksquare\)

Problem 10. Determine the existence of the inverse operator \(T^{-1}\) for the differentiation operator \(T\) as defined in section 2.6-4.

Operator Definition:

The operator \(T\) defined in section 2.6-4 is the differentiation operator acting on the vector space \(X\) of all polynomials on the interval \([a, b]\). The action of \(T\) is defined by \(T(x(t)) = x'(t)\), where \(x'(t)\) denotes the derivative of \(x(t)\) with respect to \(t\).

Inverse Operator Existence Criteria:

An operator \(T\) has an inverse \(T^{-1}\) if and only if \(T\) is bijective, which means it is both injective (one-to-one) and surjective (onto).

Injectivity Analysis:

\(T\) is injective if \(T(x) = T(y)\) implies \(x = y\). For the differentiation operator, if \(x'(t) = y'(t)\) for two polynomials \(x(t)\) and \(y(t)\), then \(x(t)\) and \(y(t)\) differ by at most a constant. Hence, for \(T\) to be injective, we must restrict our attention to a subspace of \(X\) where the constant of integration is fixed, for example by setting \(x(a) = 0\) for all \(x \in X\).

Surjectivity Analysis:

\(T\) is surjective if for every function \(y(t)\) in the codomain, there exists an \(x(t)\) in the domain such that \(T(x) = y\). The differentiation operator is surjective onto the space of all differentiable functions on \([a, b]\) that can be expressed as the derivative of a polynomial, which is again the space of all polynomials on \([a, b]\).

Existence of \(T^{-1}\):

For the differentiation operator \(T\), an inverse would correspond to the integration operator. However, since integration includes a constant of integration, \(T\) is not surjective onto \(X\), and therefore, its inverse \(T^{-1}\) does not exist as a map back into \(X\).

Conclusion:

The inverse \(T^{-1}\) of the differentiation operator \(T\) as defined in 2.6-4 does not exist within the space of all polynomials on \([a, b]\) because \(T\) is not surjective onto \(X\). The differentiation operator, without additional constraints, does not have a unique inverse that maps back to the original polynomial space due to the constant of integration involved in the antiderivative.

\(\blacksquare\)

Counterexample Illustration:

Consider the differentiation operator \(T\) on the space \(X\) of polynomials over an interval \([a, b]\). We are given a function \(y(t) = e^t\) which is not a polynomial. Our goal is to find a polynomial \(x(t)\) such that \(x'(t) = y(t)\).

Attempt to Find \(x(t)\):

The inverse operation to differentiation is integration. Thus, we integrate \(y(t)\) to find \(x(t)\):

\begin{equation*} x(t) = \int y(t) dt = \int e^t dt = e^t + C \end{equation*}

where \(C\) represents the constant of integration.

Analysis:

The result of the integration, \(x(t) = e^t + C\), is not a polynomial. Hence, it does not reside in the space \(X\) of polynomials on \([a, b]\). This shows that \(y(t)\), a non-polynomial function, does not have an antiderivative that is a polynomial in \(X\).

Conclusion:

Since the integration maps \(y(t) = e^t\) to a function outside the space of polynomials, it demonstrates that the differentiation operator \(T\) is not surjective over the space \(X\). Consequently, \(T\) does not have an inverse \(T^{-1}\) that maps back to \(X\). The function \(y(t) = e^t\) serves as a counterexample, indicating that there are functions in the codomain of \(T\) for which no polynomial in \(X\) is a pre-image, thereby confirming the non-existence of an inverse operator \(T^{-1}\) that returns to the original polynomial space \(X\).

Problem 11. Verify the linearity of the operator \(T: X \rightarrow X\) defined by \(T(x) = bx\) for a fixed \(2 \times 2\) complex matrix \(b\), and determine the condition for the existence of the inverse operator \(T^{-1}\).

Proof of Linearity: To demonstrate that \(T\) is linear, it must satisfy additivity and homogeneity.

  • Additivity:

For any \(2 \times 2\) matrices \(x\) and \(y\) in \(X\):

\begin{equation*} T(x + y) = b(x + y) = bx + by = T(x) + T(y) \end{equation*}

  • Homogeneity:

For any complex scalar \(\alpha\) and matrix \(x\) in \(X\):

\begin{equation*} T(\alpha x) = b(\alpha x) = \alpha bx = \alpha T(x) \end{equation*}

Since \(T\) satisfies both properties, we conclude that \(T\) is indeed a linear operator.

Condition for the Existence of \(T^{-1}\): The inverse operator \(T^{-1}\) exists if and only if \(T\) is bijective, which entails being both injective and surjective.

  • Injectivity:

\(T\) is injective if \(T(x) = T(y)\) implies \(x = y\). For \(T\), this condition holds if the matrix \(b\) is invertible, i.e., \(\text{det}(b) \neq 0\).

  • Surjectivity:

\(T\) is surjective if for every \(z\) in \(X\), there exists an \(x\) such that \(T(x) = z\). This is true if \(b\) is invertible, allowing us to solve \(x = b^{-1}z\) for any \(z\).

Therefore, the inverse operator \(T^{-1}\) exists if and only if the matrix \(b\) is invertible, characterized by a non-zero determinant, \(\text{det}(b) \neq 0\).

\(\blacksquare\)

Problem 12. Assess the surjectivity of the operator \(T: X \rightarrow X\), defined by \(T(x) = bx\) for a fixed matrix \(b\) in \(X\), where \(X\) is the vector space of all \(2 \times 2\) complex matrices, and \(bx\) denotes the standard product of matrices.

Surjectivity Definition:

An operator \(T\) is said to be surjective if for every matrix \(z\) in \(X\), there is a matrix \(x\) in \(X\) such that \(T(x) = z\). Formally, this means that the equation \(bx = z\) has a solution for every matrix \(z\) in \(X\).

Condition for Surjectivity:

The operator \(T\) defined by matrix multiplication is surjective if and only if the matrix \(b\) is invertible. This is equivalent to the requirement that \(\text{det}(b) \neq 0\). If \(b\) is invertible, then for every matrix \(z\) in \(X\), there exists a unique matrix \(x = b^{-1}z\) that solves the equation \(bx = z\), indicating that \(T\) maps onto the entire space \(X\).

Conclusion:

Surjectivity of the operator \(T\) hinges on the invertibility of the matrix \(b\). If \(b\) is not invertible (i.e., \(\text{det}(b) = 0\)), not all matrices \(z\) in \(X\) will have a pre-image under \(T\), and thus \(T\) will not be surjective. Conversely, if \(b\) is invertible, \(T\) is surjective, ensuring that the inverse operator \(T^{-1}\) exists and operates as \(T^{-1}(z) = b^{-1}z\) for all \(z\) in \(X\).

\(\blacksquare\)

Problem 13 Prove that if \(\{x_1, \ldots, x_n\}\) is a linearly independent set in \(\mathcal{D}(T)\), and \(T: \mathcal{D}(T) \rightarrow Y\) is a linear operator with an inverse, then the set \(\{Tx_1, \ldots, Tx_n\}\) is also linearly independent.

Proof:

Assume for contradiction that \(\{Tx_1, \ldots, Tx_n\}\) is not linearly independent. Then there exist scalars \(c_1, \ldots, c_n\), not all zero, such that:

\begin{equation*} c_1 Tx_1 + \ldots + c_n Tx_n = 0. \end{equation*}

Applying the inverse operator \(T^{-1}\) to both sides, and using the linearity of \(T^{-1}\), we obtain:

\begin{equation*} c_1 T^{-1}(Tx_1) + \ldots + c_n T^{-1}(Tx_n) = T^{-1}(0). \end{equation*}

Since \(T^{-1}T\) is the identity operator on \(\mathcal{D}(T)\), we have \(T^{-1}(Tx_i) = x_i\) for all \(i\). Knowing that the identity operator maps \(0\) to \(0\), the equation simplifies to:

\begin{equation*} c_1 x_1 + \ldots + c_n x_n = 0. \end{equation*}

This implies that \(c_1, \ldots, c_n\) must all be zero because \(\{x_1, \ldots, x_n\}\) is linearly independent, contradicting our assumption.

Conclusion:

Therefore, the set \(\{Tx_1, \ldots, Tx_n\}\) must be linearly independent, under the condition that \(T\) is invertible. This holds true due to the fundamental properties of linear transformations and their inverses in vector space theory.

\(\blacksquare\)

Problem 14. Prove that for a linear operator \(T: X \rightarrow Y\) with \(\text{dim} X = \text{dim} Y = n\), the range of \(T\), \(\mathcal{R}(T)\), is equal to \(Y\) if and only if the inverse operator \(T^{-1}\) exists.

Proof:

Forward Direction (\(\mathcal{R}(T) = Y\) implies \(T^{-1}\) exists):

If \(\mathcal{R}(T) = Y\), then \(T\) is surjective, meaning for every \(y \in Y\), there exists at least one \(x \in X\) such that \(T(x) = y\). Since \(\text{dim} X = \text{dim} Y\), \(T\) is a surjective linear map between two finite-dimensional vector spaces of equal dimension, which implies \(T\) is also injective. This is a consequence of the Rank-Nullity Theorem, which in this case implies that \(\text{nullity}(T) = 0\) because \(\text{rank}(T) = \text{dim} Y = n\) and \(\text{rank}(T) + \text{nullity}(T) = \text{dim} X\).

Being both injective and surjective, \(T\) is bijective, and therefore an inverse \(T^{-1}\) exists by definition.

Reverse Direction (\(T^{-1}\) exists implies \(\mathcal{R}(T) = Y\)):

If \(T^{-1}\) exists, then by definition, \(T\) is bijective, meaning it is both injective and surjective. The surjectivity of \(T\) immediately gives us \(\mathcal{R}(T) = Y\), because for every \(y \in Y\), the existence of \(T^{-1}\) guarantees an \(x \in X\) such that \(T(x) = y\).

Conclusion:

The range of \(T\), \(\mathcal{R}(T)\), is equal to \(Y\) if and only if \(T\) is bijective, and since \(T\) is linear, this bijectivity is equivalent to the existence of an inverse \(T^{-1}\). This holds true for finite-dimensional vector spaces \(X\) and \(Y\) of equal dimension \(n\).

\(\blacksquare\)

Detailed Explanation of the Rank-Nullity Theorem in Context:

The Rank-Nullity Theorem is pivotal in understanding the relationship between the dimensions of a linear operator's range, null space, and domain. For a linear operator \(T: X \rightarrow Y\) with \(\text{dim} X = \text{dim} Y = n\), the theorem is expressed as:

\begin{equation*} \text{rank}(T) + \text{nullity}(T) = \text{dim} X \end{equation*}

Here, \(\text{rank}(T)\) represents the dimension of the range of \(T\) (\(\mathcal{R}(T)\)), and \(\text{nullity}(T)\) signifies the dimension of the null space of \(T\) (\(N(T)\)).

Application to the Given Problem:

  1. If \(\mathcal{R}(T) = Y\):

  • The rank of \(T\) is the dimension of \(Y\), hence \(\text{rank}(T) = \text{dim} Y = n\).

  • Applying the Rank-Nullity Theorem, and knowing \(\text{dim} X = n\), we deduce that \(\text{nullity}(T) = 0\), which implies that \(T\) is injective.

  • A linear operator that is injective and surjective is bijective, indicating the existence of an inverse \(T^{-1}\).

  1. If \(T^{-1}\) Exists:

    • The existence of \(T^{-1}\) implies \(T\) is bijective. Consequently, \(T\) is injective, leading to \(\text{nullity}(T) = 0\).

    • Since \(T\) is also surjective, \(\text{rank}(T) = \text{dim} Y = n\).

    • The Rank-Nullity Theorem then confirms that \(\text{rank}(T) + \text{nullity}(T) = n\), which equals \(\text{dim} X\), thus confirming that \(\mathcal{R}(T) = Y\).

Conclusion:

The Rank-Nullity Theorem in this scenario confirms that the linear operator \(T\) is invertible if and only if it is surjective. When the domain and codomain are finite-dimensional vector spaces of equal dimension, surjectivity implies injectivity, which is integral to establishing the existence of an inverse operator \(T^{-1}\).

Problem 15. We are tasked with proving that the range \(\mathcal{R}(T)\) of a linear operator \(T\) defined on the vector space \(X\) of all real-valued functions with derivatives of all orders is the entirety of \(X\). However, we must also demonstrate that the inverse \(T^{-1}\) does not exist. This is to be contrasted with Problem 14.

Showing that \(\mathcal{R}(T)\) is all of \(X\):

Any function \(y(t)\) in \(X\) can be expressed as the derivative of another function in \(X\), as the space includes functions with derivatives of all orders. We can take an antiderivative of \(y(t)\) to find a function \(x(t)\) in \(X\) whose derivative is \(y(t)\), that is, \(x'(t) = y(t)\). Since the space of functions is closed under integration, this antiderivative \(x(t)\) is also in \(X\). This demonstrates that for every \(y(t)\) in \(X\), there exists an \(x(t)\) in \(X\) such that \(T(x(t)) = y(t)\), confirming that \(\mathcal{R}(T)\) is all of \(X\).

Showing that \(T^{-1}\) does not exist:

An inverse operator \(T^{-1}\) would map a function \(y(t)\) to a function \(x(t)\) such that \(T(x(t)) = y(t)\). However, the process of taking an antiderivative is not unique due to the constant of integration. Hence, \(T\) is not injective, as multiple functions in \(X\) can map to the same function under \(T\). Since injectivity is a necessary condition for the existence of an inverse, \(T^{-1}\) does not exist.

Comparison with Problem 14 and Comments:

Problem 14 involves a finite-dimensional vector space, where surjectivity implies invertibility. In contrast, Problem 15 deals with an infinite-dimensional vector space of smooth functions, where surjectivity is not sufficient for invertibility. The non-uniqueness of the antiderivatives prevents \(T\) from being injective, unlike in finite dimensions, where surjectivity implies injectivity due to the Rank-Nullity Theorem.

Conclusion:

Despite \(\mathcal{R}(T)\) covering all of \(X\), the non-uniqueness of the antiderivative, due to the constant of integration, prevents \(T\) from being injective, thus precluding the existence of \(T^{-1}\). This example underscores a significant distinction between linear operators in finite-dimensional spaces and those in infinite-dimensional spaces.