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.

$\blacksquare$

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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 $\xi_1 = \xi_2$.

$\blacksquare$

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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 $\xi_1 = \xi_2$.

$\blacksquare$

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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.

$\blacksquare$

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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.

$\blacksquare$

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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.

$\blacksquare$

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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.

$\blacksquare$

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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 $x(t)$:

$x(t) = \int y(t) dt = \int e^t dt = e^t + C$

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 $X$.

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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 $X$:

• Homogeneity:

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

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, $\text{det}(b) \neq 0$.

$\blacksquare$

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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 $X$.

$\blacksquare$

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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:

$c_1 Tx_1 + \ldots + c_n Tx_n = 0.$

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

$c_1 T^{-1}(Tx_1) + \ldots + c_n T^{-1}(Tx_n) = T^{-1}(0).$

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:

$c_1 x_1 + \ldots + c_n x_n = 0.$

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.

$\blacksquare$

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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:

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 $T^{-1}$.

2. 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 $T^{-1}$.

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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.