Linear Transformations

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Okay, let’s dive into a detailed tutorial on linear transformations, covering the definition, examples, matrix representation, kernel, image, and properties like injectivity and surjectivity. This is a fundamental topic in linear algebra.

6. Linear Transformations: A Detailed Tutorial

A linear transformation is a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. Let $V$ and $W$ be vector spaces over the same field $F$ (which is often the set of real numbers $\mathbb{R}$ or complex numbers $\mathbb{C}$).

Definition: A function $T: V \rightarrow W$ is a linear transformation if it satisfies the following two properties for all vectors $\mathbf{u}, \mathbf{v} \in V$ and all scalars $c \in F$:

Additivity: $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$
Homogeneity (Scalar Multiplication): $T(c\mathbf{u}) = cT(\mathbf{u})$

These two properties can be combined into a single condition: $T(c\mathbf{u} + d\mathbf{v}) = cT(\mathbf{u}) + dT(\mathbf{v})$ for all $\mathbf{u}, \mathbf{v} \in V$ and all scalars $c, d \in F$.

Key Consequences of Linearity:

$T(\mathbf{0}_V) = \mathbf{0}_W$, where $\mathbf{0}_V$ is the zero vector in $V$ and $\mathbf{0}_W$ is the zero vector in $W$. (Proof: $T(\mathbf{0}_V) = T(0\mathbf{u}) = 0T(\mathbf{u}) = \mathbf{0}_W$ for any $\mathbf{u} \in V$).
$T(-\mathbf{u}) = -T(\mathbf{u})$. (Proof: $T(-\mathbf{u}) = T((-1)\mathbf{u}) = (-1)T(\mathbf{u}) = -T(\mathbf{u})$).

Examples of Linear Transformations

Let’s look at some examples to understand the concept better.

Example 1: Transformation in $\mathbb{R}^2$ (Rotation)

Consider a transformation $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ that rotates every vector in $\mathbb{R}^2$ counterclockwise by an angle $\theta$. If $\mathbf{v} = \begin{pmatrix} x \ y \end{pmatrix}$, then $T(\mathbf{v}) = \begin{pmatrix} x\cos\theta - y\sin\theta \ x\sin\theta + y\cos\theta \end{pmatrix}$.

Let’s check if this is a linear transformation. Let $\mathbf{u} = \begin{pmatrix} x_1 \ y_1 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} x_2 \ y_2 \end{pmatrix}$, and let $c$ be a scalar.

Additivity: $T(\mathbf{u} + \mathbf{v}) = T\left(\begin{pmatrix} x_1 + x_2 \ y_1 + y_2 \end{pmatrix}\right) = \begin{pmatrix} (x_1 + x_2)\cos\theta - (y_1 + y_2)\sin\theta \ (x_1 + x_2)\sin\theta + (y_1 + y_2)\cos\theta \end{pmatrix}$ $= \begin{pmatrix} (x_1\cos\theta - y_1\sin\theta) + (x_2\cos\theta - y_2\sin\theta) \ (x_1\sin\theta + y_1\cos\theta) + (x_2\sin\theta + y_2\cos\theta) \end{pmatrix} = T(\mathbf{u}) + T(\mathbf{v})$
Homogeneity: $T(c\mathbf{u}) = T\left(\begin{pmatrix} cx_1 \ cy_1 \end{pmatrix}\right) = \begin{pmatrix} (cx_1)\cos\theta - (cy_1)\sin\theta \ (cx_1)\sin\theta + (cy_1)\cos\theta \end{pmatrix}$ $= \begin{pmatrix} c(x_1\cos\theta - y_1\sin\theta) \ c(x_1\sin\theta + y_1\cos\theta) \end{pmatrix} = c \begin{pmatrix} x_1\cos\theta - y_1\sin\theta \ x_1\sin\theta + y_1\cos\theta \end{pmatrix} = cT(\mathbf{u})$

Thus, rotation is a linear transformation.

Example 2: Transformation in $\mathbb{R}^2$ (Projection onto the x-axis)

Consider $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ defined by $T\left(\begin{pmatrix} x \ y \end{pmatrix}\right) = \begin{pmatrix} x \ 0 \end{pmatrix}$. This transformation projects every vector onto the x-axis. You can verify that this is also a linear transformation using the definition.

Example 3: Transformation in $\mathbb{R}^2$ (Translation)

Consider $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ defined by $T\left(\begin{pmatrix} x \ y \end{pmatrix}\right) = \begin{pmatrix} x + a \ y + b \end{pmatrix}$, where $a$ and $b$ are constants (not both zero).

Let’s check the first property: $T(\mathbf{u} + \mathbf{v}) = T\left(\begin{pmatrix} x_1 + x_2 \ y_1 + y_2 \end{pmatrix}\right) = \begin{pmatrix} (x_1 + x_2) + a \ (y_1 + y_2) + b \end{pmatrix} = \begin{pmatrix} x_1 + a + x_2 + a \ y_1 + b + y_2 + b \end{pmatrix}$ $T(\mathbf{u}) + T(\mathbf{v}) = \begin{pmatrix} x_1 + a \ y_1 + b \end{pmatrix} + \begin{pmatrix} x_2 + a \ y_2 + b \end{pmatrix} = \begin{pmatrix} x_1 + a + x_2 + a \ y_1 + b + y_2 + b \end{pmatrix}$

This seems correct, let’s recheck. $T(\mathbf{u} + \mathbf{v}) = \begin{pmatrix} x_1 + x_2 + a \ y_1 + y_2 + b \end{pmatrix}$ $T(\mathbf{u}) + T(\mathbf{v}) = \begin{pmatrix} x_1 + a \ y_1 + b \end{pmatrix} + \begin{pmatrix} x_2 + a \ y_2 + b \end{pmatrix} = \begin{pmatrix} x_1 + x_2 + 2a \ y_1 + y_2 + 2b \end{pmatrix}$

If $a \neq 0$ or $b \neq 0$, then $T(\mathbf{u} + \mathbf{v}) \neq T(\mathbf{u}) + T(\mathbf{v})$. Also, $T(\mathbf{0}) = T\left(\begin{pmatrix} 0 \ 0 \end{pmatrix}\right) = \begin{pmatrix} a \ b \end{pmatrix} \neq \begin{pmatrix} 0 \ 0 \end{pmatrix}$ if $a$ or $b$ is non-zero. Therefore, translation is generally not a linear transformation.

Example 4: Transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ defined by a matrix

Let $A$ be an $m \times n$ matrix. The transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ defined by $T(\mathbf{x}) = A\mathbf{x}$ (where $\mathbf{x}$ is an $n \times 1$ column vector) is a linear transformation. This is because matrix multiplication satisfies the properties of additivity and homogeneity: $A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v}$ $A(c\mathbf{u}) = c(A\mathbf{u})$

Example 5: Differentiation of Polynomials

Let $P_n$ be the vector space of polynomials of degree at most $n$. The transformation $D: P_n \rightarrow P_{n-1}$ defined by $D(p(x)) = p’(x)$ (the derivative of $p(x)$) is a linear transformation. If $p(x)$ and $q(x)$ are polynomials and $c$ is a scalar: $D(p(x) + q(x)) = (p(x) + q(x))’ = p’(x) + q’(x) = D(p(x)) + D(q(x))$ $D(cp(x)) = (cp(x))’ = cp’(x) = cD(p(x))$

Example 6: Integration of Functions

Let $C[a, b]$ be the vector space of continuous functions on the interval $[a, b]$. The transformation $I: C[a, b] \rightarrow \mathbb{R}$ defined by $I(f) = \int_a^b f(x) dx$ is a linear transformation. $I(f + g) = \int_a^b (f(x) + g(x)) dx = \int_a^b f(x) dx + \int_a^b g(x) dx = I(f) + I(g)$ $I(cf) = \int_a^b cf(x) dx = c \int_a^b f(x) dx = cI(f)$

Matrix Representation of a Linear Transformation

A fundamental result in linear algebra is that any linear transformation between finite-dimensional vector spaces can be represented by a matrix.

Let $V$ be an $n$-dimensional vector space with basis $\mathcal{B} = {\mathbf{b}_1, \mathbf{b}_2, …, \mathbf{b}_n}$ and $W$ be an $m$-dimensional vector space with basis $\mathcal{C} = {\mathbf{c}_1, \mathbf{c}_2, …, \mathbf{c}_m}$. Let $T: V \rightarrow W$ be a linear transformation.

To find the matrix representation of $T$ with respect to the bases $\mathcal{B}$ and $\mathcal{C}$ (denoted as $[T]_{\mathcal{B}}^{\mathcal{C}}$ or simply $[T]$ when the bases are understood), we need to determine the images of the basis vectors of $V$ under $T$ and express these images as linear combinations of the basis vectors of $W$.

For each $\mathbf{b}j \in \mathcal{B}$, $T(\mathbf{b}_j)$ is a vector in $W$, so it can be uniquely written as a linear combination of the basis vectors in $\mathcal{C}$: $T(\mathbf{b}_j) = a{1j}\mathbf{c}1 + a{2j}\mathbf{c}2 + … + a{mj}\mathbf{c}m = \sum{i=1}^{m} a_{ij}\mathbf{c}_i$

The coefficients of this linear combination form the $j$-th column of the matrix representation $[T]{\mathcal{B}}^{\mathcal{C}}$: $[T]{\mathcal{B}}^{\mathcal{C}} = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}
a_{21} & a_{22} & \cdots & a_{2n}
\vdots & \vdots & \ddots & \vdots
a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix}$

If $\mathbf{v} \in V$ has a coordinate vector $[\mathbf{v}]{\mathcal{B}} = \begin{pmatrix} x_1 \ x_2 \ \vdots \ x_n \end{pmatrix}$ with respect to the basis $\mathcal{B}$, then the coordinate vector of $T(\mathbf{v})$ with respect to the basis $\mathcal{C}$, denoted as $[T(\mathbf{v})]{\mathcal{C}}$, is given by the matrix product: $[T(\mathbf{v})]{\mathcal{C}} = [T]{\mathcal{B}}^{\mathcal{C}} [\mathbf{v}]_{\mathcal{B}}$

Example: Matrix Representation

Let $T: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be a linear transformation defined by $T\left(\begin{pmatrix} x \ y \end{pmatrix}\right) = \begin{pmatrix} x + y \ 2x - y \ 3y \end{pmatrix}$. Let the standard bases for $\mathbb{R}^2$ and $\mathbb{R}^3$ be $\mathcal{B} = {\mathbf{e}_1, \mathbf{e}_2} = \left{ \begin{pmatrix} 1 \ 0 \end{pmatrix}, \begin{pmatrix} 0 \ 1 \end{pmatrix} \right}$ and $\mathcal{C} = {\mathbf{f}_1, \mathbf{f}_2, \mathbf{f}_3} = \left{ \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}, \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}, \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} \right}$.

We find the images of the basis vectors of $\mathbb{R}^2$ under $T$: $T(\mathbf{e}_1) = T\left(\begin{pmatrix} 1 \ 0 \end{pmatrix}\right) = \begin{pmatrix} 1 + 0 \ 2(1) - 0 \ 3(0) \end{pmatrix} = \begin{pmatrix} 1 \ 2 \ 0 \end{pmatrix} = 1\mathbf{f}_1 + 2\mathbf{f}_2 + 0\mathbf{f}_3$ $T(\mathbf{e}_2) = T\left(\begin{pmatrix} 0 \ 1 \end{pmatrix}\right) = \begin{pmatrix} 0 + 1 \ 2(0) - 1 \ 3(1) \end{pmatrix} = \begin{pmatrix} 1 \ -1 \ 3 \end{pmatrix} = 1\mathbf{f}_1 - 1\mathbf{f}_2 + 3\mathbf{f}_3$

The matrix representation of $T$ with respect to the standard bases is: $[T]_{\mathcal{B}}^{\mathcal{C}} = \begin{pmatrix} 1 & 1 \ 2 & -1 \ 0 & 3 \end{pmatrix}$

Now, let’s take an arbitrary vector $\mathbf{v} = \begin{pmatrix} x \ y \end{pmatrix}$ in $\mathbb{R}^2$. Its coordinate vector with respect to $\mathcal{B}$ is $[\mathbf{v}]{\mathcal{B}} = \begin{pmatrix} x \ y \end{pmatrix}$. $[T(\mathbf{v})]{\mathcal{C}} = [T]{\mathcal{B}}^{\mathcal{C}} [\mathbf{v}]{\mathcal{B}} = \begin{pmatrix} 1 & 1 \ 2 & -1 \ 0 & 3 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} x + y \ 2x - y \ 3y \end{pmatrix}$ The coordinate vector with respect to $\mathcal{C}$ is indeed $\begin{pmatrix} x + y \ 2x - y \ 3y \end{pmatrix}$, which corresponds to the vector $T(\mathbf{v})$ we defined earlier.

Kernel (Null Space) of a Linear Transformation

The kernel (or null space) of a linear transformation $T: V \rightarrow W$, denoted by $\text{ker}(T)$ or $N(T)$, is the set of all vectors in $V$ that are mapped to the zero vector in $W$: $\text{ker}(T) = {\mathbf{v} \in V \mid T(\mathbf{v}) = \mathbf{0}_W}$

Properties of the Kernel:

The kernel of a linear transformation is always a subspace of the domain $V$.
- Contains the zero vector: $T(\mathbf{0}_V) = \mathbf{0}_W$, so $\mathbf{0}_V \in \text{ker}(T)$.
- Closed under addition: If $\mathbf{u}, \mathbf{v} \in \text{ker}(T)$, then $T(\mathbf{u}) = \mathbf{0}_W$ and $T(\mathbf{v}) = \mathbf{0}_W$. Thus, $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) = \mathbf{0}_W + \mathbf{0}_W = \mathbf{0}_W$, so $\mathbf{u} + \mathbf{v} \in \text{ker}(T)$.
- Closed under scalar multiplication: If $\mathbf{u} \in \text{ker}(T)$ and $c$ is a scalar, then $T(c\mathbf{u}) = cT(\mathbf{u}) = c\mathbf{0}_W = \mathbf{0}_W$, so $c\mathbf{u} \in \text{ker}(T)$.

Example: Finding the Kernel

Consider the linear transformation $T: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ defined by $T\left(\begin{pmatrix} x \ y \end{pmatrix}\right) = \begin{pmatrix} x + y \ 2x - y \ 3y \end{pmatrix}$. To find the kernel, we need to solve $T\left(\begin{pmatrix} x \ y \end{pmatrix}\right) = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}$: $\begin{pmatrix} x + y \ 2x - y \ 3y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}$

This gives the system of linear equations: $x + y = 0$ $2x - y = 0$ $3y = 0$

From the third equation, $y = 0$. Substituting this into the first equation, $x + 0 = 0$, so $x = 0$. The second equation is also satisfied: $2(0) - 0 = 0$. The only solution is $x = 0$ and $y = 0$. Therefore, $\text{ker}(T) = \left{ \begin{pmatrix} 0 \ 0 \end{pmatrix} \right}$, which is the zero subspace of $\mathbb{R}^2$.

Image (Range) of a Linear Transformation

The image (or range) of a linear transformation $T: V \rightarrow W$, denoted by $\text{im}(T)$ or $R(T)$, is the set of all vectors in $W$ that are the image of some vector in $V$: $\text{im}(T) = {\mathbf{w} \in W \mid \mathbf{w} = T(\mathbf{v}) \text{ for some } \mathbf{v} \in V}$

Properties of the Image:

The image of a linear transformation is always a subspace of the codomain $W$.
- Contains the zero vector: $T(\mathbf{0}_V) = \mathbf{0}_W$, so $\mathbf{0}_W \in \text{im}(T)$.
- Closed under addition: If $\mathbf{w}_1, \mathbf{w}_2 \in \text{im}(T)$, then there exist $\mathbf{v}_1, \mathbf{v}_2 \in V$ such that $T(\mathbf{v}_1) = \mathbf{w}_1$ and $T(\mathbf{v}_2) = \mathbf{w}_2$. Then $\mathbf{w}_1 + \mathbf{w}_2 = T(\mathbf{v}_1) + T(\mathbf{v}_2) = T(\mathbf{v}_1 + \mathbf{v}_2)$. Since $\mathbf{v}_1 + \mathbf{v}_2 \in V$, $\mathbf{w}_1 + \mathbf{w}_2 \in \text{im}(T)$.
- Closed under scalar multiplication: If $\mathbf{w} \in \text{im}(T)$ and $c$ is a scalar, then there exists $\mathbf{v} \in V$ such that $T(\mathbf{v}) = \mathbf{w}$. Then $c\mathbf{w} = cT(\mathbf{v}) = T(c\mathbf{v})$. Since $c\mathbf{v} \in V$, $c\mathbf{w} \in \text{im}(T)$.
If $V$ is finite-dimensional with a basis ${\mathbf{b}_1, \mathbf{b}_2, …, \mathbf{b}_n}$, then the image of $T$ is the span of the images of the basis vectors: $\text{im}(T) = \text{span}{T(\mathbf{b}_1), T(\mathbf{b}_2), …, T(\mathbf{b}_n)}$

Example: Finding the Image

Consider the linear transformation $T: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ defined by $T\left(\begin{pmatrix} x \ y \end{pmatrix}\right) = \begin{pmatrix} x + y \ 2x - y \ 3y \end{pmatrix}$. Using the standard basis of $\mathbb{R}^2$, ${\begin{pmatrix} 1 \ 0 \end{pmatrix}, \begin{pmatrix} 0 \ 1 \end{pmatrix}}$, we have: $T\left(\begin{pmatrix} 1 \ 0 \end{pmatrix}\right) = \begin{pmatrix} 1 \ 2 \ 0 \end{pmatrix}$ $T\left(\begin{pmatrix} 0 \ 1 \end{pmatrix}\right) = \begin{pmatrix} 1 \ -1 \ 3 \end{pmatrix}$

The image of $T$ is the span of these two vectors: $\text{im}(T) = \text{span}\left{ \begin{pmatrix} 1 \ 2 \ 0 \end{pmatrix}, \begin{pmatrix} 1 \ -1 \ 3 \end{pmatrix} \right}$ This is a subspace of $\mathbb{R}^3$. Since these two vectors are linearly independent (one is not a scalar multiple of the other), the image is a plane passing through the origin in $\mathbb{R}^3$.

Relationship between Matrix Representation and Image:

If $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is given by $T(\mathbf{x}) = A\mathbf{x}$, where $A$ is an $m \times n$ matrix, then the image of $T$ is the column space of the matrix $A$, i.e., the span of the columns of $A$.

Properties of Linear Transformations: Injectivity and Surjectivity

Injectivity (One-to-one)

A linear transformation $T: V \rightarrow W$ is injective (or one-to-one) if for every $\mathbf{w} \in W$, there is at most one $\mathbf{v} \in V$ such that $T(\mathbf{v}) = \mathbf{w}$. Equivalently, if $T(\mathbf{u}) = T(\mathbf{v})$, then $\mathbf{u} = \mathbf{v}$.

Theorem: A linear transformation $T: V \rightarrow W$ is injective if and only if its kernel is the zero subspace, i.e., $\text{ker}(T) = {\mathbf{0}_V}$.

Proof:

($\Rightarrow$) Assume $T$ is injective. If $\mathbf{v} \in \text{ker}(T)$, then $T(\mathbf{v}) = \mathbf{0}_W$. We also know that $T(\mathbf{0}_V) = \mathbf{0}_W$. Since $T$ is injective and $T(\mathbf{v}) = T(\mathbf{0}_V)$, it must be that $\mathbf{v} = \mathbf{0}_V$. Thus, $\text{ker}(T) = {\mathbf{0}_V}$.
($\Leftarrow$) Assume $\text{ker}(T) = {\mathbf{0}_V}$. Suppose $T(\mathbf{u}) = T(\mathbf{v})$ for some $\mathbf{u}, \mathbf{v} \in V$. Then $T(\mathbf{u}) - T(\mathbf{v}) = \mathbf{0}_W$. By linearity, $T(\mathbf{u} - \mathbf{v}) = \mathbf{0}_W$. This means that $\mathbf{u} - \mathbf{v} \in \text{ker}(T)$. Since $\text{ker}(T) = {\mathbf{0}_V}$, we have $\mathbf{u} - \mathbf{v} = \mathbf{0}_V$, which implies $\mathbf{u} = \mathbf{v}$. Therefore, $T$ is injective.

Example: Checking for Injectivity

Surjectivity (Onto)

A linear transformation $T: V \rightarrow W$ is surjective (or onto) if for every $\mathbf{w} \in W$, there exists at least one $\mathbf{v} \in V$ such that $T(\mathbf{v}) = \mathbf{w}$. In other words, the image of $T$ is equal to the codomain $W$, i.e., $\text{im}(T) = W$.

Theorem (Rank-Nullity Theorem): For a linear transformation $T: V \rightarrow W$, where $V$ is a finite-dimensional vector space, $\text{dim}(\text{ker}(T)) + \text{dim}(\text{im}(T)) = \text{dim}(V)$ Here, $\text{dim}(\text{ker}(T))$ is called the nullity of $T$, and $\text{dim}(\text{im}(T))$ is called the rank of $T$.

Relationship between Surjectivity and Dimensions:

If $T: V \rightarrow W$ is a linear transformation between finite-dimensional vector spaces, then:

If $\text{dim}(V) < \text{dim}(W)$, $T$ cannot be surjective. (By Rank-Nullity Theorem, $\text{dim}(\text{im}(T)) \leq \text{dim}(V) < \text{dim}(W)$).
If $\text{dim}(V) > \text{dim}(W)$, $T$ cannot be injective (because $\text{dim}(\text{ker}(T)) = \text{dim}(V) - \text{dim}(\text{im}(T)) \geq \text{dim}(V) - \text{dim}(W) > 0$, so the kernel is not just the zero vector).
If $\text{dim}(V) = \text{dim}(W)$, then $T$ is injective if and only if it is surjective. (If $T$ is injective, $\text{dim}(\text{ker}(T)) = 0$, so $\text{dim}(\text{im}(T)) = \text{dim}(V) = \text{dim}(W)$, meaning $\text{im}(T) = W$, so $T$ is surjective. Conversely, if $T$ is surjective, $\text{dim}(\text{im}(T)) = \text{dim}(W) = \text{dim}(V)$, so $\text{dim}(\text{ker}(T)) = 0$, meaning $T$ is injective).

Example: Checking for Surjectivity

For the transformation $T: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ defined by $T\left(\begin{pmatrix} x \ y \end{pmatrix}\right) = \begin{pmatrix} x + y \ 2x - y \ 3y \end{pmatrix}$, we found that $\text{im}(T) = \text{span}\left{ \begin{pmatrix} 1 \ 2 \ 0 \end{pmatrix}, \begin{pmatrix} 1 \ -1 \ 3 \end{pmatrix} \right}$. The dimension of the image (rank of $T$) is 2, as the two spanning vectors are linearly independent. The dimension of the domain is $\text{dim}(\mathbb{R}^2) = 2$. By the Rank-Nullity Theorem, $\text{dim}(\text{ker}(T)) + \text{dim}(\text{im}(T)) = 2$, so $\text{dim}(\text{ker}(T)) + 2 = 2$, which gives $\text{dim}(\text{ker}(T)) = 0$, consistent with our earlier finding.

Since the dimension of the image (2) is less than the dimension of the codomain (3), the image is a proper subspace of the codomain, and thus the transformation is not surjective. There are vectors in $\mathbb{R}^3$ that are not in the image of $T$. For example, $\begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix}$ cannot be expressed as a linear combination of $\begin{pmatrix} 1 \ 2 \ 0 \end{pmatrix}$ and $\begin{pmatrix} 1 \ -1 \ 3 \end{pmatrix}$.

Isomorphism

A linear transformation $T: V \rightarrow W$ is called an isomorphism if it is both injective and surjective. If there exists an isomorphism between two vector spaces $V$ and $W$, then $V$ and $W$ are said to be isomorphic, denoted by $V \cong W$. Isomorphic vector spaces have essentially the same algebraic structure.

For finite-dimensional vector spaces, two vector spaces are isomorphic if and only if they have the same dimension. If $\text{dim}(V) = \text{dim}(W) = n$, then a linear transformation $T: V \rightarrow W$ is an isomorphism if and only if it is either injective or surjective.

Summary of Key Concepts:

Linear Transformation: A function between vector spaces that preserves vector addition and scalar multiplication.
Matrix Representation: A way to represent a linear transformation between finite-dimensional vector spaces as a matrix.
Kernel (Null Space): The set of vectors in the domain that are mapped to the zero vector in the codomain. It is a subspace of the domain.
Image (Range): The set of vectors in the codomain that are the image of some vector in the domain. It is a subspace of the codomain.
Injectivity (One-to-one): A linear transformation is injective if and only if its kernel is the zero subspace.
Surjectivity (Onto): A linear transformation is surjective if and only if its image is equal to the codomain.
Rank-Nullity Theorem: For a linear transformation $T: V \rightarrow W$ with finite-dimensional $V$, $\text{dim}(\text{ker}(T)) + \text{dim}(\text{im}(T)) = \text{dim}(V)$.
Isomorphism: A linear transformation that is both injective and surjective.

This detailed tutorial should provide a solid foundation for understanding linear transformations. Remember to practice with various examples to solidify your understanding of these concepts. Good luck with your studies!

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