Question

Verify that the kernel of a transformation \(\mathbf{T}: \mathcal{V} \rightarrow \mathcal{W}\) is a subspace of \(V\), and that \(\mathbf{T}(\mathcal{V})\) is a subspace of \(\mathcal{W}\).

Short answer

After demonstrating that the kernel and image satisfy the properties required of a subspace (including the zero vector, and being closed under addition and scalar multiplication), we can conclude that the kernel of a transformation \(\mathbf{T}\) is a subspace of \(\mathcal{V}\) and the image of transformation \(\mathbf{T}\) is a subspace of \(\mathcal{W}\).

Step-by-step solution

Kernel as a Subspace

Let's first consider the kernel, which is the set of all vectors that get mapped to the zero vector in \(\mathcal{W}\). We need to prove three main properties for the kernel: (i) The zero vector (\(0\)) is included in the Kernel. Applying \(\mathbf{T}(0) = 0\), the zero vector must be a part of Kernel. (ii) It is closed under addition. If we have two vectors \(v_1, v_2 \in \mathcal{V}\) such that \(\mathbf{T}(v_1) = 0\) and \(\mathbf{T}(v_2) = 0\), then \(\mathbf{T}(v_1 + v_2) = \mathbf{T}(v_1) + \mathbf{T}(v_2) = 0+0 = 0\). Hence, \(v_1 + v_2\) is also in the kernel of \(\mathbf{T}\). (iii) It is also closed under scalar multiplication. For any scalar \(c\), if \(v\) is in the kernel, then \(\mathbf{T}(c v) = c \mathbf{T}(v) = c .0 = 0\). Therefore, kernel is a subspace of \(\mathcal{V}\).

Image as a Subspace

Now, let's analyze the image, which is the set of all vectors in \(\mathcal{W}\) that \(\mathbf{T}\) maps to. Again, we need to verify three properties: (i) \(\mathbf{T}(0)\) is in the image. (ii) The image of \(\mathbf{T}\) is closed under vector addition. If \(w_1\) and \(w_2\) are in the image, then there exist \(v_1\), \(v_2\) in \(\mathcal{V}\) such that \(\mathbf{T}(v_1)=w_1\) and \(\mathbf{T}(v_2)=w_2\). So, \(\mathbf{T}(v_1 + v_2) = \mathbf{T}(v_1) + \mathbf{T}(v_2) = w_1 + w_2\), hence the image is closed under addition. (iii) The image of \(\mathbf{T}\) is closed under scalar multiplication. If \(w\) is in the image, then there is a vector \(v\) in \(\mathcal{V}\) such that \(\mathbf{T}(v)=w\). So for any scalar \(c\), \(\mathbf{T}(c v) = c \mathbf{T}(v) = c w\), hence the image is closed under scalar multiplication. Therefore, the image is a subspace of \(\mathcal{W}\).

Key concepts

Kernel of a Transformation

The kernel of a transformation, often denoted as \(\ker(T)\) for a linear transformation \(\mathbf{T}: \mathcal{V} \rightarrow \mathcal{W}\), is the set of all vectors in \(\mathcal{V}\) that are mapped to the zero vector in \(\mathcal{W}\). Understanding this concept is fundamental in linear algebra.

• Contains the Zero Vector: For a subspace, the zero vector must be part of it. Since the transformation of zero, \(\mathbf{T}(0)\), naturally yields zero, it implies that the zero vector is in \(\ker(T)\).
• Closure under Addition: If two vectors \(v_1\) and \(v_2\) satisfy \(\mathbf{T}(v_1) = 0\) and \(\mathbf{T}(v_2) = 0\), their sum \(v_1 + v_2\) will also be in the kernel, since \(\mathbf{T}(v_1 + v_2) = 0 + 0 = 0\).
• Closure under Scalar Multiplication: If a vector \(v\) is in the kernel, then any scalar multiple of it will map to the zero vector as well, because \(\mathbf{T}(c \cdot v) = c \cdot \mathbf{T}(v) = 0\).

These properties ensure that the kernel forms a subspace of \(\mathcal{V}\). Understanding these fundamental properties helps in deeper studies like solving systems of linear equations more effectively.

Subspace Verification

A subspace is a set of vectors that is closed under addition and scalar multiplication, and includes the zero vector. Verifying whether a particular set of vectors is a subspace involves checking these specific requirements one by one.

• First, ensure that the zero vector is present. Without it, the set fails to be a subspace.
• Verify closure under addition. This means adding any two vectors from the set should result in another vector that is also part of the set.
• Confirm closure under scalar multiplication. When any vector in the set is multiplied by a scalar, the resulting vector must reside within the set.

Subspace verification is a crucial skill in understanding vector spaces and involves applying these principles methodically. By applying these steps, you can confidently identify sets that are true subspaces, thus facilitating accurate mathematical modeling.

Vector Space Properties

Vector spaces are foundational structures in algebra, characterized by specific properties that dictate their behavior. These properties must be satisfied for any set of vectors to be considered a vector space.

• Vector Addition and Scalar Multiplication: These two operations must be defined and consistent in the space.
• Exists a Zero Vector: Every vector space contains a zero vector that, when added to any vector in the space, yields the same vector.
• Commutative and Associative Laws: Vector addition is both commutative, i.e., \(u + v = v + u\), and associative, i.e., \(u + (v + w) = (u + v) + w\).
• Distributive Properties: These hold for scalar multiplications, ensuring consistency and linkage between operations.
  For example, \(c(v + w) = cv + cw\) and \((c + d)v = cv + dv\).

These axioms form the backbone of linear algebra, ensuring that we can predictably manipulate and combine vectors, providing a robust framework for exploring mathematical constructs.

Image of a Transformation

The image, or range, of a linear transformation \(\mathbf{T}: \mathcal{V} \rightarrow \mathcal{W}\), denoted by \(\text{Im}(T)\), consists of all vectors in \(\mathcal{W}\) that can be expressed as \(\mathbf{T}(v)\) for some vector \(v\) in \(\mathcal{V}\).

• Includes \(\mathbf{T}(0)\): The zero vector in \(\mathcal{V}\), when transformed, results in the zero vector in \(\mathcal{W}\), thus \(\mathbf{T}(0) \in \text{Im}(T)\).
• Closed under Addition: If vectors \(w_1\) and \(w_2\) are in the image of \(T\), their sum \(w_1 + w_2\) remains in the image because they can be traced back to the sum of preimages, namely vectors \(v_1\) and \(v_2\) in \(\mathcal{V}\).
• Closed under Scalar Multiplication: Any vector \(w\) in the image, when multiplied by a scalar, \(c\), will also remain in the image. This is because \(w\) can be written as \(\mathbf{T}(v)\) for some \(v\), and \(c \cdot w = \mathbf{T}(c \cdot v)\).

These properties ensure that the image of \(T\) behaves just like a subspace within \(\mathcal{W}\), enabling extensive applications in diverse scenarios such as differential equations, computer graphics and data transformations.