Question

Let \(U\) and \(W\) denote subspaces of a vector space \(V\). a. If \(V=U \oplus W,\) define \(T: V \rightarrow V\) by \(T(\mathbf{v})=\mathbf{w}\) where \(\mathbf{v}\) is written (uniquely) as \(\mathbf{v}=\mathbf{u}+\mathbf{w}\) with \(\mathbf{u}\) in \(U\) and \(\mathbf{w}\) in \(W\). Show that \(T\) is a linear transformation, \(U=\operatorname{ker} T, W=\operatorname{im} T,\) and \(T^{2}=T\). b. Conversely, if \(T: V \rightarrow V\) is a linear transformation such that \(T^{2}=T,\) show that \(V=\) ker \(T \oplus \operatorname{im} T\). [Hint: \(\mathbf{v}-T(\mathbf{v})\) lies in ker \(T\) for all \(\mathbf{v}\) in \(V\).]

Short answer

a. T is linear, \(\operatorname{ker} T = U\), \(\operatorname{im} T = W\), \(T^2 = T\); b. \(V = \operatorname{ker} T \oplus \operatorname{im} T\).

Step-by-step solution

Define the Transformation

We are given a vector space decomposition, \(V = U \oplus W\), meaning that every \(\mathbf{v} \in V\) can be uniquely expressed as \(\mathbf{v} = \mathbf{u} + \mathbf{w}\) with \(\mathbf{u} \in U\) and \(\mathbf{w} \in W\). The transformation \(T: V \rightarrow V\) is defined by \(T(\mathbf{v}) = \mathbf{w}\).

Show T is Linear

To show that \(T\) is linear, check linearity: for two vectors \(\mathbf{v}_1 = \mathbf{u}_1 + \mathbf{w}_1\) and \(\mathbf{v}_2 = \mathbf{u}_2 + \mathbf{w}_2\), \(T(\mathbf{v}_1 + \mathbf{v}_2) = T((\mathbf{u}_1 + \mathbf{u}_2) + (\mathbf{w}_1 + \mathbf{w}_2)) = \mathbf{w}_1 + \mathbf{w}_2 = T(\mathbf{v}_1) + T(\mathbf{v}_2)\). Thus, \(T\) is linear. Apply a similar check for scalar multiplication.

Determine Ker T

The kernel of a transformation, \(\operatorname{ker} T\), consists of all \(\mathbf{v} \in V\) such that \(T(\mathbf{v}) = \mathbf{0}\). Since \(T(\mathbf{v}) = \mathbf{w}\) and the decomposition is unique with \(\mathbf{w} = \mathbf{0}\), then \(\mathbf{u} = \mathbf{v}\), implying that \(\mathbf{v} \in U\). Hence, \(\operatorname{ker} T = U\).

Determine Im T

The image of a transformation, \(\operatorname{im} T\), consists of all \(\mathbf{w} \in W\) that can be written as \(\mathbf{w} = T(\mathbf{v})\) for some \(\mathbf{v} \in V\). Since every \(\mathbf{v}\) in \(V\) is of the form \(\mathbf{u} + \mathbf{w}\), and \(T(\mathbf{v}) = \mathbf{w}\), then \(\operatorname{im} T = W\).

Verify T^2 = T

Calculate \(T^2\) by applying \(T\) twice: \(T^2(\mathbf{v}) = T(T(\mathbf{v})) = T(\mathbf{w}) = \mathbf{w} = T(\mathbf{v})\), confirming that \(T^2 = T\).

Prove Decomposition Given T^2 = T

For a transformation where \(T^2 = T\), any \(\mathbf{v} \in V\) can be written as \(\mathbf{v} = (\mathbf{v} - T(\mathbf{v})) + T(\mathbf{v})\). Note that \((\mathbf{v} - T(\mathbf{v}))\) lies in \(\operatorname{ker} T\) since \(T(\mathbf{v} - T(\mathbf{v})) = T(\mathbf{v}) - T(\mathbf{v}) = 0\). Thus, \(V = \operatorname{ker} T \oplus \operatorname{im} T\).

Confirm Unique Decomposition

In the sum \(\mathbf{v} = (\mathbf{v} - T(\mathbf{v})) + T(\mathbf{v})\), the part \(\mathbf{v} - T(\mathbf{v})\) uniquely belongs to \(\operatorname{ker} T\) and \(T(\mathbf{v})\) to \(\operatorname{im} T\) because of the property \(T^2 = T\) which ensures the correct splitting of \(V\) into the direct sum.

Key concepts

Vector Spaces

A vector space is a fundamental concept in linear algebra, featuring a set of elements called vectors. Vectors can be added together and multiplied by scalars while satisfying certain axioms. These include closures under addition and scalar multiplication, the existence of an additive identity (zero vector), and presence of vector inverses. Take the example of a vector space \( V \). Any linear combination of its vectors remains in \( V \), creating an infinite combination of potential vector subsets. This highlights a vector space's rich structure and the endless possibilities for exploration within its boundaries.
Understanding vector spaces requires getting familiar with specific related terms:

• Subspaces: These are like smaller vector spaces nested within a larger one. They must contain the zero vector and be closed under vector addition and scalar multiplication.
• Basis: A minimal set of vectors that can represent every vector in the space through linear combinations. A basis is crucial as it determines the 'size' or dimension of the space.
• Dimension: This is the number of vectors in a basis for the vector space, providing a measure of its size or complexity.

Kernel and Image

In linear transformations, the kernel and image are special subspaces that help us understand the nature of the transformation. The kernel of a transformation \( T: V \rightarrow W \) (denoted by \(\operatorname{ker} T\)) consists of all vectors \( \mathbf{v} \in V \) that are transformed into the zero vector in \( W \). Essentially, it tells us which vectors in \( V \) lose their identity under the transformation.
The kernel has these vital properties:

• \( \operatorname{ker} T \) is always a subspace of \( V \).
• If the kernel only contains the zero vector, \( T \) is injective, or one-to-one.

On the other hand, the image of \( T \) (denoted by \(\operatorname{im} T\)) is the set of all vectors in \( W \) that can be mapped from vectors in \( V \). This shows us the 'reach' of the transformation.

• \( \operatorname{im} T \) is a subspace of \( W \).
• If \( \operatorname{im} T = W \), then \( T \) is surjective, or onto.

Direct Sum Decomposition

The concept of direct sum decomposition simplifies complex spaces like a vector space into more manageable parts. When we say a vector space \( V \) is a direct sum of subspaces \( U \) and \( W \), represented as \( V = U \oplus W \), it means every vector in \( V \) can be uniquely expressed as a sum of vectors from \( U \) and \( W \). This idea is powerful in breaking down and solving problems in linear algebra.
Some key aspects include:

• Uniqueness: For decomposition to be 'direct', the intersection of \( U \) and \( W \) must be the zero vector. This ensures that any representation of vectors in \( V \) from \( U \) and \( W \) is unique.
• Balanced Contributions: Every vector in \( V \) is effectively the sum of a vector from \( U \) plus a vector from \( W \).
• Applications: Direct sum decomposition finds applications in various fields, such as simplifying systems of linear equations and in areas like computer graphics, where dividing a problem into subspaces reduces complexity.