Question

Suppose that \(V\) is finite dimensional and \(\mathbf{T} \in \mathcal{L}(\mathcal{V}, \mathcal{W})\) is not zero. Prove that there exists a subspace \(\mathcal{U}\) of \(V\) such that \(\operatorname{ker} \boldsymbol{T} \cap \mathcal{U}=\\{0\\}\) and \(\mathbf{T}(v)=\mathbf{T}(\mathcal{U})\)

Short answer

The property holds as desired, demonstrated by defining a new subspace \(U\) of \(V\) spanned by the extension part of the basis of \(V\) beyond the basis for ker(T). It is verified that this \(U\) satisfies the given properties with respect to T.

Step-by-step solution

Find a basis for the Null Space

By the assumption that T is not zero, the image space of T, denoted by \(\mathbf{T}(v)\), is not null. This implies that its null space or kernel, denoted by ker(T), is a proper subspace of \(V\), not equal to \(V\) itself, as the dimension of image is non-zero. Hence, by the basis theorem, there exists a basis for the kernel of T, say \(\mathbf{v_1, v_2, ..., v_k}\).

Extend the Basis to V

Next, this basis can be extended to form a basis for \(V\), say \(\mathbf{v_1, v_2, ..., v_k, w_1,..., w_m}\). We know that any vector \(v \in V\) can be written uniquely as a linear combination of these basis vectors.

Define the Subspace U

Now, define the subspace \(\mathcal{U}\) to be the subspace spanned by the vectors \(\mathbf{w_1, ..., w_m}\). Hence, any vector in this subspace can be written as a combination of these vectors and not include any part of the null space of T.

Verify the Properties

Firstly, by definition, since \(U\) does not include the vectors forming the kernel, the intersection of ker(T) and U will only be the zero vector. Secondly, for any vector \(v \in V\), it can be written as a combination of the basis vectors of both ker(T) and U. But since T sends all vectors in its kernel to zero, \(\mathbf{T}(v)\) will only depend on the \(U\)-component of \(v\). This means that \(\mathbf{T}(U) = \mathbf{T}(V)\), which completes the proof.

Key concepts

Finite Dimensional Vector Spaces

In the realm of linear algebra, a finite dimensional vector space is a pivotal concept. It refers to a vector space that has a finite basis. This means the space can be spanned by a finite number of vectors. A basis is a set of vectors within a vector space that is both linearly independent (none of the vectors can be written as a linear combination of the others) and spanning (any vector in the space can be expressed as a linear combination of these basis vectors).

When it comes to problem-solving, this notion is crucial since it assures that every vector in the space can be uniquely represented by a linear combination of the basis vectors. When T is a transformation from one such space, V, to another, W, the solution utilizes this property to constructively prove the existence of a particular subspace.

Null Space

The null space, also known as the nullity of a linear transformation, is the set of all vectors that are mapped to the zero vector by that transformation. In symbols, if T is a linear transformation from V to W, then the null space is given by \( \text{null space}(\boldsymbol{T}) = \{ \textbf{v} \text{ in } \text{V} \text{ | } \boldsymbol{T}(\textbf{v}) = \textbf{0} \text{ in } \text{W} \} \).

This concept is significant because it tells us about the 'kernel' of the transformation—the 'heart' where the transformation annihilates vectors. In our exercise, the null space is used to argue the existence of a complementary subspace in which the transformation is non-null.

Kernel of a Linear Transformation

The kernel of a linear transformation, often denoted as \( \text{ker}(\boldsymbol{T}) \), is synonymous with the null space. It includes all vectors from the domain that are mapped to the zero vector in the codomain. The kernel is a critical concept as it is always a subspace of the domain vector space. This means it has the properties of closure under addition and scalar multiplication.

Understanding the kernel is pivotal when solving for transformations, such as in the provided exercise. If the kernel is non-trivial (not just {0}), it indicates that the transformation is not injective (one-to-one).

Basis Theorem

The basis theorem has great implications in understanding vector spaces. It states that every finite-dimensional vector space has a basis and that all bases of a vector space have the same number of elements, known as the dimension of the vector space. This theorem provides assurance that for any linearly independent set of vectors, there exists an extension to a basis for the full space.

Applying this to our exercise, the initial basis of the kernel (a subspace of V) can be extended to a basis for the entire space V. This concept is harnessed to construct a subspace that, in conjunction with the kernel, spans the original vector space.

Subspaces in Linear Algebra

In linear algebra, subspaces are subsets of vector spaces that themselves are vector spaces, satisfying the required axioms, notably closure under addition and scalar multiplication. Key properties of subspaces include containing the zero vector, and any linear combination of vectors in the subspace also being within the subspace.

In solving linear transformation problems, understanding subspaces allows us to decompose vector spaces, much like in the step-by-step solution provided. The subspace U, as constructed, serves to complement the kernel, depicting that while kernels show us where transformations are 'inactive,' subspaces like U can display where transformations are 'active.'