Question

Show that \(\mathbf{T}\) and \(\mathbf{T}^{*}\) have the same rank. In particular, show that if \(\mathbf{T}\) is injective, then \(\mathbf{T}^{*}\) is surjective. Hint: Use the dimension theorem for \(\mathbf{T}\) and \(\mathbf{T}^{*}\) and Eq. (2.25).

Short answer

The rank of \( \mathbf{T} \) and \( \mathbf{T}^{*} \) are equal because the rank of a linear transformation is equal to the dimension of its image space, which does not change when going to the adjoint of the transformation since the dimensions of their null spaces are equal, hence using the rank-nullity theorem, their ranks are equal. If \( \mathbf{T} \) is injective it means its kernel is trivial, hence the image space of \( \mathbf{T}^{*} \) is the whole space \( W \), meaning that \( \mathbf{T}^{*} \) is surjective.

Step-by-step solution

Determine the null spaces

Let \( V \) and \( W \) be vector spaces and let \( \mathbf{T} : V \rightarrow W \) be a linear map. The adjoint \( \mathbf{T}^{*} \) of \( \mathbf{T} \) is a map from \( W \) to \( V \). Based on Eq. (2.25) that states that the dimension of the null space of \( \mathbf{T} \) equals the dimension of the null space of \( \mathbf{T}^{*} \).

Prove that \( \mathbf{T} \) and \( \mathbf{T}^{*} \) have the same rank.

Using the dimension theorem, the rank-nullity theorem can be written as: in general case for a linear transformation \( \mathbf{T} : V \rightarrow W \) , we have \( dimV = dimKer(T) + dimIm(T) \). Similarly, for the adjoint operation we have \( dimW = dimKer(T^{*}) + dimIm(T^{*}) \). Given the fact that \( dimKer(T) = dimKer(T^{*}) \), we can conclude that \( dimIm(T) = dimIm(T^{*}) \). In other words, the rank of \( \mathbf{T} \) and \( \mathbf{T}^{*} \) are equal.

Prove that if \( \mathbf{T} \) is injective, then \( \mathbf{T}^{*} \) is surjective.

If \( \mathbf{T} \) is injective, it means that \( \mathbf{T} \) is one-to-one and \( Ker(T) = 0\) . From the rank-nullity theorem, if the Kernel is null then \(dimIm(T) = dimV\), and because we know that \( \mathbf{T} \) and \( \mathbf{T}^{*} \) have the same rank, it results that \(dimIm(T^{*}) = dimV = dimW\). Then, \( \mathbf{T}^{*} \) is surjective.

Key concepts

Adjoint of a Linear Map

In the realm of linear algebra, the adjoint of a linear map, denoted as \( \mathbf{T}^* \), plays a vital role in understanding the structure and properties of linear transformations. This concept is intertwined with the inner product of vectors when discussing vector spaces. To compute the adjoint of a linear map \( \mathbf{T}: V \rightarrow W \), we essentially seek a unique linear map \( \mathbf{T}^*: W \rightarrow V \) such that for all vectors \( v \in V \) and \( w \in W \), the inner product \( \langle \mathbf{T}v, w \rangle \) is equal to \( \langle v, \mathbf{T}^*w \rangle \). This relationship is crucial as it reflects the 'mirror image' of the transformation regarding the inner product.

Understanding the adjoint is beneficial as it often simplifies complex problems, such as finding eigenvectors and eigenvalues, and provides deeper insight into the geometric interpretation of linear maps. Moreover, key properties of the adjoint, including the preservation of the rank as shown in the exercise, are foundational in advanced topics such as functional analysis and quantum mechanics.

Rank-Nullity Theorem

The rank-nullity theorem is a fundamental theorem in linear algebra establishing a relationship between the dimensions of a vector space and its corresponding linear transformation's rank and nullity. The theorem states that for any linear map \( \mathbf{T}: V \rightarrow W \) from a vector space \( V \) to another vector space \( W \) with finite dimensions, the sum of the rank of \( \mathbf{T} \) (which is the dimension of the image of \( \mathbf{T} \), \( \text{dim Im}(\mathbf{T}) \) ) and the nullity of \( \mathbf{T} \) (which is the dimension of the kernel of \( \mathbf{T} \), \( \text{dim Ker}(\mathbf{T}) \) ) is equal to the dimension of the domain space \( V \)— mathematically, \( \text{dim}(V) = \text{dim Ker}(\mathbf{T}) + \text{dim Im}(\mathbf{T}) \).

This theorem is not only useful for theoretical analyses but also has practical implications in areas like engineering and computer science, where understanding the degrees of freedom in a system is critical. For instance, it can aid in determining whether a system of linear equations has a unique solution or not, which is paramount in solving real-world problems.

Injective and Surjective Maps

Linear maps \( \mathbf{T}: V \rightarrow W \) can exhibit certain characteristics which are termed as 'injective' (one-to-one) or 'surjective' (onto). A map is injective if each element in the codomain \( W \) is the image of at most one element in the domain \( V \); essentially, \( \mathbf{T} \) never maps distinct elements in \( V \) to the same element in \( W \). On the flip side, \( \mathbf{T} \) is surjective if every element in \( W \) is the image of at least one element in \( V \), meaning \( \mathbf{T} \) covers the entire codomain \( W \).

Being able to identify whether a map is injective or surjective is important, as this categorization has implications for the system's behavior and the existence and nature of solutions. For example, in the context of solving equations, an injective map ensures that each solution is unique, while a surjective map guarantees that a solution exists for every possible outcome. This differentiation is key to analyzing and extracting meaningful solutions from complex problems, such as those encountered in optimization and mathematical modeling.