Question

Let \(0 \longrightarrow V_{1} \longrightarrow \cdots \longrightarrow V_{n} \longrightarrow 0\) be an exact sequence of finite-dimensional vector spaces. Show that \(\sum(-1)^{i} \operatorname{dim}\left(V_{i}\right)=0\).

Short answer

Question: Prove that for an exact sequence of finite-dimensional vector spaces, the alternating sum of their dimensions is equal to zero. Answer: By applying the rank-nullity theorem to the given exact sequence, we derived the formula: $$\operatorname{dim}(V_i) = \operatorname{dim}(\operatorname{Im}(f_{i-1})) + \operatorname{dim}(\operatorname{Im}(f_i)).$$ This allowed us to rewrite the alternating sum of the dimensions and observe that most terms cancel out because of the alternating signs. The remaining two terms are $$(-1)^n \operatorname{dim}(\operatorname{Im}(f_{n-1})) - \operatorname{dim}(\operatorname{Im}(f_0)).$$ Since \(\operatorname{Im}(f_{n-1}) = V_n\) and \(\operatorname{Im}(f_0) = 0\), the alternating sum equals to zero.

Step-by-step solution

Recall the definition of an exact sequence

An exact sequence is a sequence of linear maps between vector spaces with the property that the image of one map is equal to the kernel of the next map. That is, $$\operatorname{Im}(f_{i-1}) = \operatorname{Ker}(f_i) \quad \text{ for all } 1 \leq i \leq n-1,$$ where \(f_i : V_i \rightarrow V_{i+1}\) are linear maps.

Rank-Nullity theorem

The rank-nullity theorem states that for any linear map \(f : U \rightarrow W\) between finite-dimensional vector spaces \(U\) and \(W\), we have: $$\operatorname{dim}(U) = \operatorname{dim}(\operatorname{Ker}(f)) + \operatorname{dim}(\operatorname{Im}(f)).$$

Apply the Rank-Nullity theorem to the exact sequence

We will now use the rank-nullity theorem for each of the maps \(f_i : V_i \rightarrow V_{i+1}\) in the given exact sequence. We have $$\operatorname{dim}(V_i) = \operatorname{dim}(\operatorname{Ker}(f_i)) + \operatorname{dim}(\operatorname{Im}(f_i)).$$ By the definition of the exact sequence given in Step 1, we know \(\operatorname{Im}(f_{i-1}) = \operatorname{Ker}(f_i)\). Thus, we can rewrite the previous equation as $$\operatorname{dim}(V_i) = \operatorname{dim}(\operatorname{Im}(f_{i-1})) + \operatorname{dim}(\operatorname{Im}(f_i)).$$

Compute the alternating sum of the dimensions

We want to compute the alternating sum $$\sum (-1)^i \operatorname{dim}(V_i) \; \text{ for } \; i = 1, 2, \dots, n.$$ Using the equation from Step 3, we can rewrite the alternating sum as $$\sum (-1)^i (\operatorname{dim}(\operatorname{Im}(f_{i-1})) + \operatorname{dim}(\operatorname{Im}(f_i))).$$ Now, we split the sum into two separate sums: $$\sum (-1)^i \operatorname{dim}(\operatorname{Im}(f_{i-1})) + \sum (-1)^i \operatorname{dim}(\operatorname{Im}(f_i)).$$ Notice that both sums have the same number of terms, and all terms except the first in the first sum and the last in the second sum cancel out with each other, due to the alternating signs. Therefore, the alternating sum reduces to $$(-1)^n \operatorname{dim}(\operatorname{Im}(f_{n-1})) - \operatorname{dim}(\operatorname{Im}(f_0)).$$ Since \(V_n \to 0\) and \(0 \to V_1\) are part of the exact sequence, we have \(\operatorname{Im}(f_{n-1}) = V_n\) and \(\operatorname{Im}(f_0) = 0\). So the alternating sum becomes $$(-1)^n \operatorname{dim}(V_n) - \operatorname{dim}(0) = 0,$$ as desired.

Key concepts

Rank-Nullity Theorem

In the realm of linear algebra, the Rank-Nullity Theorem is a fundamental concept that helps us understand the relationship between the dimensions of different aspects of a linear map. For any linear map, denoted as \( f: U \rightarrow W \), where \( U \) and \( W \) are finite-dimensional vector spaces, the theorem provides the following equation:\[\text{dim}(U) = \text{dim}(\text{Ker}(f)) + \text{dim}(\text{Im}(f)).\]
The equation reflects the fact that the dimension of the domain space \( U \) is partitioned into two parts:

• \( \text{dim}(\text{Ker}(f)) \): This is the dimension of the kernel, the part of the domain that gets mapped to the zero vector in \( W \).
• \( \text{dim}(\text{Im}(f)) \): The image's dimension, representing the elements of \( W \) that are the result of applying \( f \) to elements of \( U \).

Understanding this theorem is crucial for unpacking problems that deal with the structure of vector spaces and their linear transformations. It's a tool that gives insight into how the dimensions of these spaces interrelate.

Linear Maps

Linear maps, or linear transformations, are functions between vector spaces that preserve vector addition and scalar multiplication. Consider a linear map \( f: V \rightarrow W \), where \( V \) and \( W \) are vector spaces. The function \( f \) is linear if for any vectors \( x, y \in V \) and any scalar \( c \), the following two conditions hold:

1. \( f(x + y) = f(x) + f(y) \) (Additivity)2. \( f(cx) = c f(x) \) (Homogeneity of degree 1)

Linear maps are essential in bridging between different vector spaces, allowing us to transform vectors while maintaining the inherent algebraic structure. Such mappings are core to the study of solutions involving vector spaces and help in extending geometrical notions like rotations and projections to higher dimensions.

Kernel and Image

The kernel and image are two critical subsets of a vector space associated with any linear map. For a linear map \( f: V \rightarrow W \), they hold significant positions in understanding how \( f \) functions.

• Kernel: The kernel of \( f \), denoted \( \text{Ker}(f) \), is the set of all vectors in \( V \) that map to the zero vector in \( W \). In symbols, \( \text{Ker}(f) = \{ v \in V \mid f(v) = 0\} \). The dimension of this set is a measure of how many vectors are completely nullified by the mapping \( f \).


• Image: The image of \( f \), denoted \( \text{Im}(f) \), is the set of all images under \( f \) of vectors in \( V \). In other words, \( \text{Im}(f) = \{ f(v) \mid v \in V \} \). This subset represents the outputs that can be reached by applying \( f \) to \( V \).

These concepts are foundational in linear algebra and are often explored using the Rank-Nullity Theorem to bridge the dimensions of the kernel and image with the original vector space.

Finite-Dimensional Vector Spaces

Finite-dimensional vector spaces are vector spaces that have a finite basis. This means the set of vectors that can span the entire space is finite. A classic example is \( \mathbb{R}^n \), where any vector can be expressed as a combination of \( n \) basis vectors. Such spaces are crucial in both theoretical and applied mathematics for several reasons:

• They provide a framework where computation and visualization are feasible, since working with a finite number of basis vectors simplifies many problems.
• Many physical and real-world systems can be modeled using finite-dimensional vector spaces, making them essential in engineering, physics, and data science.
• Operations such as finding linear maps or computing dimension via algorithms become tractable.

Being finite-dimensional implies that techniques like the Rank-Nullity Theorem are applicable, ensuring that the algebra of the space is well-understood and manageable.