For each of the digraphs in Exercise 2.6.3, see whether you can predict a collection of independ

For each of the digraphs in Exercise 2.6.3, see whether you...

Key Concepts

Cycle Space Basis

The cycle space of a graph is the set of all cycles (or circuits) that can be formed in the graph, and a basis for this space is a minimal set of independent cycles from which all other cycles can be linearly generated. In directed graphs, identifying such a basis involves determining a set of independent circuits that respect the direction of edges. This concept is pivotal when connecting combinatorial properties of the graph with linear algebraic structures represented by the incidence matrix and its cokernel.

Cokernel of a Matrix

The cokernel of a matrix, particularly in the context of the incidence matrix, is the quotient space obtained when the codomain of a matrix is divided by the image of the matrix. In graph theory, the cokernel can be used to identify relationships among edges, particularly in the study of cycles. Analyzing the cokernel helps in determining dependencies in a network, and in this context, its basis elements can correspond to independent circuits that span the cycle space.

Incidence Matrix

The incidence matrix of a digraph is a matrix that represents the relationship between vertices and edges. Each row typically corresponds to a vertex and each column to an edge, with entries that indicate whether an edge enters, leaves, or is not incident to a vertex. This matrix provides a powerful algebraic representation of the graph, making it possible to study properties like connectivity, flow, and the existence of circuits through linear algebraic methods.

Circuit in Graph Theory

A circuit, or cycle, in graph theory is a closed path that visits a sequence of vertices starting and ending at the same vertex without necessarily repeating any edge. In digraphs, circuits must respect the edge orientations, meaning that traversing the cycle must follow the designated direction of each edge. Identifying independent circuits is crucial, as these form the building blocks for the cycle space of a graph, allowing for systematic analysis of all cycles in the network.

Directed Graph

A directed graph, or digraph, is a set of nodes connected by edges that have a specific direction. This structure is fundamental in representing relationships that have an inherent orientation, such as one-way streets or dependency orders. In the context of circuits, digraphs provide the framework in which paths and cycles are considered with respect to their direction, enabling the study of directional dependencies and flows within the network.

Transcript

00:01 Here we need to determine whether the vectors are linearly independent or are linearly dependent in rq.

00:11 So the first question is, the solution for part a is we have given a vectors v1 is equal to minus 304.

00:24 This is our first vector.

00:27 The second vector is v2 is equal to 5 minus 1 2.

00:32 5 minus 1 to the third vector v3 is equal to 1 1 1 and 3 so here we need to determine here we determine whether whether the set s is equal to v1 v2 and v3 is linearly independent or linearly dependent set so let let k1 k2 and k3 be any scalar be any scalers then k1 into v1 plus k2 plus k3 into v3 is equal to zero zero which implies that k1 into the vector so it will be minus 3 is 0 and 4 plus k2 the vector is 5 minus 1 and 2 plus k 3 so k 3 into v3 is 1 1 is equal to 0 vector 0 and 0 so we need to solve this so it will be the first component of the vector is minus 3 k 1 plus 5 k2 plus k 3 second component is minus k2 plus k3 the third component is 4k1 plus 2 k2 plus 3 k3 which is equal to 0 0 and 0 now we need to form a homogeneous linear equation from this so we have which imply minus 3 k1 plus 5 k2 plus k3 is equal to 0 then minus k2 and minus k2 plus k3 is equal to 0 4k1 plus 2k2 plus 3 k3 is equal to 0 so this is a system of homogeneous linear equation now we need to solve this and find the value of k1 k2 and k3 so now we solve the above system as follow so we have so writing it in matrix form we get we write the system in augmented matrix form and then augmented matrix form and then solve it then we solve it so we have so the entry in the first row is minus 3 510 the second row is 0 minus 1 110 the third row is 4 2 3 3 3 3 3 3 now we need to solve this so we have the operation elementary row operation is so r1 is replaced by so elementary operation is r1 is replaced by sorry r3 is replaced by that is replaced by row 3 minus r3 minus minus 4 by 3 times of so the output will be minus 351 then the second row is 0 minus 1 1 0 3rd row is 0 26 by 3 so after doing the calculation we get 26 by 3 then it will be 13 by 3 and 0 now again we apply elementary row operations so the operation is r3 is replaced by row 3 that is r3 minus minus times of 26 by 3 of r2 that is root 2 so after doing this we get so we have the row first is minus 351 then 0 the second row is 0 minus 1 1 0 the third row is 0 0 13 and 0 from this we get our equation is so the linear equation is minus 3 k1 plus 5k2 plus k3 is equal to 0 minus k2 plus k3 is equal to 0 and 13 times of k3 is equal to 0 now by back substitution we can evaluate the value of this k1 k2 and k3 scalers so it will be so the value of k3 is k3 is equal to 0 by 13 that is 0 now substituting the value of k 3 in an above equation we get k2 minus k2 plus 0 is equal to 0 which imply k2 is equal to 0 now substituting the value of k2 and k 3 in our equation here we get minus 3 times of k1 is equal to 0 which imply k1 is equal to 0 thus we have a trivial solution for this system of linear equation here we get we get a trivial solution trivial solution for the system for the system of linear equation that is k1 is equal to k2 is equal to k3 is equal to 0 hence hence the vectors are linearly independent...

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