Bar Graphs: Practical Examples & Insights

Recent questions in Bar Graphs
Descriptive StatisticsAnswered question
Dylan Benitez Dylan Benitez 2022-11-04

If G is a bipartite Euler and Hamiltonian graph, prove that complement of G, G ¯ is not Eulers.
I would like to know if my proof of the statement in the title is correct.
So, I started like this:
As G is a bipartite graph, it has two sets X and Y. Using the condition G is Hamiltonian, then |X|=|Y|.
As G is also Eulerian, stands d G ( v ) is even v V ( G ), where V(G) is set of vertices of the graph G.
Now, let's look at some vertex v X ( G ). As stated above, it's degree is even. Let's look at two possible cases:
If |X|=|Y| is even too, then in G ¯ , v will be connected with all the remaining vertices from X. That vertex v will also be connected with remaining vertices from Y, with which it wasn't connected in the graph G. That is, d G ¯ ( v ) = | X | 1 + m, where m represents number of remaining vertices from Y. As |X| is even, then |X|−1 is odd, and m is also even, because d G ¯ ( v ) is even and |Y| is even, so the remaining number of vertices, m is even too. Sum of an even and an odd number is odd, so d G ¯ ( v ) is odd. That means G ¯ is not Eulerian;
If | X | = | Y | is odd, in G ¯ , v will be connected with all the remaining vertices from X and all the remaining vertices from Y, and let's call the number of Y remaining vertices m. As | Y | is odd and d G ( v ) is even, then m is odd. Degree of v in G ¯ is once again d G ¯ ( v ) = | X | 1 + m, but this time | X | 1is even, and m is odd. d G ¯ ( v ) is odd and that means G ¯ is not Eulerian.
Thank You!

Jaydan Ball Jaydan Ball 2022-08-21

Existence of solution for matrix equation ( I α A ) x ¯ = b ¯
This is my first question in here and I would be really thankful if someone could help me with understanding the matter.
I am solving a matrix equation ( I α A ) x ¯ = b ¯ for a positive vector x ¯ . I don't know anything about the signs of the scalar α and vector b ¯ (but I can always split the solution into several cases). Matrix A is a symmetric matrix with tr(A)=0 (basically it is an adjacent matrix of an undirected graph so it is built only of 0 and 1 with 0 on its diagonal).
My approach would be just to write it down as x ¯ = ( I α A ) 1 b ¯ whenever ( I α A ) is nonsingular (as far as I understand that happens for a finite number of α and thus Lebesgue measure of this set of "inappropriate" α is 0).
In most of the literature, however, it is required that 1 > α λ m a x ( A ) for the convergence of I + α A + α 2 A 2 + and for the solution x ¯ to exist (where λ m a x ( A ) is a maximum eigenvalue of A) . Moreover, then d e t ( I α A ) = 1 α d e t ( A ) = 1 α i λ i > 0 and hence it is also invertible.
So I have two questions here:
Q1: Why do I need convergence of I + α A + α 2 A 2 + in here? I understand that if it converges then k = 0 + α k A k = ( I α A ) 1 , but why do I need that for the existence of a solution x¯? For instance, if we take a simple example:
A = | 0 1 1 0 | α = 30
λ m a x ( A ) = 1 and hence, convergence condition is violated: 30>1. However, I can still calculate x ¯ . Inverse of ( I α A ) exists and is equal to 1 899 | 1 30 30 1 | . If, for instance, b ¯ was negative, then it would give me a positive x ¯ . Or is the condition 1 > α λ m a x ( A ) needed to guarantee that inverse will give a positive solution with both positive α and b ¯ ?
Q2: What will happen if α < 0 (and potentially b ¯ < 0)? What are the conditions for existence of solution x ¯ ? What are the conditions for it to be positive?

Statistical bar graphs stand for the categorical variables, discrete variables, or continuous ones that should be grouped in class intervals. There will be other types of examples that you can find in our list of answers to the statistical questions. They will help you build the diagrams in statistics. If you are dealing with equations, you will use different types of bar graphs like multiple or stacked, depending on what kind of statistical data must be applied. Ensure that you check how the graphical sections have been answered. It will let you connect the variables and see how they function.