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comment Existence of a $d$-regular graph such that $|N_G(x) \cap N_G(y)| = \lambda$.
The Latin square graph constructed from a $6\times6$ Latin square has parameter $(36,15;6,6)$.
1d
comment Cut Space of Vertices without Orthogonal Complement of Cycle Space?
You are confusing "orthogonal complement", which is a linear algebra concept and which applies here, withe the graph theory "complement", which does not apply here.
2d
reviewed Close List the partitions of the set $S = \{1, 2, 3\}$.
2d
reviewed Close Give me your opinion about those books
2d
reviewed Close Probability Help with finding mean and variance of estimators
2d
reviewed Close Particle locating/collision prediction in bounded (two-dimensional) environments
2d
reviewed Close A question involving logarithms; How to solve?
2d
reviewed Close Is it unitary matrix or not?
2d
reviewed Close How can I convert this second order equation into a first order equation?
2d
reviewed Close Multiplying boths sides of an equation by $\frac{1}{x}$
2d
reviewed Close Cardinality of set of infinite subsets of $\mathbb{N}$
2d
reviewed Close Confusion of intersection of two 2-d planes in 4-d
2d
reviewed Close Using Serre's Theorem, and showing lie algebra is 1-D
Apr
29
reviewed Close How to calculate the total number of dissimilar terms (terms having different powers in x)…
Apr
29
reviewed Close Example of a non-measurable set for which the outer measure is known
Apr
29
reviewed Close Find the number of ways of coloring pentagonal faces of the dihedral in three colors.
Apr
28
comment Do $A$ and $B$ have the same eigenvalues?
@M.S.: If Hermitian matrices commute then they have the same eigenvectors (more precisely, they can be simultaneously diagonalized). Their eigenvalues will not be the same unless the matrices are equal.
Apr
28
comment Do $A$ and $B$ have the same eigenvalues?
@M.S.: My last claim does not require $A$ and $B$ to be Hermitian. I included it because I thought that the false claim might be a mixed up version of this statement.
Apr
28
comment Do $A$ and $B$ have the same eigenvalues?
$AA^2- A^2A= A^3-A^3 = 0$.
Apr
28
comment undirected unweighted graphs having $1$ as an eigenvalue
In a word: No. But since they're not classified, it is difficult to provide a reference.