Therefore, using the first theorem, the characters of irreducible representations of G form an orthonormal set on \ C _ { \ text { class } } ( G ) with respect to this inner product.
12.
We will prove the same statement for " T " : the operator " T " can be diagonalized by an orthonormal set of eigenvectors, each of which corresponds to a real eigenvalue.
13.
I was reading about the singular value decomposition of a matrix and it said that given an orthonormal set of vectors { u1, . . ., ur } you had to extend it to an orthonormal basis of R ^ n.
14.
The matrix " P " is Hermitian, therefore diagonalizable, so it is the identity matrix in other words the columns of " M " are an orthonormal set and the columns of " N " are an orthogonal set.
15.
Given a pre-Hilbert space " H ", an orthonormal basis for " H " is an orthonormal set of vectors with the property that every vector in " H " can be written as an infinite linear combination of the vectors in the basis.
16.
Where f _ 1, f _ 2, \ ldots and g _ 1, g _ 2, \ ldots are orthonormal sets ( not necessarily complete ), and \ lambda _ 1, \ lambda _ 2, \ ldots is a sequence of positive numbers with limit zero, called the accumulate only at zero.
17.
This is related to the fact that the only vector orthogonal to a dense linear subspace is the zero vector, for if " S " is any orthonormal set and " v " is orthogonal to " S ", then " v " is orthogonal to the closure of the linear span of " S ", which is the whole space.
18.
You said that : " But these basis functions don't seem to be orthogonal ( the inner product of two distinct basis functions, over R, doesn't seem to be 0 . " If the " " u " " you quote is allowed to range over all integers, then the fact that the basis functions form a ( countable ) orthonormal set is an easy computation.
