On Wed, Aug 30, 2023 at 10:45 Undescribed Horrific Abuse, One Victim & Survivor of Many <gmkarl@gmail.com> wrote:
== Formal definition == If {{mvar|T}} is a linear transformation from a vector space {{mvar|V}} over a [[Field (mathematics)|field]] {{mvar|F}} into itself and {{math|'''v'''}} is a [[zero vector|nonzero]] vector in {{mvar|V}}, then {{math|'''v'''}} is an eigenvector of {{mvar|T}} if {{math|''T''('''v''')}} is a scalar multiple of {{math|'''v'''}}.<ref>{{harvnb|Roman|2008|loc=p. 185 §8}}</ref> Thi
brainstem boss i read it by laughing like you said and i understand it ;-} it says an eigenvector is a vector that a matrix has a scaling operation on. (i could be wrong :S i think it means scaling matrices have every vector as an eigenvector :S ?) anyway the laughing-to-read was really nice :) i understand (should do more (ish(?
can be written as <math display=block>T(\mathbf{v}) = \lambda \mathbf{v},</math> where {{mvar|λ}} is a scalar in {{mvar|F}}, known as the '''eigenvalue''', '''characteristic value''', or '''characteristic root''' associated with {{math|'''v'''}}.
There is a direct correspondence between ''n''-by-''n'' [[Square matrix|square matrices]] and linear transformations from an [[Dimension|''n''-dimensional]] vector space into itself, given any [[Basis (linear algebra)|basis]] of the vector space. Hence, in a finite-dimensional vector space, it is equivalent to define eigenvalues and eigenvectors using either the language of [[Matrix (mathematics)|matrices]], or the language of linear transformations.{{sfn|Herstein|1964|pp=228, 229}}{{sfn|Nering|1970|p=38}}
If {{mvar|V}} is finite-dimensional, the above equation is equivalent to{{sfn|Weisstein|n.d.}} <math display=block>A\mathbf{u} = \lambda \mathbf{u}.</math>
where {{mvar|A}} is the matrix representation of {{mvar|T}} and {{math|'''u'''}} is the [[coordinate vector]] of {{math|'''v'''}}.