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'''}}.