Question

Is there any proof or theorem that says "The dimension of eigenvector space of a matrix A is <= n " for a n x n matrix A ??

Answer 1

do you have an online linear algebra book?

Answer 2

no .. but i have gilbert strangs linear algbra book

Answer 3

oh, in real life?

Answer 4

ya :D

Answer 5

@perl could u refer me a online proof for this please

Answer 6

regrettably, i dont know one

Answer 7

@Loser66 ya... but how ?? is there any proof of it or any informal arguments that i can use to prove it

Answer 8

@Loser66 not nessarily sometimes an eigenvalue can have 2 independent eigenvectors

Answer 9

\[\left[\begin{matrix}2 & 0 \\ 0& 2\end{matrix}\right]\] see this matix has eigenvalue 2 and has 2 independent eigenvectors

Answer 10

ya

Answer 11

oh..i didnt know that

Answer 12

url link?

Answer 13

i dont really know what eigenspace or eigenvector means, i have gaps in my linear algebra

Answer 14

^ WUT :o

Answer 15

I thought it was in the definition of the eigenspace that the eigenspace was a subset of \(\large \mathbb{R}^n\)

Answer 16

what @kirbykirby said. That's part of the definition, and so it must be a subspace of \(\mathbb{R}^n\), and therefore has dimension \(\le n\).

Answer 17

but prof. gilbert strang never mentioned it .do u have an acticle that says it is a definition ??

Answer 18

@KingGeorge

Answer 19

http://www.ms.uky.edu/~lee/amspekulin/eigenvectors.pdf check the 4th page

Answer 20

@kirbykirby thank you i will check that now

Answer 21

you could also check here, http://www.math.caltech.edu/~2010-11/2term/ma001b-pr/10Ma1bPracEvalues1.pdf

Answer 22

@kirbykirby reg first article you showed , it is stated as a theorem so it must have a proof ,but the proof is not given there , i will check the 2nd one now

Answer 23

the second one states it like a definition. As a nullspace of \(A - \lambda I\), and \(A-\lambda I\) is an \(n \times n\) matrix, and realize that the nullity + rank = n by the rank-nullity theorem, so \(null(A-\lambda I) \le n\)

Answer 24

yes that implies each \[\lambda \] has eigenspace less than or equal to n , but i am asking a proof of the whole eigenspace has dimension <= n

Answer 25

The eigenspace \(\large E_\lambda=\{\vec{v}\in \mathbb{R}^n|A\vec{v}=\lambda\vec{v} \}=\{\vec{v}\in \mathbb{R}^n|(A-\lambda I)\vec{v}=\vec{0} \}=null(A-\lambda I)\)

Answer 26

https://www.math.ku.edu/~mandal/math290/m290NotesChSEVEN.pdf here they show it's a subspace of R^n using the Subspace test