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Prove A is Diagonalizable (Actual Question)

  • Thread starter pezola
  • Start date
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[SOLVED] Prove A is Diagonalizable (Actual Question)

1. Homework Statement


Suppose that A [tex]\in[/tex] M[tex]^{nxn}[/tex](F) and has two distinct eigenvalues, [tex]\lambda[/tex][tex]_{1}[/tex] and [tex]\lambda[/tex][tex]_{2}[/tex], and that dim(E(subscript [tex]\lambda[/tex][tex]_{1}[/tex] ))= n-1. Prove that A is diagonalizable.


3. The Attempt at a Solution

So far, I know that dim(E subscript [tex]\lambda[/tex]2) [tex] \geq1[/tex]
and that
dim(E subscript [tex]\lambda[/tex]1) + dim(E subscript [tex]\lambda[/tex]2) [tex] \leq[/tex] n.
So dim(E subscript [tex]\lambda[/tex]2) = 1.

I am not exactly sure how this helps me to show A is digonalizable. Maybe I am thinking of something else and don't need this to prove A is diagonalizable. Please help.

(Also, sorry about my prevous blank post; I am new)
 

Answers and Replies

tiny-tim
Science Advisor
Homework Helper
25,789
249
Welcome to PF!

So far, I know that dim(E subscript [tex]\lambda[/tex]2) [tex] \geq1[/tex]
and that
dim(E subscript [tex]\lambda[/tex]1) + dim(E subscript [tex]\lambda[/tex]2) [tex] \leq[/tex] n.
So dim(E subscript [tex]\lambda[/tex]2) = 1.
Hi pezola! Welcome to PF! :smile:

Your reasoning seems fine. :smile:

Can't you now use proof by induction - that is, assume the theorem is true for all numbers up to n - 1, and then prove it for n?
(Also, sorry about my prevous blank post; I am new)
:smile: … no problemo! … :smile:
 
StatusX
Homework Helper
2,563
1
A matrix is diagonalizable iff its eigenvectors form a basis for the space (do you understand why?). You've pretty much shown this, maybe just another word or too on why the eigenvectors are independent.
 

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