Linear algebra - Diagonalizable matrix

  • Thread starter BitterX
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  • #1
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This question has a lot data but I don't really know how to connect it all together


Homework Statement



Let A be a 5x5 matrix over R

3 eigenvectors of A are:
[itex]u_1=(1,0,0,1,1)[/itex]
[itex]u_2=(1,1,0,0,1)[/itex]
[itex]u_3=(-1,0,1,0,0)[/itex]

also:
[itex]\rho (2I-A)>\rho (3I-A)[/itex]

and: [itex]A(1,2,2,1,3)^t=(0,4,6,2,6)^t[/itex]

prove that A is diagonalizable and find a diagonal matrix similiar to it.

Homework Equations





The Attempt at a Solution



What I can make of this is:
[itex](1,2,2,1,3)^t=u_1 + 2u_2 + 2u_3[/itex]

(then maybe I can say that P has the eigenvectors as columns and
[itex]AP(1,2,2,0,0)^t=(0,4,6,2,6)^t[/itex]

but then what?)

and because
[itex]\rho (2I-A)>\rho (3I-A)[/itex]
[itex]\rho (2I-A)\leq 5[/itex]
it means that
[itex]5>\rho (3I-A)[/itex]
and so
[itex]det(3I-A)=0[/itex]
and 3 is an eigenvalue of A.

I would like some hints/suggestions on what to do.

Thanks!
 

Answers and Replies

  • #2
HallsofIvy
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A 5 by 5 matrix is diagonalizable if and only if it has 5 independent eigenvectors. You are given three independent eigenvectors. Can you find two more?
 
  • #3
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I know how to get the eigenvectors from an existing matrix
but not from an eigenvalue or anything they gave me... :/
 
  • #4
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I think I found one eigenvector,

if 3 is an eigenvalue of A, then A is similar to a matrix which it's first column is
(3,0,0,0,0)t

so an eigenvector of this matrix is (1,0,0,0,0) and thus it's an eigenvector of A as well

I only need to find one more :D (assuming I'm right).
 

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