Linear algebra: eigenvalues, kernel

  • Thread starter Felafel
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  • #1
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Homework Statement


I've tried to solve the following exercise, but I don't have the solutions and I'm a bit uncertain about result. Could someone please tell if it's correct?
Given the endomorphism ##\phi## in ##\mathbb{E}^4## such that:
##\phi(x,y,z,t)=(x+y+t,x+2y,z,x+z+2t)## find:
A) ## M_{\phi}^{\epsilon \epsilon}##
B)##ker(\phi)##
C)eigenvalues and multiplicities
D)eigenspaces
E)is ##\phi## self-adjoint or not? explain

The Attempt at a Solution



A)
( 1 1 0 1 )
( 1 2 0 0 )
( 0 0 1 0 )
( 1 0 1 2 )

B) ker(phi)=> solutions of MX=0
x+y+t=0
x+2y=0
z=0
x+z+2t=0 => ker= v=(-2,1,0,1)

C) to find the eigenvalues i calculate the determinant from the characteristical polynomial:

( 1-T 1 0 1 )
( 1 2-T 0 0 )
( 0 0 1-T 0 ) => ## T^4-3T^3+4T^2-6T+4=0 ##
( 1 0 1 2-T ) where the only eigenvalue is 1 with multiplicity 1.

D) eigenspace:
i substitute T with 1:
( 0 1 0 1 )
( 1 1 0 0 )
( 0 0 0 0 ) => v=(-1,1,-2,-1)
( 1 0 1 1 )
which is self-adjoint because the multiplicity equals the dimension of the eigenspace
 

Answers and Replies

  • #2
HallsofIvy
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Homework Helper
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Homework Statement


I've tried to solve the following exercise, but I don't have the solutions and I'm a bit uncertain about result. Could someone please tell if it's correct?
Given the endomorphism ##\phi## in ##\mathbb{E}^4## such that:
##\phi(x,y,z,t)=(x+y+t,x+2y,z,x+z+2t)## find:
A) ## M_{\phi}^{\epsilon \epsilon}##
B)##ker(\phi)##
C)eigenvalues and multiplicities
D)eigenspaces
E)is ##\phi## self-adjoint or not? explain

The Attempt at a Solution



A)
( 1 1 0 1 )
( 1 2 0 0 )
( 0 0 1 0 )
( 1 0 1 2 )

B) ker(phi)=> solutions of MX=0
x+y+t=0
x+2y=0
z=0
x+z+2t=0 => ker= v=(-2,1,0,1)
Yes, that is correct. And the fact that the kernel is one-dimensional tells you that there exist an eigenvector with eigenvalue 0.

C) to find the eigenvalues i calculate the determinant from the characteristical polynomial:

( 1-T 1 0 1 )
( 1 2-T 0 0 )
( 0 0 1-T 0 ) => ## T^4-3T^3+4T^2-6T+4=0 ##
( 1 0 1 2-T ) where the only eigenvalue is 1 with multiplicity 1.
That's impossible. This is a 4 by 4 matrix the algebraic multiplicities must add to 4. Further, you have the polynomial wrong. As I said before, the fact that the the kernel is one-dimensional tells you that 0 is an eigenvalue (of geometric multiplicity 1) so T= 0 must be a single root of the polynomial. There is no "+4" term.

In fact, the eigenvalues are 0, 1, 2, and 3.

D) eigenspace:
i substitute T with 1:
( 0 1 0 1 )
( 1 1 0 0 )
( 0 0 0 0 ) => v=(-1,1,-2,-1)
( 1 0 1 1 )
which is self-adjoint because the multiplicity equals the dimension of the eigenspace
IF it were true that 1 were the only eigenvalue, this would have proven that the this matrix is NOT self-adjoint. The "multiplicity" of the eigenvalue ("algebraic multiplicity") would be 4 while the dimension of the subspace ("geometric multiplicity") would be 1.

However, that is not true. Here, there are four distinct eigenvalues so four independent eigenvectors. That means that each eigenspace is one dimensional.
 
  • #3
171
0
now i've understood, thank you :)!
just one more question: what if the kernel were multi-dimensional? woukd there still exist an eigenvector with eigenvalue 0 or should i assume something else?
 

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