Let A=((5,1,2),(0,6,0),(1,-1,4)) (**) A=((1),(0),(1))=((3),(0),(-3)), A=((1),(1),(0))=((6),(6),(0)), A((1),(-1),(1))=((6),(-6),(6)), ((1),(1),(0))+((1),(-1),(1))=((2),(0),(1)) Which of the following is true?

Rene Nicholson

Rene Nicholson

Answered question

2022-10-16

Is (2,0,1) an eigenvector of A?
Let A = ( 5 1 2 0 6 0 1 1 4 )
(**) A ( 1 0 1 ) = ( 3 0 3 ) , A ( 1 1 0 ) = ( 6 6 0 ) , A ( 1 1 1 ) = ( 6 6 6 ) , ( 1 1 0 ) + ( 1 1 1 ) = ( 2 0 1 )
Which of the following is true?
(1) (2,0,1) is an eigenvector associated to 3
(2) (2,0,1) is an eigenvector associated to 6
(3) (2,0,1) is not an eigenvector of A
I found that only (2) is true by using the characteristic polynomial, but how do I prove that (2) is true using (**), as intended by the question?

Answer & Explanation

Annabella Ferguson

Annabella Ferguson

Beginner2022-10-17Added 15 answers

Well, you have
A ( 2 0 1 ) = A ( ( 1 1 0 ) + ( 1 1 1 ) ) = ( 6 6 0 ) + ( 6 6 6 ) = ( 12 0 6 ) = 6 ( 2 0 1 ) .
Deborah Proctor

Deborah Proctor

Beginner2022-10-18Added 3 answers

Note that if v 1 and v 2 are two eigenvectors of a matrix M with respect to the eigenvalue λ, i.e. M v 1 = λ v 1 and M v 2 = λ v 2 , then any linear combination of v 1 and v 2 , is also an eigenvector of M corresponding to the eigenvalue λ, since
M ( a v 1 + b v 2 ) = a M v 1 + b M v 2 = a λ v 1 + b λ v 2 = λ ( a v 1 + b v 2 )
Now for v 1 = ( 1 1 0 ) and v 2 = ( 1 1 1 ) , we have A v 1 = 6 v 1 and A v 2 = 6 v 2 . Hence v 1 and v 2 are eigenvectors of A corresponding to the eigenvalue 6. Hence v 1 + v 2 = ( 2 0 1 ) is an eigenvector of A corresponding to the eigenvalue 6

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