eigen vector

[bhuvaneshnick]

here the eigen vector is (0,0,0) isnt it? though its not given in options
because we have all eigen value equal to (0,0,0) so it leadind to eigen vectors zero

 

[Ishuda]

If you look at the definition of an eigen vector of the square matrix A, it is defined as a non-zero vector such that
A c = $\displaystyle \lambda$ c
for a column vector c or, equivalently
r A = $\displaystyle \lambda$ r
for a row vector r. Thus r = [0 0 0] is not an eigen vector.

As an aside we note that if
A c = $\displaystyle \lambda$ c
then
(A c)^T = c^T A^T = $\displaystyle \lambda$ c^T
where the superscript T represents the transpose. Thus r and c are related as eigen vectors of a square matrix and its transpose.

 

[bhuvaneshnick]

If you look at the definition of an eigen vector of the square matrix A, it is defined as a non-zero vector such that
A c = $\displaystyle \lambda$ c
for a column vector c or, equivalently
r A = $\displaystyle \lambda$ r
for a row vector r. Thus r = [0 0 0] is not an eigen vector.

As an aside we note that if
A c = $\displaystyle \lambda$ c
then
(A c)^T = c^T A^T = $\displaystyle \lambda$ c^T
where the superscript T represents the transpose. Thus r and c are related as eigen vectors of a square matrix and its transpose.

small change the final vector is (0,y,z)' isnt it?

 

[Ishuda]

small change the final vector is (0,y,z)' isnt it?

You end up with
$\displaystyle -\lambda x_1 + a x_3 = 0$
$\displaystyle -\lambda x_2 = 0$
$\displaystyle -\lambda x_3 = 0$
and that is correct. However, previously you had shown that the eigen values were zero so that those equations become
$\displaystyle a x_3 = 0$
$\displaystyle 0 = 0$
$\displaystyle 0 = 0$
That is, x₁ and x₂ are arbitrary (but not both zero) and x₃ is zero, i.e. the eigen vectors are [x₁ x₂ 0]^T , not both x₁ and x₂ zero.

 

[bhuvaneshnick]

You end up with
$\displaystyle -\lambda x_1 + a x_3 = 0$
$\displaystyle -\lambda x_2 = 0$
$\displaystyle -\lambda x_3 = 0$
and that is correct. However, previously you had shown that the eigen values were zero so that those equations become
$\displaystyle a x_3 = 0$
$\displaystyle 0 = 0$
$\displaystyle 0 = 0$
That is, x₁ and x₂ are arbitrary (but not both zero) and x₃ is zero, i.e. the eigen vectors are [x₁ x₂ 0]^T , not both x₁ and x₂ zero.

yes its big mistake.Though the vector we got(x1,x2,0) does not match with any of option given in the question.what do you think about that

 

[HallsofIvy]

Seriously? Your matrix is the 0 matrix. Is it not obvious that the 0 matrix times any vector is 0? Yes, $\displaystyle <x_1, x_2, 0>$ is an eigenvector for any $\displaystyle x_1$ or $\displaystyle x_2$ but that is because any vector, of the form $\displaystyle <x_1, x_2, x_3>$, for any $\displaystyle x_1$, $\displaystyle x_2$, $\displaystyle x_3$, is an eigenvector!

It looks to me like you have fallen into the trap of memorizing formulas rather than the learning the concepts. You should have been able to look at the 0 matrix and, from the basic definitions of "eigenvalues" and "eigenvectors" have realized that "0" is the only eigenvalue and that all vectors are eigenvectors.

(Most text books define an eigenvalue by "$\displaystyle \lambda$ is an eigenvalue of A if there exist a non-zero vector v, such that $\displaystyle Av= \lambda v$" and then define an eigenvector, corresponding to eigenvalue $\displaystyle \lambda$, as any such non-zero vector. I, and a few textbooks prefer to include the "non-zero" part in the definition of "eigenvalue" but not in the definition of "eigenvalue". That way we can say that "the set of all eigenvectors corresponding to a given eigenvalue form a subspace" rather than having to say that "the set of all eigenvectors corresponding to a given eigenvalue, together with the 0 vector, form a subspace". If you use the more common definition of "eigenvector", you would have to say that "any vector except the 0 vector is an eigenvector for the 0 matrix".)

 

[Ishuda]

Seriously? Your matrix is the 0 matrix. ...

HallsofIvy,
That last entry in row 1 is an a, not a zero. I had to look twice myself.

 

[Ishuda]

yes its big mistake.Though the vector we got(x1,x2,0) does not match with any of option given in the question.what do you think about that

You are mistaken, it does match two of the answers (B with x1=a and x2=0 and D with x1=0 and x2=a). In fact, it would match any non-trivial answer where x₃=0 since x1 and x2 can be chosen as anything (except both zero). BTW: Technically, it should be mentioned that a is not zero.