True or false eigenvalue problem

[achuthan1988]

Q)Different eigenvectors corresponding to an eigenvalue of a matrix must be linearly dependant?
Is the above statement true or false.Give reasons.

 

[micromass]

Q)Different eigenvectors corresponding to an eigenvalue of a matrix must be linearly dependant?
Is the above statement true or false.Give reasons.

What did you try already?? If we know what you tried then we'll know where to help??

Begin by looking at some examples of matrices. Pick some arbitrary (easy) matrices and calculate its eigenvectors.

 

[HallsofIvy]

I would NOT look at specific matrices. This can be done better more abstractly.

Suppose u and v are non-zero eigenvectors, for linear operator A, corresponding to distinct eigenvalues [itex]\lambda_1[/itex] and [itex]\lambda_2[/itex] respectively. Then [itex]Au= \lambda_1 u[/itex] and [itex]Av= \lambda_2 v[/itex].

Suppose they are not independent. Then [itex]v= \mu u[/itex] for some non-zero scalar [itex]\mu[/itex]. That gives [itex]Av= \mu Au[/itex] or [itex]\lambda_1 v= \mu \lambda_2 v[/itex] so that [itex] v= (\mu \lambda_2)/\lambda_1)u[/itex].

But that means [itex](\mu \lambda_2)/\lambda_1= \mu[/itex] so that [itex]\lambda_2= \lambda_1[/itex], a contradiction.

(Of course, this requires [itex]\lambda_1\ne 0[/itex] so we can divide by it. If [itex]\lambda_1= 0[/itex], just reverse the [itex]\lambda_1[/itex] and [itex]\lambda_2[/itex]. They cannot both be 0 because they are distinct.)

Now, can you extend that to any number of independent eigenvectors?