Eigenspace of A for Eigenvalue 3: Basis Find

[roam]

Homework Statement

A = \left[\begin{array}{ccccc} 2&2&-1 \\ 1&3&-1 \\ 0&2&1 \end{array}\right]

Given that \lambda = 3 is an eigenvalue of A, find a basis for the eigenspace corresponding to the eigenvalue 3.

Homework Equations

The Attempt at a Solution

Is this question asking for the corresponding eigenvector to the eigenvalue \lambda = 3 is??

I already found that the corresponding eigenvector to the eigenvalue 3 is: {1, 1, 1}

So, what do I need to write as a basis for the eigenspace corresponding to \lambda = 3 ?

 

[Pengwuino]

Remember, an eigenvector is defined by the property Ax = \lambda x.

However, for example, it is also true that A(3x) = \lambda (3x). Thus every constant multiple of your eigenvector is also an eigenvector of A. Your basis is the eigenvector.

 

[HallsofIvy]

Homework Statement

A = \left[\begin{array}{ccccc} 2&2&-1 \\ 1&3&-1 \\ 0&2&1 \end{array}\right]

Given that \lambda = 3 is an eigenvalue of A, find a basis for the eigenspace corresponding to the eigenvalue 3.

Homework Equations

The Attempt at a Solution

Is this question asking for the corresponding eigenvector to the eigenvalue \lambda = 3 is??

I already found that the corresponding eigenvector to the eigenvalue 3 is: {1, 1, 1}

So, what do I need to write as a basis for the eigenspace corresponding to \lambda = 3 ?

I presume that what you found was that the equations reduce to x= z and y= z. That is, that any eigenvector can be written as, say, {x, x, x}= x{1, 1, 1}. Penguino said "Your basis is the eigenvector." I would say, rather, that the basis is the singleton set containing that vector" but I doubt the distinction is important.

 

[roam]

I presume that what you found was that the equations reduce to x= z and y= z. That is, that any eigenvector can be written as, say, {x, x, x}= x{1, 1, 1}. Penguino said "Your basis is the eigenvector." I would say, rather, that the basis is the singleton set containing that vector" but I doubt the distinction is important.

Yes, that's right. Thanks very much :)