# Eigenspace of A for Eigenvalue 3: Basis Find

• roam
In summary, the question is asking for a basis for the eigenspace corresponding to the eigenvalue 3 of the matrix A, where A is given as A = \left[\begin{array}{ccccc} 2&2&-1 \\ 1&3&-1 \\ 0&2&1 \end{array}\right]. The eigenvector corresponding to the eigenvalue 3 is {1, 1, 1}, and this can be written as a basis for the eigenspace. It is also important to note that any constant multiple of this eigenvector is also an eigenvector, making it a suitable basis for the eigenspace.

## 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.

## 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 $$\lambda = 3$$ ?

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.

roam said:

## 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.

## 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.

HallsofIvy said:
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 :)