How Do You Determine the Free Variable in Linear Systems for Eigenvectors?

[roam]

Homework Statement

For a few problems dealing with eigenvectors, I substituted my eigenvalues into the characteristic equations. I got systems of linear equations. I need to find the general solution to the systems in order to find the corresponding eigenvectors.
For example in one problem I have to solve:

A = \left[\begin{array}{ccccc} -3&-3 \\ -4&-4 \end{array}\right] \left[\begin{array}{ccccc} x\\ y \end{array}\right] = \left[\begin{array}{ccccc} 0\\ 0 \end{array}\right]

-3x-3y=0
-4x-4y=0

It looks as if x,y are equal. I think I might need to write one in terms of the other but I'm no sure which.

or for example the system of equations:

-x-y-z=0
-x-y-z=0
-x-y-z=0

We have 3 variables and 3 equations which are exactly the same. How do we decide which variable to use as the "free" variable?

Thanks.

 

[dx]

Both these are easy to solve by inspection. For example, in the first one, set x = 1 and y = -1.

 

[Pengwuino]

It doesn't matter. In the 2nd example, you can pick x = -y-z or y = -x-z or z = -x-y. You will still have the same solution space even though the eigenvectors don't look the same.

 

[HallsofIvy]

Remember that the set of all eigenvectors corresponding to a given eigenvalue for a vector space- there is not one single solution. In fact, the condition that A-\lambda I NOT have an inverse guarentees that your set of equation not have a single solution.

For the first, -3x-3y= 0, -4x- 4y= 0, it is not the case that "x,y are equal". Both equations are equivalent to -x-y= 0 and then y= -x, not x. Every eigenvector is of the form <x, -x> = x<1, -1>. The vector space of all eigenvectors is spanned by the single vector <1, -1>.

For the second set where every equation is of the form -x-y-z= 0 you have one equation in three variables. That means you can solve for any one in terms of the other two- it doesn't matter which you choose. For example, z= -x- y. Taking x=1, y= 0, z= -1 so <1, 0, -1> is in the space. Taking x= 0, y= 1, z= -1 so <0, 1, -1> is also in the space. Those two vectors form a basis for the space of eigenvectors.

If you had chosen instead to solve for y, y= -x- z. Now taking x=1, z= 0 gives <1, -1, 0> and taking x=0, z= 1 gives <0, -1, 1> . Those form a different basis for the same space.

If you had chosen to solve for x, x= -y- z giving <-1, 1, 0> and <-1, 0, 1>, yet another basis for the same space.