Homework Help: Finding eigenvectors of [[1,-1,-1],[-1,1,-1],[-1,-1,1]]

1. Jul 15, 2012

thetrystero

he eigenvalues of the 3x3 matrix [[1,-1,-1],[-1,1,-1],[-1,-1,1]] are 2,2, and -1.
how can i compute the eigenvectors?
for the case lambda=2, for example, i end up with an augmented matrix [[-1,-1,-1,0],[-1,-1,-1,0],[-1,-1,-1,0]] so i'm stuck at this point.

much appreciated.

2. Jul 15, 2012

Ray Vickson

So, you need to solve the linear system
$$x_1 + x_2 + x_3 = 0\\ x_1 + x_2 + x_3 = 0\\ x_1 + x_2 + x_3 = 0$$
There are lots of solutions. In fact, you should be able to find two linearly independent solution vectors, corresponding to the double eigenvalue 2.

RGV

3. Jul 15, 2012

HallsofIvy

I prefer to work from the basic definitions (perhaps I just never learned these more sophisticated methods!):
Saying that 2 is an eigenvalue of this matrix means there exist a non-zero vector such that
$$\begin{bmatrix}1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}x - y- z\\ -x+ y- z \\ -x- y+ z\end{bmatrix}= \begin{bmatrix}2x \\ 2y\\ 2z\end{bmatrix}$$
which gives the three equations x- y- z= 2x, -x+ y- z= 2y, -x- y+ z= 2z which are, of course, equivalent to -x- y- z= 0, -x- y- z= 0, -x- y- z= 0. Those three equations are the same. We can, for example, say that z= -x- y so that any vector of the form <x, y, -x- y>= <x, 0, -x>+ <0, y, -y>= x<1, 0, -1>+ y<0, 1, -1> is an eigenvector. Notice that the eigenvalue, 2, not only has algebraic multiplicity 2 (it is a double root of the characteristic equation) but has geometric multiplicity 2 (the space of all corresponding eigenvalues is 2 dimensional).

Similarly, the fact that -1 is an eigenvalue means there are x, y, z, satisfying x- y- z= -x, -x+ y- z= -y, -x- y+ z= -z which are, of course, equivalent to 2x- y- z= 0, -x+ 2y- z= 0, -x- y+ 2z= 0. If we subtract the second equation from the first, we eliminate z to get 3x- 3y= 0 so y= x. Putting that into the third equation, 2x+ 2z= 0 so z= -x.
Any eigenvector corresponding to eigenvalue -1 is of the form <x, x, -x>= x<1, 1, -1>.

4. Jul 15, 2012

thetrystero

by that reasoning, can i not have <1,-1,0> and <-1,1,0> as my two solutions for eigenvalue 2? but wolframalpha says i need to have the case where y=0.

also, i think the solution for eigenvalue -1 is <1,1,1>

5. Jul 16, 2012

Ray Vickson

Your two listed vectors (for eigenvalue 2) are just multiples of each other. You need two *linearly independent* eigenvectors, such as <1,-1,0> and <1,0,-1>, or <0,-1,1> and <1,-1/2, -1/2>, etc. There are infinitely many possible pairs of vectors <x1,y1,z1> and <x2,y2,z2> that are linearly independent and satisfy the equation x+y+z=0. Any such pair will do.

RGV

Last edited: Jul 16, 2012
6. Jul 16, 2012

thetrystero

yes, i had thought of that, but found it uncomfortable that of all the many possibilities, both my professor and wolframalpha chose the cases y=0 and z=0 as solutions, so i was wondering what made these two special compares to the others.

7. Jul 17, 2012

Ray Vickson

There is nothing special about these choices, except for the fact that they both have one component = 0 so are, in a sense, the simplest possible. However, you could equally take x=0 and y=0 or x=0 and z=0.

RGV

8. Jul 17, 2012

HallsofIvy

If you have a vector that depends upon parameters, say, <x, y, -x- y> as I have above, then choosing x= 0, y= 1 gives you <0, 1, -1> and choosing x= 1, y= 0 gives <1, 0, -1>. That is, in effect, the same as writing <x, y, -x- y>= <x, 0, -x>+ <0, y, -y>= x<1, 0, -1>+ y<0, 1, -1>, showing that any such vector is a linear combination of <1, 0, -1> and <0, 1, -1>.