Finding eigenvectors of [[1,-1,-1],[-1,1,-1],[-1,-1,1]]

[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.

 

[Ray Vickson]

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.

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

 

[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>.

 

[thetrystero]

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>.

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>

 

[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

 

[thetrystero]

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

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.

 

[Ray Vickson]

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.

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

 

[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>.