# How do i build the third vector of this operator?

1. Mar 5, 2008

### transgalactic

how do i build the third vector of this operator???

http://img212.imageshack.us/my.php?image=img86161am0.jpg

i have found two vectors of the operator
but it defined as R3
so i dont know by what laws should i go in order to build the third one

i also got dim ker differ =0 so it has something to do with that i guess

???????

2. Mar 5, 2008

### HallsofIvy

Staff Emeritus
The problem asks you to construct a linear operator that
1. Is NOT invertible.
2. Has eigenvalue 1 with corresponding eigenvector (1, 1, 1)
3. Has eigenvalue 2 with corresponding eigenvector (0, 1, 1)

Yes, saying that the operator is NOT invertible means that is CANNOT be one-to-one and so its kernel is not just the 0 vector. The fact that there exist non-zero v such that Tv= 0= 0v also means this operator must have eigenvalue 0. (0, 1, -1) is orthogonal to both (1, 1, 1) and (0, 1, 1) so you can take that to be an eigenvector corresponding to eigenvalue 0.

3. Mar 6, 2008

### transgalactic

the solution takes
S(0,0,1)=(0,0,0)

why is that??
0,0,1 is not orthogonal to the others

???????

4. Mar 6, 2008

### HallsofIvy

Staff Emeritus
No, but it is independent of the others, and that is enough.

5. Mar 6, 2008

### transgalactic

so you ar saying that the vector that you porposed
is independant too
thats why its valid
ok
but why they put (0,0,0) vector on the right side??

ohhh and by the way good morning in your hemisphere

Last edited: Mar 6, 2008
6. Mar 6, 2008

### HallsofIvy

Staff Emeritus
What does "independent" mean? What does "orthogonal" mean? It should be very simple for you to convince your self that if a vector is orthogonal to another vector, they are independent. That was the idea I was using.

It is also true, and fairly easy to prove, that if two vectors are eigenvectors of a linear transformation, corresponding to different eigenvalues, then they are idependent. This linear transformation has 3 eigenvalues: 1, 2, and 0. You were given eigenvectors corresponding to eigenvalues 1 and 2. You could be sure of getting a vector independent of the other 2 by using an eigenvector corresponding to the third eigenvalue, 0. Of course, if v is an eigenvector corresponding to eigenvalue 0, then it must satisfy A v= 0v= 0. They were showing that the given vector had that property.

7. Mar 6, 2008

### transgalactic

ok i was told a whole other thing
i was told that we make an independant vector from the left
in order to make a basis and we have the (0,0,0) because its dim ker differs one
it must have at least one all zeros vector

but what if we knew that the matrix is invertable
how should i constract the right side??

8. Mar 6, 2008

### HallsofIvy

Staff Emeritus
No, that's not a "whole other thing"- it sounds like just what I said- except that I don't know what you mean by "from the left". Also "construct" what "right side"?

9. Mar 6, 2008

### transgalactic

on the left i ment the left vector of the third equation
on the right i ment the right vector of the third equation
S(0,0,1)=(0,0,0)

was told that we make an independant vector from the left
in order to make a basis and we have the (0,0,0) because its dim ker differs one
it must have at least one all zeros vector

but what if we knew that the matrix is invertable
how should i constract the right side??