# Eigen value 0 prove

1. Mar 6, 2009

### transgalactic

prove that a matrix cannot be invertible if it has eigen value of 0.

not invertible mean that the determinant equals 0.

x(x-2)=0
??

2. Mar 6, 2009

### Staff: Mentor

Where does x(x -2) come in?

You have an n x n matrix A that has 0 as an eigenvalue. How do you find the eigenvalues of a matrix? How are eigenvalues defined?

3. Mar 6, 2009

### transgalactic

x(x -2) its just an example for a polynomial which is not inertible
i thought of proving like this

that if an operator T is not invertible then det T=0 then dim ker differs 0
(we have a row of zeros) so i have eigen value 0

correct
??

Last edited: Mar 6, 2009
4. Mar 6, 2009

### transgalactic

"If an equation containing a variable parameter possesses nontrivial solutions only for certain special values of the parameter, these solutions are called eigenfunctions and the special values are called eigenvalues."

how to use it
??

5. Mar 6, 2009

### Office_Shredder

Staff Emeritus
Not the kind of eigenvalue you want (well, sort of is, but let's not complicate things). An eigenvalue $$\lambda$$ of a matrix A is a number such that there exists v in the vector space, v non-zero, with

$$Av = \lambda v$$

6. Mar 6, 2009

### transgalactic

so if i have eigen value zero
Av=0

how does it show that its not invertible
i know that now v plays the role of kernel A
but i dont know what is the link between the kernel of an operator and its
ability to be invertible
??

7. Mar 6, 2009

### Office_Shredder

Staff Emeritus
A matrix is invertible if it has... an inverse. To have an inverse, it must be 1-1. Think about what having a non-trivial kernel means with regards to this

8. Mar 6, 2009

### transgalactic

1-1 means that for each x i have a y value and vise versa
i really dont know what is the affect of kernel
i know how to find it in a matrix
i know
dim Im + dim Ker =V

??

9. Mar 6, 2009

Remember that in order for a vector $$v$$ to be an eigenvector, it must be true that $$v \ne 0$$. If that is the case, yet $$A v = 0 [tex], what can you conclude? 10. Mar 6, 2009 ### transgalactic zero vector is the product of multiplication of the operarator by the kernel vectors i dont know ?? 11. Mar 6, 2009 ### statdad Well, remember that if the matrix [tex] A$$ is invertible, the solution to

$$Ax = b$$

is always given by

$$x = A^{-1}b$$

Now, suppose that $$v \ne 0$$ is an eigenvector that corresponds to the eigenvalue $$\lambda = 0$$

Now assume that $$A$$ is invertible. What does the equation

$$Av = \lamba \, v = 0$$

tell you about $$v$$? What does this contradict, and what assumption led to that contradiction?

12. Mar 6, 2009

### Dr.D

Can you use the fact that the inverse is the matrix of cofactors divided by the determinant, and that division would be undefined if you had a zero eigenvalue because the determinant is the product of the eigenvalues?

13. Mar 6, 2009

### maze

Consider
Ax = 0
A0 = 0

What is A-1(0)?

14. Mar 7, 2009

### transgalactic

??

15. Mar 7, 2009

### transgalactic

why if dim ker differs 0
then its not invertible
??

16. Mar 7, 2009

### Staff: Mentor

Let's get organized. You're trying to prove this statement, which is a paraphrase of your first post in this thread:

If matrix A has 0 as an eigenvalue, A does not have an inverse.​

If A has 0 as an eigenvalue, Ax = 0x = 0, for some nonzero vector x (the eigenvector associated with eigenvalue 0).

So Ax = 0, where x is not the zero vector.
From this equation, is it possible to say either that A definitely has an inverse or that A definitely doesn't have an inverse?

17. Mar 7, 2009

### transgalactic

i can say that definitely we have a kernel vector in A
but i dont know how it affects the invertiblity of A
??

18. Mar 7, 2009

### Staff: Mentor

There is always at least one vector in ker(A), regardless of whether A has an inverse.

19. Mar 7, 2009

### transgalactic

yes 0,0,0,

20. Mar 7, 2009

### HallsofIvy

Staff Emeritus
This has already been said, but I'll try again. If Ax= 0 and A has an inverse, what do you get if you multiply both sides of that equation by A-1? What does that tell you about the kernel of A? How is that a contradiction?