Eigenvalues of an Invertible Matrix

1. Mar 20, 2009

1. The problem statement, all variables and given/known data

Prove that a square matrix is invertible if and only if no eigenvalue is zero.

2. Relevant equations

3. The attempt at a solution

If a matrix has an inverse then its determinant is not equal to 0. Eigenvalues form pivots in the matrix. If any of the pivots are zero, then the determinant will be 0?...Is this correct logic? If so, how do I write it as a formal proof?

2. Mar 20, 2009

Feldoh

How do we find eigenvalues? t is an eigenvector if det(A-tI) = 0, right?

Does this help?

3. Mar 20, 2009

I'm still not really sure, sorry. Is it because we set up the problem so that there are non-trivial solutions, but if one of the eigenvalues is 0 there is a trivial solution?

4. Mar 20, 2009

gabbagabbahey

What is det(A-tI) if t=0?

5. Mar 20, 2009

Oh, if t=0, then det(A-tI)=det(A)=0, which means A is not invertible.

6. Mar 20, 2009

Dick

If 0 is an eigenvalue with eigenvector v then Av=0. Right? And A(2v)=2A(v)=0. Forget determinants. Just think about what 'invertible' means.

7. Mar 20, 2009

maze

A function is invertible if it is 1-1 and onto. Here is a sketch of a possible proof (you will have to fill in the details)

Let M be a n x n matrix with no zero eigenvalues. (M:Rn->Rn)

(1-1) Suppose for the sake of contradiction that M is not 1-1. Then there are distinct vectors x and y such that Mx = My.

... (fill in this part) ...

This is a contradiction. Therefore M is 1-1.

(onto) Let {b1, b2, ..., bn} be a basis for Rn. By the definition of a basis, the bi's are linearly independent, so a1 b1 + a2 b2 + ... + an bn =/= 0 for any nontrivial choice of scalars ai (ai's can be anything except for all a1=a2=...=an=0).

... (fill in this part) ...

Therefore {Mb1, Mb2, ..., Mbn} are a basis for Rn, so M is onto.

8. Mar 20, 2009

Dick

That is WAY too complicated. Even my example was too complicated. If A(0)=0 and A(v)=0, then A is not invertible. Period.

9. Mar 20, 2009

maze

you also need to show that it is onto. For example, if the matrix was not square then it could be 1-1 and still noninvertible. (consider [1 0; 0 1; 0 0])

10. Mar 20, 2009

schlynn

And if det(A)=0 there's no eigenvalues right? Because the matrix isn't invertible. Does this mean that you can't write a characteristic polynomial of the matrix?

11. Mar 20, 2009

Dick

The problem said that the matrix was square.

12. Mar 20, 2009

Dick

If det(A)=0 then there is a zero eigenvalue. That's all. The matrix is NOT invertible, and yes, it does have a characteristic polynomial. They ALL do. Why would you think not?

13. Mar 20, 2009

maze

Right exactly, but you still need to show that a 1-1 square matrix is also onto.

14. Mar 20, 2009

Dick

Right you are as well. I was sort of forgetting the 'if and only if' part.

15. Sep 30, 2009

kdmckale

While thinking of a matrix inverse as an inverse of a function, there are many things to be aware of with algebraic structure (matrix multiplication not being commutative for one). Admissible but not recommended. There are a lot more tools that can make this proof much easier. I applaud the one side where you observed that the det(A) would be zero if lambda=0. This proves that if you have a zero eigenvalue then your matrix is singular and hence, does not have an inverse.

Now, you must assume (hint think of the other proof backwards) you have a singular matrix (non-invertible--which means what about the determinant?) I think you have it from this point.

Last edited: Sep 30, 2009
16. Sep 30, 2009

kdmckale

In fact, if you're clever enough you can write the whole proof as an if and only if proof instead of a two-sided proof.

17. Sep 30, 2009

HallsofIvy

This appears to be your basic difficulty. All of the responses here, as well as the problem itself, assume you know what an "eigenvalue" is! Do you?

18. Sep 30, 2009

The determinant of a square matrix equals the product of the eigenvalues of the matrix. That alone is enough to prove A is invertible if and only if there are no eigenvalues equal to zero.

19. Sep 30, 2009

kdmckale

Ahhh...an even slicker way...I should have thought of that last night when I was using that exact fact on my ODE homework...huge kudos :)

20. Sep 30, 2009

kdmckale

I would propose the answer is "no" after reading most of the posts :) This really is a very simple problem (as statdad has shown)...there's just so many way to show it.

I suspect the issue is lack of enrollment in a linear algebra course. As important as it is, so few people take the course. *sigh*