1. Apr 8, 2007

daniel_i_l

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

2. Relevant equations

3. The attempt at a solution
Intuitivly it looks right and the examples that i tried worked out but is it always true?
Thanks.

2. Apr 8, 2007

Hurkyl

Staff Emeritus
What kind of object is A, and which definition of adjoint are you studying?

3. Apr 8, 2007

daniel_i_l

A is a nxn matrix and the definition that I'm using is the transpose of the cofactor matrix.

4. Apr 8, 2007

matt grime

Adj(A) is the matrix that satisfies

From this we see that (det(Adj(A))/det(A))*A is Adj(Adj(A)), when det(A) is non-zero. I leave it to you to determine what the det of Adj(A) is.

5. Apr 8, 2007

Hurkyl

Staff Emeritus
As I recall, one of the principal properties of the adjoint is
A * adj(A) = det(A) * I​
which should allow you to easily settle your question, at least for invertible matrices.

(P.S. I too have heard this called the adjoint before)

6. Apr 8, 2007

daniel_i_l

matt grime: The det of adj(A) is det(A)^n-1 if A is invertible. So that means that adj(adj(A)) = det(A)^n-2 * A =/= A. But what happens is A is singular?

Hurkyl: As i said above, this would mean that it isn't true for invertible matrices but what about the singular ones?
Thanks!

7. Apr 8, 2007

Hurkyl

Staff Emeritus
If you just want a specific example, there's the zero matrix.

8. Apr 8, 2007

daniel_i_l

That would be a case where it is true, is there a case where it isn't true?
Thanks.

9. Apr 8, 2007

Hurkyl

Staff Emeritus
You haven't proven they're unequal. In fact, you said yourself you know of examples where they are equal.

10. Apr 8, 2007

Hurkyl

Staff Emeritus
Well, have you really tried many examples? The formula you derived suggests that most matrices do not satisfy the given equation... so the fact that you did not find a counterexample suggests that you really haven't tried much!

11. Apr 8, 2007

daniel_i_l

Well, they're only equal when det(A) = 1 and when A is invertible, not in general, right?
Thanks.

12. Apr 8, 2007

Hurkyl

Staff Emeritus
You have shown that, if you assume A is invertible, then they're equal iff det(A)^(n-2) = 1.

In particular, it says nothing about the noninvertible case, and you missed some cases in the invertible case.

Wait, is your arithmetic right? Shouldn't that be det(A)^(2n-2)?

Last edited: Apr 8, 2007
13. Apr 8, 2007

daniel_i_l

What invertible cases did I miss? And how do I approach the non-invertible cases?
And I think that the arithmetic is right:
|adjA|/|A| = |A|^(n-1) / |A| = |A|^(n-1-1) = |A|^(n-2)
Thanks!

Last edited: Apr 8, 2007
14. Apr 8, 2007

matt grime

In the non-invetible cases, there are differences depending on what the rank of A. It is easy to find cases where Adj(A) is the zero matrix. As Hurkyl has said, the number of cases where youir conjecture is true is vanishingly small, so you really can't have tried that many cases.

15. Apr 8, 2007

Hurkyl

Staff Emeritus
How did you deduce det(A) = 1? Is that something you're always allowed to do?

16. Apr 9, 2007

daniel_i_l

Your're right:) My problem was that I only tried 2x2 matrices, but for this invertable 3x3 matrix:
[1 0 0]
[0 1 0]
[0 0 2]
The equation isn't true. and for the non-invertable matrix where all the numbers are 1 it also isn't true.

17. Apr 9, 2007

matt grime

It isn't true for many 2x2 matrices either.