Is Adjoint of an Adjoint Always Equal to the Original? - General Question

[daniel_i_l]

Homework Statement

I have a general question about adjoints, does:
adj(adj(A)) = A?

Homework Equations

The Attempt at a Solution

Intuitivly it looks right and the examples that i tried worked out but is it always true?
Thanks.

 

[Hurkyl]

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

 

[daniel_i_l]

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

 

[matt grime]

That isn't the adjoint, that is the adjugate.

Adj(A) is the matrix that satisfies

A*Adj(A)=det(A)I

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.

 

[Hurkyl]

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)

 

[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!

 

[Hurkyl]

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

 

[daniel_i_l]

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

 

[Hurkyl]

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?

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

 

[Hurkyl]

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

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!

 

[daniel_i_l]

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

 

[Hurkyl]

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

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)?

 

[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!

 

[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.

 

[Hurkyl]

What invertible cases did I miss?

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

 

[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.

 

[matt grime]

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