Show that det(A) is the product of all the roots of the characteristic

In summary, the conversation discusses the proof that the determinant of an n x n matrix is the product of all the roots of its characteristic polynomial. The definition of the determinant is stated as the product of all the eigenvalues, and it is shown that any matrix can be transformed into a diagonal or Jordan matrix, making the determinant equal to the product of the eigenvalues on the main diagonal. The difficulty in proving this without showing any attempt is highlighted.
  • #1
k10sstar
1
0
Let A be an n x n matrix. Show that det(A) is the product of all the roots of the characteristic polynomial of A.
 
Physics news on Phys.org
  • #2


State you definitions of det(A) and the characteristic polynomial of A, and show what you have done so far. If your definition of det(A) is the product of all the roots of the characteristic polynomial of A, you are done.
 
  • #3


The quick proof is that any matrix is similar to either a diagonal matrix or a Jordan matrix with its eigenvalues on the main diagonal. In either case, the determinant is just the product of the numbers on the main diagonal- the product of the eigenvalues.

The difficulty with you not showing any attempt yourself is that we don't know what definitions and theorems you have available to prove this.
 

1. How is the determinant related to the characteristic polynomial?

The determinant of a square matrix A is equal to the product of all the eigenvalues of A, which are the roots of the characteristic polynomial.

2. Can you give an example of how the determinant is the product of roots of the characteristic polynomial?

Consider a 2x2 matrix A with eigenvalues λ1 and λ2. The characteristic polynomial is given by (λ-λ1)(λ-λ2), and the determinant of A is (A11 * A22) - (A12 * A21). When we substitute the eigenvalues, we get (λ1 * λ2) - (0 * 0) = λ1 * λ2, which is the product of the roots.

3. How can we use the determinant to find the characteristic polynomial?

By expanding the determinant of a square matrix A, we can get a polynomial expression in terms of λ, which is the characteristic polynomial of A. This polynomial can then be factored to find the eigenvalues, which are the roots of the characteristic polynomial.

4. Is the relationship between the determinant and characteristic polynomial valid for all matrices?

Yes, the relationship holds for all square matrices, regardless of their size or whether they are invertible or not.

5. Can the determinant be used to find all the roots of a characteristic polynomial?

No, the determinant only gives the product of the roots, not the individual roots themselves. To find all the roots, the characteristic polynomial needs to be factored or solved using other methods.

Similar threads

I Prove that the limit of this matrix expression is 0
    • Linear and Abstract Algebra
Replies
5
Views
1K
I Question regarding fundamental region of a lattice
    • Linear and Abstract Algebra
Replies
20
Views
1K
I About permutation acting on the Identity matrix
    • Linear and Abstract Algebra
Replies
7
Views
549
MHB Relations between map and matrix
    • Linear and Abstract Algebra
Replies
10
Views
1K
I Looking for a creative or quick method for finding roots in GF(p^n)
    • Linear and Abstract Algebra
Replies
3
Views
741
I Finding a polynomial that has solution (root) as the sum of roots
    • Linear and Abstract Algebra
Replies
13
Views
1K
I Determinant of a specific, symmetric Toeplitz matrix
    • Linear and Abstract Algebra
Replies
1
Views
897
I I think I've hit upon a very easy proof of impossibility of quintic
    • Linear and Abstract Algebra
Replies
2
Views
950
I Proof that if T is Hermitian, eigenvectors form an orthonormal basis
    • Linear and Abstract Algebra
Replies
3
Views
2K
I Question Regarding Definition of Tensor Algebra
    • Linear and Abstract Algebra
Replies
10
Views
331

Hot Threads

Recent Insights

Back
Top