- #1

ltrane2003

- 1

- 0

## Homework Statement

Prove that the product of the diagonal entries of an nxn matrix A equals the product of the eigenvalues of A.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter ltrane2003
- Start date

- #1

ltrane2003

- 1

- 0

Prove that the product of the diagonal entries of an nxn matrix A equals the product of the eigenvalues of A.

- #2

AstroRoyale

- 110

- 0

You have to show at least some attempt first before anyone will help.

- #3

DavidWhitbeck

- 351

- 1

(a) this is a standard textbook result, see if it's in your book first

(b) examine the coefficients of the characteristic polynomial carefully

- #4

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 971

The very quickest would involve using the "Jordan Normal Form"- do you know what that is?

- #5

Dick

Science Advisor

Homework Helper

- 26,263

- 621

- #6

rock.freak667

Homework Helper

- 6,223

- 31

## Homework Statement

Prove that the product of the diagonal entries of an nxn matrix A equals the product of the eigenvalues of A.

That is not true for all matrices...only in certain types of matrices is that true.

- #7

AstroRoyale

- 110

- 0

- #8

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 971

- #9

Vid

- 402

- 0

Both the eigenvalues are zero.

For a diagonal matrix the determinant is just the product of the diagonals.

The eiganvalues are |xI-A|...

- #10

- #11

Vid

- 402

- 0

Diagonal matrices have 0 everywhere but the diagonal. Your matrix is symmetric.

- #12

DavidWhitbeck

- 351

- 1

Yeah I totally screwed up, the OP never returned but I'll just say what I was thinking-- the constant term in the characteristic polynomial is known to be the determinant of the matrix, call it A, but also if you factor the polynomial knowing that it's roots are the eigenvalues [tex]\lambda_i[/tex] then the constant term is also

[tex](-1)^N\prod_{i=1}^{N}\lambda_i[/tex]

and there you have it--

[tex]\prod_{i=1}^{N}\lambda_i = (-1)^N\det A[/tex]

where A is an N by N matrix.

- #13

Dick

Science Advisor

Homework Helper

- 26,263

- 621

Yeah I totally screwed up, the OP never returned but I'll just say what I was thinking-- the constant term in the characteristic polynomial is known to be the determinant of the matrix, call it A, but also if you factor the polynomial knowing that it's roots are the eigenvalues [tex]\lambda_i[/tex] then the constant term is also

[tex](-1)^N\prod_{i=1}^{N}\lambda_i[/tex]

and there you have it--

[tex]\prod_{i=1}^{N}\lambda_i = (-1)^N\det A[/tex]

where A is an N by N matrix.

Now you have to explain away that (-1)^N. It shouldn't be there. If I is the NxN identity. det(I)=1 and the product of the eigenvalues is certainly 1.

- #14

DavidWhitbeck

- 351

- 1

[tex]\prod_{i=1}^{N}\lambda_i = \det A[/tex]

Share:

- Last Post

- Replies
- 1

- Views
- 646

- Replies
- 5

- Views
- 341

- Replies
- 15

- Views
- 861

- Replies
- 8

- Views
- 803

- Last Post

- Replies
- 8

- Views
- 207

- Last Post

- Replies
- 4

- Views
- 418

- Last Post

- Replies
- 6

- Views
- 546

- Replies
- 2

- Views
- 670

- Replies
- 18

- Views
- 503

- Last Post

- Replies
- 8

- Views
- 943