Linear Algebra Proof Help

  • Thread starter ltrane2003
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  • #1
ltrane2003
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Homework Statement


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



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The Attempt at a Solution

 

Answers and Replies

  • #2
AstroRoyale
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You have to show at least some attempt first before anyone will help.
 
  • #3
DavidWhitbeck
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Since you haven't tried the problem I won't provide much help, but I will say a couple of vague hints

(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
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The problem with you not showing any work is that we have no idea what kind of facts you have to work with. I can think of several different ways of proving that, depending upon how "sophiticated" you want to be.

The very quickest would involve using the "Jordan Normal Form"- do you know what that is?
 
  • #5
Dick
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But it's not even true!? Take A=[[0,1],[1,0]]. Product of the eigenvalues is -1. Product of the diagonal entries is 0. Am I missing something?
 
  • #6
rock.freak667
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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
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The product of the eigenvalues is the determinant, right. The statement would be true for a diagonal matrix for sure :)
 
  • #8
HallsofIvy
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Yes, it is true (and is trivial) for every diagonal matrix. But that is a very small subset of all matrices.
 
  • #9
Vid
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But it's not even true!? Take A=[[0,1],[1,0]]. Product of the eigenvalues is -1. Product of the diagonal entries is 0. Am I missing something?

Both the eigenvalues are zero.

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

The eiganvalues are |xI-A|...
 
  • #10
Dick
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Both the eigenvalues are zero.

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

The eiganvalues are |xI-A|...

A*[1,1]=[1,1]. That doesn't look like a zero eigenvector to me.
 
  • #11
Vid
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Yea, my statement about your matrix was wrong.

Diagonal matrices have 0 everywhere but the diagonal. Your matrix is symmetric.
 
  • #12
DavidWhitbeck
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The product of the eigenvalues is the determinant, right. The statement would be true for a diagonal matrix for sure :)

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
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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
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Ah yeah the constant term in the characteristic polynomial is not [tex]\det A[/tex] it's [tex](-1)^N\det A[/tex] and then we have instead

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

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