Is Det(A+B) Equal to Det(A) + Det(B)?

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Discussion Overview

The discussion revolves around the relationship between the determinant of the sum of two matrices, specifically whether det(A+B) is equal to det(A) + det(B). Participants explore this question in the context of mathematical reasoning and proofs.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that det(A+B) might be greater than det(A) + det(B) but seeks a proof to confirm this assumption.
  • Another participant provides counterexamples showing that neither det(A+B) = det(A) + det(B) nor det(A+B) = det(A) + det(B) holds in general, using specific matrices A and B.
  • A later reply mentions the equality det(xA) = x^n det(A) for a scalar x and matrix A, indicating that the relationship between det(xA) and xdet(A) does not hold under certain conditions.
  • Another counterexample is presented where both matrices A and B have determinants equal to zero, while their sum has a non-zero determinant.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are multiple competing views regarding the relationship between the determinants of the matrices involved. The discussion remains unresolved.

Contextual Notes

Limitations include the dependence on specific examples and the lack of a general proof or disproof for the original question posed. The discussion highlights the complexity of determinant properties without resolving the underlying mathematical questions.

phyin
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det(A+B) ?? det(A) + det(B)

I'm guessing greater than but I'm not too sure. I need a proof on this so I can be assured of it and then use the statement to prove something else.

any hint (or link to proof) would be much appreciated.

edit:

x*det(A) ?? det(x*A)
what's the relation there?
 
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Neither holds, take

[tex]A=B=\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)[/tex]

for one equality and


[tex]A=\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right),~B=-A[/tex]

for the other.

For the last question, we do have the equality

[tex]det(xA)=x^n detA[/tex]

where A is the rank of the matrix. From this we can deduce that neither of the inequalities between det(xA) and xdet(A) holds true. Just take x>1 and x<1.
 


thanks. i better find an alternative way to my proof ;s
 


Another simple counter-example is
[tex]A = \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} \qquad <br /> B = \begin{pmatrix} 0 & 0 \\ 0 & b \end{pmatrix}[/tex]

det(A) = det(B) = 0, det(A+B) = ab
 

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