- #1

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Show that det(A + B^T) = det(A^T + B)

det(A + B^T) = det(A^T + B)

det(A) + det(B^T) = det(A^T) + det(B)

det(A) + det(B) = det(A) + det(B)

Thanks.

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- Thread starter KataKoniK
- Start date

- #1

- 168

- 0

Show that det(A + B^T) = det(A^T + B)

det(A + B^T) = det(A^T + B)

det(A) + det(B^T) = det(A^T) + det(B)

det(A) + det(B) = det(A) + det(B)

Thanks.

- #2

HallsofIvy

Science Advisor

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det(A+B) is **not**, in general, equal to det(A)+ det(B).

What**is** true is that (A+B)^{T}= A^{T}+ B^{T} so that, in particular (A^{T}+ B)^{T}= A+ B^{T}.

What

Last edited by a moderator:

- #3

AKG

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The determinant is not a linear function, however transposition is linear. Use that fact plus the trivial fact that matrix addition is commutative, plus the fact that the det(A

- #4

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det(A + B^T) = det(A^T + B)

det(A + B^T)^T = det(A^T + B)^T

det(A^T + B) = det(A + B^T)

I can't do this, correct? Thanks for the hints though.

- #5

TD

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det(A + B^T) = det(A^T + B)

det(A + B^T)^T = det(A^T + B)

det(A^T + B) = det(A^T + B)

- #6

AKG

Science Advisor

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det(A^{T} + B) = det((A^{T} + B)^{T}) = det(A^{TT} + B^{T}) = det(A + B^{T})

- #7

HallsofIvy

Science Advisor

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det(A + B^T) = det(A^T + B)

det(A + B^T)^T = det(A^T + B)

det(A^T + B) = det(A^T + B)

which starts with what you want to show and arrives at a true statement is valid

But AKG's suggestion:

det(A

is much nicer!

- #8

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Thanks everyone!

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