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

[KataKoniK]

Is this proof correct?

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.

 

[HallsofIvy]

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.

 

[AKG]

Well your "proof" is backwards for one, and second it's invalid. It seems you're suggesting that det(X + Y) = det(X) + det(Y). What if X = I₂ (the 2x2 identity matrix) and Y = -I₂?

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^T) = det(A), to get the desired result.

 

[KataKoniK]

Hmm, so is this the correct method of doing it?

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.

 

[TD]

Well, I think it will be enough if you do it on one side

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)

 

[AKG]

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

 

[HallsofIvy]

This proof, as TD wrote:
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 if you make sure each step is reversible! (That's often called "synthetic proof".)

But AKG's suggestion:
det(A^T + B) = det((A^T + B)^T) = det(A^TT + B^T) = det(A + B^T)
is much nicer!

 

[KataKoniK]

Thanks everyone!