# Determinant proof

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.

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

What is true is that (A+B)T= AT+ BT so that, in particular (AT+ B)T= A+ BT.

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Homework Helper
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(AT) = 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.

Homework Helper
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)

Homework Helper
det(AT + B) = det((AT + B)T) = det(ATT + BT) = det(A + BT)

Homework Helper
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(AT + B) = det((AT + B)T) = det(ATT + BT) = det(A + BT)
is much nicer!

KataKoniK
Thanks everyone!