Thread: Proof with transpose - I think this is right

1. Proof with transpose - I think this is right

Hello everyone. I've got a proof done. May someone please just do a scan of the proof to see if it's correct?

Prove:

$(AB)^T$ is nonsingular

and

$((AB)^T)^{-1} = (A^{-1})^T(B^{-1})^T$

The problem says (in its "given") that A and B are invertible. (There's nothing else "given.") Hence,

$(AA^{-1}) = I$ and $(BB^{-1}) = I$

$(BB^{-1})(AA^{-1}) = I$

$((BB^{-1})(AA^{-1}))^T = I^T$

$(((B^{-1})^T)B^T)(((A^{-1})^T)A^T) = I$ by the property that $(AB)^T = B^TA^T$

$((B^{-1})^T)((A^{-1})^T)(AB)^T = I$ by the property that $(AB)^T = B^TA^T$ and the associativity of matrix multiplication.

Then, multiply both sides by $((AB)^T)^{-1}$ to obtain

$((AB)^T)^{-1} = (A^{-1})^T(B^{-1})^T$

since the inverses cancel out to get I. This also proves that there exists an inverse for the AB transposed matrix.

(On an actual problem I'd be more explicit in my steps, though, perhaps splitting up the associativity and (AB)T => BTAT into two steps.)

2. If you have these facts available, they will give you a much shorter proof of the first identity:

det(AB)=det(A)det(B)
det(A^t)=det(A)