If A and B are both invertible, then $[(A^{-1})B]^T$ is invertible.

Question

If $A$ and $B$ are both invertible, then $[A^{-1}B]^T$ is invertible.

Is this true or false? I have tried coming up with a proof but I don't really know where to start.

Answer 1

$A, B, AB$ are invertable if their respective determinants don't equal 0.

Now what ruled do you know about determinants, products, inverses and transposes.

i.e. $det(AB) = det(A)det(B)$ and $(A^{-1} B)^T = B^T(A^{-1})^T$

etc.

You should have enough to show that $det (A^{-1} B)^T \ne 0$

Answer 2

Hint: Multiply by $[B^{-1}A]^T$, remembering that $[XY]^T=Y^TX^T$.

Answer 3

The product of invertible matrices is invertible; the transpose of an invertible matrix is invertible.

Answer 4

If $A$ and $B$ are both invertible, $(AB)^t=B^tA^t$, then the inverse of $(AB)^t$ is $(B^{-1}A^{-1})^t=(A^{-1})^t(B^{-1})^t$ since

\begin{align} (AB)^t\cdot (B^{-1}A^{-1})^t&=(B^tA^t)\cdot((A^{-1})^t(B^{-1})^t)\\ &=B^t(A^t\cdot (A^{-1})^t)(B^{-1})^t\\ &=B^t(A^{-1}A)^t(B^{-1})^t\\ &=B^t I^t (B^{-1})^t\\ &=B^t I (B^{-1})^t\\ &=B^t \cdot (B^{-1})^t\\ &=(B^{-1}B)^t\\ &=I^t\\ &=I \end{align} Now for your question amend the above with $A$ replaced by $A^{-1}$. Also note a matrix $A$ is invertible iff its determinant, $\det(A)\neq0$, and then $\det(A^{-1})=(\det(A))^{-1}$. Plus $\det(A)=\det(A^t)$ and $\det(AB)=\det(A)\det(B)$.

Answer 5

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Assume $\ds{C \equiv\pars{A^{-1}B}^{T}}$ is $\color{#f00}{not}$ invertible. It means $\ds{\det\pars{C} = 0}$ Then, you'll arrive to a contradiction because

\begin{align} AC^{T} & = B\quad\implies\quad\det\pars{A}\det\pars{C} = \det\pars{B} = 0 \end{align}