Is the transpose of an invertible matrix also invertible?

Oct 27, 2015

Yes

Explanation:

In general

${A}^{T} {B}^{T} = {\left(B A\right)}^{T}$

So in particular, if $A$ is invertible:

${A}^{T} {\left({A}^{-} 1\right)}^{T} = {\left({A}^{-} 1 A\right)}^{T} = {I}^{T} = I$