# Why must the transpose of an invertible matrix be invertible?

Nov 7, 2015

If $A$ has inverse ${A}^{- 1}$ then ${A}^{T}$ has inverse ${\left({A}^{- 1}\right)}^{T}$

#### Explanation:

If you are happy to accept that ${A}^{T} {B}^{T} = {\left(B A\right)}^{T}$ and ${I}^{T} = I$, then the proof is not difficult:

Suppose $A$ is invertible with inverse ${A}^{- 1}$

Then:

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

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

So ${\left({A}^{- 1}\right)}^{T}$ satisfies the definition for being an inverse of ${A}^{T}$