# If A^T is invertible, is A invertible? What about A^TA?

Nov 7, 2015

Yes and yes

#### Explanation:

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

For any square matrices $A$ and $B$, ${A}^{T} {B}^{T} = {\left(B A\right)}^{T}$

Then:

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

And:

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

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

Then we find:

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

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

And:

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

$= {A}^{- 1} I A = {A}^{- 1} A = I$

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