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Here's the question in two parts.

Part One.

Suppose $A$ is an invertible matrix.
Explain why $A^{T}A$ is also invertible.

Part Two.

Then show that $A^{-1} = (A^{T}A)^{-1}A^{T}.$

I understand part one. Because of the fact that $det(A^{T}A) = det(A^{T}) * det(A)$,

and because the determinant of the a matrix's transpose is equal to the determinant of the original matrix, you can determine that $A^{T}A$ is transposable.

I don't understand where exactly to begin with the second part.

Ryan Moeller
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3 Answers3

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Hint: Show that $A^{-1} A = I$ where $A^{-1}$ is as given in the question.

$A^{-1} A = ((A^\top A)^{-1} A^\top) A = (A^\top A)^{-1} (A^\top A) = I$.

angryavian
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Because if $A$ and $B$ are invertible $n\times n$ matrices then $AB$ is also invertible and $(AB)^{-1}=B^{-1}A^{-1}$. The proof is very simple, just multiply: $(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AA^{-1}=I$.

So now you have $(A^TA)^{-1}A^T=A^{-1}(A^T)^{-1}A^T=A^{-1}$.

Mark
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The basic fact (cf. If $AB=I$ then $BA=I$) is that if $X$ is a square matrix whose entries are taken from a field, then $X^{-1}=Y$ iff $YX=I$. That is, $X$ has a two-sided inverse $Y$ if and only if $Y$ is a left-inverse of $X$.

So, to show that $\underbrace{A^{-1}}_{X^{-1}} = \underbrace{(A^TA)^{-1}A^T}_Y$, it suffices to prove that $\underbrace{(A^TA)^{-1}A^T}_Y\underbrace{A}_X=I$.

user1551
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