When is A transpose A invertible?

A rectangular matrix can produce a square matrix when we multiply it by its transpose. The order determines the size of that square matrix. Let’s find one of those products and see how independence tells us whether it has an inverse.

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Check the sizes of the two products

If $A$ is $m\times n$, then $A^{T}A$ is $n\times n$ and $A{A}^T$ is $m\times m$. Both products are symmetric.

$A=\left[\begin{array}{rr}1& 1\\ 2& 0\\ -1& -1\end{array}\right]$

Multiply the transpose by the original matrix

The columns of $A$ are $(1,2,-1)$ and $(1,0,-1)$. Each entry in $A^{T}A$ is the dot product of a pair of these columns.

$A^{T}A=\left[\begin{array}{rr}6& 2\\ 2& 2\end{array}\right]$

$\det \left(A^{T}A\right)=6\left(2\right)-2\left(2\right)=8$

The determinant is nonzero, so this product is invertible.

Connect invertibility with column independence

The identity $x^T{A}^{T}Ax=\Vert Ax{\Vert }^2$ shows that $A^{T}Ax=0$ exactly when $Ax=0$. Thus $A^{T}A$ is invertible exactly when the columns of $A$ are independent.

For the other order, $A{A}^T$ is invertible exactly when the rows of $A$ are independent. Here $A$ has three rows and two columns, so $A{A}^T$ is singular.

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