condition on the dimensions of a matrix for its inverse to exist
Let $A$ be an $m\times n$ matrix, what condition on the dimensions $m$ and $n$ is necessary for the quantity $(A^t\times A)^{-1}$ to exist? Please kindly provide your explanation.
Many thanks!
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$\begingroup$Let $A$ be a real matrix of any size. (This also works with complex matrices if transpose is replaced with conjugate transpose. The proof is the same.) Then $A^TA$ is a bijection (has an inverse) if and only if $A$ is an injection.
Suppose $A$ is an injection. If $A^TAx=0$ for some $x$, then $$ 0 = x^T0 = x^TA^TAx = (Ax)^T(Ax) = |Ax|^2 $$ and so $Ax=0$. By injectivity of $A$ this implies $x=0$. Therefore $A^TA$ is injective. But $A^TA$ is a square matrix, so injectivity implies bijectivity.
Suppose $A$ is not injective. Then there is an $x\neq0$ so that $Ax=0$. But then also $A^TAx=A^T0=0$, so $A^TA$ is not injective and thus not bijective.
$\endgroup$ 9 $\begingroup$For the inverse of any matrix to exist is that the matrix must be square. As long as [A(Transpose)*A] is a matrix of m x n dimensions where m = n than an inverse can exist.
Hope this helps.
$\endgroup$ 6 $\begingroup$If $A$ has size $m\times n$ then $A^T\cdot A$ has size $n\times n$ and rank at most $\min(m,n)$. Since being invertible means having rank$~n$, a necessary condition for invertibility is $n=\min(m,n)$ or equivalently $m\geq n$. This condition is not sufficient. One has $\def\rk{\operatorname{rk}}\rk(A^T\cdot A)\leq\rk A=\rk(A^T)$, so it is also necessary that $A$ has (full column) rank$~n$, which means it (or the associated linear map) is injective.
Over arbitrary fields that condition is still not sufficient, since even if $A$ is injective one has $\dim(\ker(A^T\cdot A))=\dim(\operatorname{Im}(A)\cap\ker(A^T))$ and that intersection might be nonzero making $A^T\cdot A$ non-invertible. However when the field is $\Bbb Q$ or $\Bbb R$, the standard bilinear form $x^T y$ is positive definite, and $\ker(A^T)$ is the orthogonal complement for that form of $\operatorname{Im}(A)$, which forces the intersection to be zero. So if the base field is $\Bbb Q$ or $\Bbb R$, then the condition $\rk(A)=n$ (which implies $m\geq n$) is not only necessary but also sufficient for $A^T\cdot A$ to be invertible.
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