When will $\operatorname{det}\left(A\cdot A^{\top}\right)=0$?

Question

I am writing a small computer program to solve certain linear algebra equations as part of a larger program. For two of my functions I need to evaluate $\left(A\cdot A^{\top}\right)^{-1}$. This got me wondering. Is there any significance to $\operatorname{det}\left(A\cdot A^{\top}\right)=0$ ? When will this happen, and is it common in the real world? In my particular case it is unlikely that $A$ itself would ever be a square matrix. Just throwing that out there.

Answer 1

If $A$ is $m\times n$ real matrix and $\operatorname{rank}A = k\le\min(m,n)$, then $AA^T$ is $m\times m$ matrix of rank $k$. Thus $AA^T$ is invertible if and only if $\operatorname{rank}A = m$ (in particular: if $m > n$, then $AA^T$ must be singular.)

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For complex matrices it does not hold in general, as baharampuri suggests. However for complex matrix $A$, $\operatorname{rank} A = \operatorname{rank} AA^*$ holds where $A^*$ is the conjugate transpose of $A$.

Answer 2

For A being a n x n matrix : det(A * A^T)=det(A)*det(A^T)=det(A)^2 and det(A)=0 implies that A is not invertible.

Answer 3

In the case that $A$ is a $n \times m$ matrix with $n \neq m$, then $A \cdot A^T$ is a $n \times n$ matrix. It is invertible if and only if its rank is $n$.

If $m < n$ then $$\operatorname{rank}(A \cdot A^T) \le m < n$$ so you have a singualr matrix.

If $n <m$ then $$\operatorname{rank}(A \cdot A^T) \le n$$ and equality holds if and only if $\operatorname{rank}(A) = n$.

Answer 4

Let $A_{m,n}$ be our matrix, think of $A$ as a map from $\mathbb{R}^{m} \rightarrow \mathbb{R}^n$, Now $A^{t}A$ could be thought of as a map from $\mathbb{R}^m \rightarrow \mathbb{R}^m$. This map is singular if and only if the determinant $det(AA^t)=0$. This determinant is zero if and only if the rank of this matrix is less than $m$. Which happens if and only if $rank(A)- dim( Im(A) \cap ker (A^t)) < m$. But $rank(A) \leq m$, so the previous condition happens if and only if $dim(Im(A) \cap Ker(A^t)) \neq 0$ if $rank(A)=m$. That is $Im(A) \cap Ker(A^t) $ is a non-trivial subspace if $rank(A)=m$. Or $rank(A) <m$.