# When this hold: $\det(AA^{T})=0?$

I know if A is a column vector, the equality holds. Any comments or suggestions would be greatly appreciated.

• Let $A$ be a square real matrix. $det(AA^T)=\left(det(A)\right)^2=0$ iff $det(A)=0$ – llllllllllllllllllllllllllllll Dec 24 '13 at 20:54

The determinant of a square matrix is nonzero if and only if the matrix is invertible (or, in other terminology, nonsingular).

If $A$ is a generic matrix with real coefficients, not necessarily square, we know that $AA^T$ is invertible if and only if $A$ has a right inverse. So the condition for having $\det(AA^T)=0$ is that the matrix doesn't have a right inverse, which can also be stated in the form “the matrix $A$ has rank less than the number of its rows”.

I will assume we're working over the real numbers. I will prove egreg's answer in that case.

$AA^T$ is invertible if and only if $YAA^T = 0$ implies $Y = 0$ for every row matrix $Y$.

Obviously if $YA = 0$ for a particular row matrix $Y$, then we will have $YAA^T = 0$. I claim that the converse is true.

For if $YAA^T = 0$, then $(YA)(YA)^T = YAA^T Y^T = 0$. Thus the row vector $YA$ is orthogonal to itself. (Its dot product with itself is zero.) Over $\mathbf{R}$, this implies $YA = 0$.

Thus the condition for $AA^T$ being invertible can be rephrased as: $YA = 0$ implies $Y=0$. This amounts to saying that the rows of $A$ are linearly independent. If $A$ is $m \times n$, this is the same as saying that $A$ has rank $m$.

Thus $\det(AA^T) = 0$ if and only if the rank of $A$ is less than the number of rows.