Prove that $\det(A A^t) =0$ for every $4 \times 3$ matrix $A$ I try to prove that the determinant of $A A^t$ equals $0$.  $A^t=(A \text{ transpose})$.
I tried to prove it by contradiction. if $\det(A A^t )\neq 0$ so there exists an inverse, so I tried to multiply the equation but it didn't go well.
 A: Since $A$ has $3$ columns, it has a rank of at most $3$.
Therefore the matrix $A A^T$ also has a rank of at most $3$.
As $A A^T$ is a $4\times 4$ matrix, it must be singular.
Thus its determinant is $0$.
A: If $A$ is a $4\times3$ matrix, then $A^T$ is a $3\times4$ matrix. So, you can see $A^T$ is the matrix of a linear map from $\mathbb{R}^4$ into $\mathbb{R}^3$ and you can see $A$ is the matrix of a linear map from $\mathbb{R}^3$ into $\mathbb{R}^4$. The range of the composition of these linear maps is $3$, at most, and therefore $\det(A.A^T)=0$.
A: It can be also visible from the vector representation of the matrix $A$ (suppose general case of dimension $n \times 3$ where $n > 3$) and block operations. 
Let $A=\begin{bmatrix} v_1 & v_2 & v_3\end{bmatrix}$ and
$A^T=\begin{bmatrix} v_1^T \\ v_2^T \\ v_3^T\end{bmatrix}$.
Then $AA^T=    v_1v_1^T + v_2v_2^T + v_3v_3^T $ which as a sum of scaled projections on three vectors has at most rank $3$.
Notice that maximal   rank of $AA^T$ is independent from dimension of the vector $v_i$ if only this dimension   $n \geq 3$.
Dimension  of $AA^T$ is $n \times n$ hence if $n>3$ matrix $AA^T$ must be singular.
