\[A\]
is \[(det(A^T A))^{1/2}\]
If
\[A\]
is a square matrix then since \[det A = det A^T\]
the absolute determinant of \[A\]
is equal to its determinant. If \[A\]
is a vector then the absolute value of the determinant is just the length of the vector.Suppose
\[A= \left( \begin{array}{ccc} 2 & 3 & 1 \\ 4 & 1 & 2 \end{array} \right) \]
.Then
\[A^T= \left( \begin{array}{cc} 2 & 4 \\ 3 & 1 \\ 1 & 2 \end{array} \right) \]
.\[\begin{equation} \begin{aligned} (det(A^A))^{1/2} &= (det(\left( \begin{array}{ccc}
2 & 4 \\ 3 & 1 \\ 1 & 2 \end{array} \right) \left( \begin{array}{cc} 2 & 3 & 1 \\ 4 & 1 & 2 \end{array} \right) )^{1/2} \\ &= (det (\left( \begin{array}{ccc}
2 \times 2 + 4 \times 4 & 2 \times 3 + 4 \times 1 & 2 \times 1 + 4 \times 2 \\
3 \times 2 + 1 \times 4 & 3 \times 3 + 1 \times 1 & 3 \times 1 + 1 \times 2 \\
1 \times 2 + 2 \times 4 & 1 \times 3 + 2 \times 1 & 1 \times 1 + 2 \times 2 \end{array} \right))^{1/2} \\ &= ( \left| \begin{array}{ccc} 20 & 10 & 10 \\ 10 & 10 & 5 \\ 10 & 5 & 5 \end{array} \right| )^{1/2} \\ &= (20(10 \times 5 - 5 \times 5) -10(10 \times 5 -5 \times 20)+10(10 \times 5 - 10 \times 10))^{1/2} \\ &= 0 \end{aligned} \end{equation}\]