I have seen that the $k$-dimensional volume of an parallelepiped in $\mathbb{R}^n$, i.e., $$P(v_1, \ldots, v_k) = \{t_1v_1 + \dotsb + t_kv_k : 0 \le t_i \le 1 \}$$ is $\sqrt{\det(T^{\top}T)}$, where $T$ is the $n\times k$ matrix with columns $v_1, \ldots, v_k$.

How do we know that $\det(T^{\top}T)$ is non-negative?

  • 4
    $\begingroup$ The Cauchy-Binet identity shows that $\det\left(T^T T\right)$ is the sum of the squares of all $k\times k$ minors of $T$. Sums of squares are nonnegative. $\endgroup$ Apr 25, 2016 at 3:24
  • $\begingroup$ Look at SVD decomposition, which always exist: $T = U\Sigma V^T$ where $\Sigma$ is diagonal and $U, V$ are unitray you get $T^\perp T = V\Sigma^T U^T U\Sigma V^T = V\Sigma^T \Sigma V^T = V \Sigma^2 V^T$ (using the fact that $U^TU = I$).Therefor $\det(T^\perp T) = \det(V) \det(\Sigma^2) \det(V^T) = \det(\Sigma^2) $ using the fact that determinat of unitary matrix is $1$. The result follows since $\Sigma^2$ is diagonal matrix with non-negative values. $\endgroup$
    – them
    Apr 25, 2016 at 12:08
  • $\begingroup$ You may also be interested in this question: matheducators.stackexchange.com/q/10395/117 $\endgroup$ Apr 25, 2016 at 13:56

1 Answer 1


$T^{\top}T$ is positive semidefinite, so all the eigenvalues are non-negative.

The determinant of $T^{\top}T$ is the product of the eigenvalues; hence, it is non-negative.

  • $\begingroup$ So as it is positive semi-definite $det(T^TT)$ can be equal to zero? Is this true over $\mathbb{C}^n$ also? $\endgroup$
    – josh
    Apr 25, 2016 at 9:17
  • 4
    $\begingroup$ @josh Over $\mathbb{C}$ you can do it with the Hermitian transpose. $\endgroup$
    – egreg
    Apr 25, 2016 at 10:30

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