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. $\endgroup$ – darij grinberg Apr 25 '16 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 '16 at 12:08
  • $\begingroup$ You may also be interested in this question: matheducators.stackexchange.com/q/10395/117 $\endgroup$ – Steven Gubkin Apr 25 '16 at 13:56

$T^{\top}T$ 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 '16 at 9:17
  • 4
    $\begingroup$ @josh Over $\mathbb{C}$ you can do it with the Hermitian transpose. $\endgroup$ – egreg Apr 25 '16 at 10:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.