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My question is simple (and maybe I am wrong asking this question even) what is the geometrical interpretation of determinant of a matrix in general ? I could not think anything.


marked as duplicate by Hans Lundmark, choco_addicted, user91500, Joel Reyes Noche, Watson May 31 '16 at 11:16

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In Lax's Linear Algebra and it's Applications, he furnishes an interpretation of determinants as closely related to the volume of a "simplex".

Kind of like how an n-cube is a generalization of a square and a cube, and an n-ball is a generalization of circles and spheres, a simplex is a generalization of triangles and pyramids.

If you have a simplex of dimension $n$ and one of its sides lies in the space of the other sides, it's a "degenerate" simplex, and it has zero volume, just like that of a triangle and a pyramid. So, if you view the columns of your matrix as vectors, and one of your vectors is linearly dependent, the simplex constructable through the vectors is degenerate, so the matrix has determinant $0$.

If you calculate the volume of a triangle from a base lying on an "axis", but then decide to use a different side as the base, then you have to rotate it, which places it beneath the previous axis and turns the volume negative, just as how switching columns in a matrix causes the determinant to be negative.

Once you manage to wrap your head around of the columns of a matrix being the sides of a triangle/n-dimensional simplex, it generally makes the determinant make much more sense.


$n\times n$ matrix consist of $n$ vectors in $\Bbb{R}^n$. These $n$ vectors form a parallelogram, the determinant of the matrix is the volume of that parallelogram. For example, $$\begin{pmatrix} 1&0&0\\ 0&5&0\\ 0&0&9\end{pmatrix}$$ gives us three vectors in $\Bbb{R}^3$, one is on the $x$ axes of length $1$, the other is on the $y$ axes of length $5$ and the last is on the $z$ axes of length $9$. These three vectors defined a box of dimensions $1\times5\times9$, whose volume is equal to the determinant $45$.

  • $\begingroup$ maybe the (relative) volume of the image of the unit ball by the linear operator is even clearer, since it is obvious that this is independent of the choice of (orthonormal) basis ? $\endgroup$ – reuns May 31 '16 at 5:00
  • $\begingroup$ (and there is the problem of the oriented volume, since what I said is only true for $|det(A)|$) $\endgroup$ – reuns May 31 '16 at 6:19
  • $\begingroup$ I'm aware of all these problems. But, OP asked for a simple geometric intuition and the image of the unit ball by the linear operator is not particularly straight-forward to me. $\endgroup$ – Emre May 31 '16 at 7:24

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