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for $n = 2,$ I can visualize that the determinant $n \times n$ matrix is the area of the parallelograms by actually calculate the area by coordinates. But how can one easily realize that it is true for any dimensions?

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If the column vectors are linearly dependent, both the determinant and the volume are zero. So assume linear independence. The determinant remains unchanged when adding multiples of one column to another. This corresponds to a skew translation of the parallelepiped, which does not affect its volume. By a finite sequence of such operations, you can transform your matrix to diagonal form, where the relation between determinant (=product of diagonal entries) and volume of a "rectangle" (=product of side lengths) is apparent.

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    $\begingroup$ thanks. from an constructive view of point, can the skew translation does not change volume plus multilinear condition fully determine the form of determinant? $\endgroup$ – ahala Jun 23 '13 at 15:34
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    $\begingroup$ Yes. In abstract language: The vector space of alternating $n$-forms is one-dimensional $\endgroup$ – Hagen von Eitzen May 2 '14 at 11:58
  • $\begingroup$ I don't think the relation is apparent, even in two dimensions. The proof by picture in 2d takes some (pretty little) movement of equal areas around for the second shear. $\endgroup$ – Mitch Oct 19 '18 at 19:52
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Here is the same argument as Muphrid's, perhaps written in an elementary way.

Apply Gram-Schmidt orthogonalization to $\{v_{1},\ldots,v_{n}\}$, so that \begin{eqnarray*} v_{1} & = & v_{1}\\ v_{2} & = & c_{12}v_{1}+v_{2}^{\perp}\\ v_{3} & = & c_{13}v_{1}+c_{23}v_{2}+v_{3}^{\perp}\\ & \vdots \end{eqnarray*} where $v_{2}^{\perp}$ is orthogonal to $v_{1}$; and $v_{3}^{\perp}$ is orthogonal to $span\left\{ v_{1},v_{2}\right\} $, etc.

Since determinant is multilinear, anti-symmetric, then \begin{eqnarray*} \det\left(v_{1},v_{2},v_{3},\ldots,v_{n}\right) & = & \det\left(v_{1},c_{12}v_{1}+v_{2}^{\perp},c_{13}v_{1}+c_{23}v_{2}+v_{3}^{\perp},\ldots\right)\\ & = & \det\left(v_{1},v_{2}^{\perp},v_{3}^{\perp},\ldots,v_{n}^{\perp}\right)\\ & = & \mbox{signed volume}\left(v_{1},\ldots,v_{n}\right) \end{eqnarray*}

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    $\begingroup$ Very nice explanation. Just an addition for the last equality. If $V$ is the matrix with columns $v_{1},v_{2},v_{3},\ldots,v_{n}$ then, because the columns are othogonal to each other, $U U^T = diag(||v_1||^2, \ldots, ||v_2||^2) $. Thus $|U U^T| = ||v_1||^2 \ldots ||v_n||^2$, which, given that $|U U^T| = |U| |U^T| = |U|^2$ yields $ |U| = ||v_1|| \ldots ||v_n|| = \mbox{signed volume}\left(v_{1},\ldots,v_{n}\right) $. $\endgroup$ – Sotiris Mar 9 at 22:47
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In 2d, you calculate the area of a parallelogram spanned by two vectors using the cross product. In 3d, you calculate the volume of a parallelepiped using the triple scalar product. Both of these can be written in terms of a determinant, but it's probably not clear to you what the proper generalization is to higher dimensions.

That generalization is called the wedge product. Given $n$ vectors $v_1, v_2, \ldots, v_n$, the wedge product $v_1 \wedge v_2 \wedge \ldots \wedge v_n$ is called an $n$-vector, and it has as its magnitude the $n$-volume of that $n$-parallelepiped.

What is the relationship between the wedge product and the determinant? Quite simple, actually. There is a natural generalization of linear maps to work on $k$-vectors. Given a linear map $\underline T$ (which can be represented as a matrix), the action of that map on a $k$-vector is defined as

$$\underline T(v_1 \wedge v_2 \wedge \ldots \wedge v_k) \equiv \underline T(v_1) \wedge \underline T(v_2) \wedge \ldots \wedge \underline T(v_k)$$

When talking about $n$-vectors in an $n$-dimensional space, it's important to realize that the "vector space" of these $n$-vectors is in fact one-dimensional. That is, if you think about volume, there is only one such unit volume in a given space, and all other volumes are just scalar multiples of it. Hence, when we talk about the action of a linear map on an $n$-vector, we can see that

$$\underline T(v_1 \wedge v_2 \wedge \ldots \wedge v_n) = \alpha [v_1 \wedge v_2 \wedge \ldots \wedge v_n]$$

for some scalar $\alpha$. In fact, this is a coordinate system independent definition of the determinant!

When you build a matrix out of $n$ vectors $f_1, f_2, \ldots, f_n$ as the matrix's columns, what you're really doing is the following: you're saying that, if you have a basis $e_1, e_2, \ldots, e_n$, then you're defining a map $\underline T$ such that $\underline T(e_1) = f_1$, $\underline T(e_2) = f_2$, and so on. So when you input $e_1 \wedge e_2 \wedge \ldots \wedge e_n$, you get

$$\underline T(e_1 \wedge e_2 \wedge \ldots \wedge e_n) = (\det \underline T) e_1 \wedge e_2 \wedge \ldots \wedge e_n= f_1 \wedge f_2 \wedge \ldots \wedge f_n$$

This is how you can use a matrix determinant to calculate volumes: it's just an easy way of constructing something that automatically computes the wedge product.


Edit: how one can see that the wedge product accurately gives the volume of a parallelepiped. Any vector can be broken down into perpendicular and parallel parts with respect to another vector, to a plane, and so on (or to any $k$-vector). As such, if I have two vectors $a$ and $b$, then the wedge product $a \wedge b = a \wedge b_\perp$, where $b_\perp$ is effectively the height of the parallelogram. Similarly, if I construct a parallelepiped with a vector $c$, then the wedge product $a \wedge b \wedge c = (a \wedge b_\perp) \wedge c_\perp$, where $c_\perp$ lies entirely normal to $a \wedge b_\perp$. So we can recursively do this for any $k$-vector, looking at orthogonal vectors instead, which is much simpler to see the volumes from.

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  • $\begingroup$ Thanks. Actually my question raised exactly from learning about wedge product. How can one see that the wedge product has its magnitude the n-volume of that n-parallelepiped? I guess it is equivalent to ask in term of determinant as in my question. $\endgroup$ – ahala Jun 23 '13 at 15:39
  • $\begingroup$ I've added a section to this effect. $\endgroup$ – Muphrid Jun 23 '13 at 16:27
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The determinant of a matrix A is the unique function that satisfies:

  1. $\det(A)=0$ when two columns are equal
  2. the determinant is linear in the columns
  3. if A is the identity $\det(A)=1$.

You can easily convince yourself that the oriented volume $\operatorname{vol}(v_1,v_2,\ldots,v_n)$ between $v_1, v_2,\ldots, v_n$ vectors is a function that satisfies exactly the same properties if we place the vectors as the columns of a matrix $A=(v_1,\ldots,v_n)$. Hence $\operatorname{vol}(v_1,v_2,\ldots,v_n)=\det(A)$.

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    $\begingroup$ one has to be convinced why such function is unique to reach the conclusion. $\endgroup$ – ahala Dec 12 '16 at 14:37
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    $\begingroup$ True. You can prove that the determinant is unique by constructing it using Gaussian elimination as the signed product of all the pivot of a matrix. $\endgroup$ – jacopoviti Dec 13 '16 at 14:59
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Determinant involves a cross-product of the first two vectors and a dot of the result with the third. The result of a cross product is a vector whose magnitude is the area of its null space. Said simply, any plane in 3D is the null space of its normal.The size of the plane is defined by the length of the normal. The volume is found by projecting this normal onto the third vector.

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  • $\begingroup$ This question is not about 3D case, it is about nD cases. $\endgroup$ – ahala Dec 12 '16 at 14:39
  • $\begingroup$ Yes, my bad. Assuming from nD we get an nxn matrix. Remove one row and then find the null space of the remaining n-1xn hyperplane which is inevitably a vector perpendicular to that hyperplane. Dot this vector with the one removed and I believe this amounts to the volume in nD. $\endgroup$ – Shadi Dec 26 '16 at 3:14
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You can also invoke the change-of-variable theorem in higher dimensions. A $n$-dimensional parelellopiped $\mathcal P=\mathcal P(a_1,\dots,a_n)$ in $\mathbb R^n$ (where the $a_i$ are independent vectors in $\mathbb R^n$) is the set of all $x$ such that: $$ x=c_1a_1+\dots+c_ka_n, $$ with $0\leq c_i\leq 1$. We can define the linear transformation $h(x)=A\cdot x$, where $A$ is the $n\times n$ matrix with $a_i$ as its columns. This gives us $\mathcal P=h([0,1]^n)$. The volume of $h([0,1]^n)$ is equal to $h((0,1)^n))$ (those sets are equal modulo a set of measure zero), so we can apply the change-of-variable theorem: $$ v(\mathcal P)=\int_{h((0,1)^n)}1=\int_{(0,1)^n}\vert\det Dh\vert=\vert\det A\vert. $$

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    $\begingroup$ Isn't the change of variables theorem based on what we are asked to prove? $\endgroup$ – Theorem Aug 16 '18 at 14:56

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