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Math Help - Determinants as scalings factors and cross products

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    Determinants as scalings factors and cross products

    In two dimensions, it is easy to explain why the determinant of a linear transformation works as a scaling factor of parallellograms, by showing that the formula is the same as the cross product.

    Likewise, for three dimensions, the determinant could be shown to be a scaling factor by showing how it relates to the scalar triple vector product.

    In 4 dimensions, I've read that the determinant still works as a scaling factor, though I have no idea why.

    My problem with using the cross product as an explanation, is that the cross product "itself" is written as a determinant, so the proof just appears circular to me.

    What I'm wondering, is if there is some deeper connection between the formula for the determinant(as a scaling factor) and the cross product, or if it's just a coincidence that they're the same.
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    Re: Determinants as scalings factors and cross products

    Right after posting this, I got an idea. If it is possible to show that the area/volume/hypervolume spanned by the row vectors (and their opposite corners) is indeed the same as the formula for the determinant, without any reference to cross products, it would explain everything.

    Not sure if this is possible, or where to look if it's been done somewhere though.
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    Re: Determinants as scalings factors and cross products

    It turns out that the n-dimensional *oriented* content (also called "volume", but clearly reduces to " oriented area" in 2 dimensions, and "oriented length" in 1 dimension) is an alternating n-linear form on a space of n dimensions.

    The number of k-linear alternating forms on a space of n dimensions is known to be n choose k (the binomial coefficient), which means that there is, up to a scalar factor, only ONE such alternating n-linear form. The alternating form that returns the value 1 on the "positively oriented" unit n-cube, is called "the volume element" (or dV in multi-variable calculus) and is easily seen to be the determinant function.

    In 3 dimensions, given that the volume of a unit cube (with the usual "right-hand (positive) orientation) is 1, we see from trilinearity, that a box of a units along the x-axis, b units along the y-axis, and c units along the z-axis has volume: abc.

    In fact, the cross-product can be written as a "formal determinant":

    \mathbf{u} \times \mathbf{v} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\u  _1&u_2&u_3\\v_1&v_2&v_3 \end{vmatrix}

    It is possible to define a "cross-product" of n-1 vectors in n space to give an n-vector, but this only works for PAIRS of vectors, when n-1 = 2, that is to say when n = 3, which is why it is sometimes said: "the cross-product can only be defined in 3 dimensions" (we usually think of a product as a binary operation).

    The fact that you have noticed some of this already puts you way ahead of the game. The main ingredient you are missing is that det is a SIGNED function, exchanging any two orthogonal basis vectors in a coordinate system reverses the sign of the "signed n-dimensional content". What we typical think of as "volume" is thus the magnitude of the determinant (its absolute value). This is represented in the cross-product by the identity:

    \mathbf{u} \times \mathbf{v} = -\mathbf{v} \times \mathbf{u}

    The area of math that deals with all this is called "multi-linear algebra" which has its applications in finding such things as surface area, length of curves, and volumes of solids. The basic idea is that we form a "tangent space" (a local linear space where we approximate a portion of a curve by a line segment, a portion of a surface by a plane, or a portion of a solid by a cube), and calculate the "distortion" of these linear spaces by the derivative, and then sum over all these tiny "volume elements".

    It turns out the the derivative itself can be viewed as a matrix (in the one-dimensional case, it is a 1x1 matrix, a number), and the "bit of volume" we are after is found by taking the determinant of this matrix (the matrix is called the Jacobian), which measures the "scaling" of the distortion of the volume element from the unit cube (or square, or interval). For example, when finding the surface area of a sphere, we add up a bunch of tiny parallelograms (we can do this because it is possible to define a "consistent" orientation of the tangent planes that "kiss" the sphere).

    The actual math gets messy, calculating surface areas can be quite difficult.

    The short answer is yes: there IS a connection, and it is no coincidence.
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    Re: Determinants as scalings factors and cross products

    great stuff. I knew about the idea of area and volume elements from my tiny delving into differential geometry but this is new to me. Any recommended sources on multi-linear algebra?
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    Re: Determinants as scalings factors and cross products

    Thanks alot for your input! The reason I started wondering about this, was because I were never satisfied with the way my textbook explained the jacobian when introducing variable changes in multiple integrals.

    Some of the stuff in your post I couldn't quite follow though. I've had an undergrad-level course in abstract algebra (and one in linear algebra), but I've never heard of k-linear alternating forms or of multilinear algebra. I will look it up.

    But all right, back to the cross product. As you wrote, it is only a "formal determinant". It's absolute value should be:

    1) the area of the parallellogram spanned by the two vectors in question.

    The actual formula seems to compute:

    2) the sum of the areas spanned by the vector components in the xy, xz and yz-planes, multiplied by a unit vector in the perpendicular direction in each case.

    My question now is, is there an easy intuitive way to show why those two things are the same thing? Or is there only a messy non-intuitive way with pages of hairy algebra?

    Thank you so much again!
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    Re: Determinants as scalings factors and cross products

    I was also thinking about what you said about "cross products" in dimensions other than R3. By taking the determinant:

    |i j|
    |x y|

    I've found that I actually get a new vector perpendicular to (x,y), with the same magnitude! So I could imagine doing the same thing in 4 dimensions with 3 vectors. Fascinating!
    Last edited by gralla55; February 4th 2014 at 11:30 AM.
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    Re: Determinants as scalings factors and cross products

    Quote Originally Posted by gralla55 View Post
    Thanks alot for your input! The reason I started wondering about this, was because I were never satisfied with the way my textbook explained the jacobian when introducing variable changes in multiple integrals.

    Some of the stuff in your post I couldn't quite follow though. I've had an undergrad-level course in abstract algebra (and one in linear algebra), but I've never heard of k-linear alternating forms or of multilinear algebra. I will look it up.

    But all right, back to the cross product. As you wrote, it is only a "formal determinant". It's absolute value should be:

    1) the area of the parallellogram spanned by the two vectors in question.

    The actual formula seems to compute:

    2) the sum of the areas spanned by the vector components in the xy, xz and yz-planes, multiplied by a unit vector in the perpendicular direction in each case.

    My question now is, is there an easy intuitive way to show why those two things are the same thing? Or is there only a messy non-intuitive way with pages of hairy algebra?

    Thank you so much again!
    The easiest way I can think of to explain it is, imagine expanding by minors along the first row.

    Another way to look at it: we can form a NEW product between vectors, called the exterior product or wedge. For example, the exterior product of two vectors is the (oriented) parallelogram spanned by them. This gives us:

    \mathbf{i}\wedge\mathbf{j} the (oriented) unit square in the xy-plane.

    The main algebraic property of this product is that:

    \mathbf{u} \wedge \mathbf{v} = -\mathbf{v} \wedge \mathbf{u}, which implies that:

    \mathbf{u} \wedge \mathbf{u} = 0.

    The wedge product of two vectors is called (by various authors) a "blade", "2-blade" or "2-vector".

    OK, but what exactly, is the wedge product of two arbitrary vectors? In R2 we have:

    \mathbf{u} \wedge \mathbf{v} = (u_1\mathbf{i} + u_2\mathbf{j})\wedge(v_1\mathbf{i} + v_2\mathbf{j})

     = (u_1v_1)(\mathbf{i}\wedge\mathbf{i}) + (u_1v_2)(\mathbf{i}\wedge\mathbf{j}) + (u_2v_1)(\mathbf{j}\wedge\mathbf{i}) + (u_2v_2)(\mathbf{j}\wedge\mathbf{j})

     = (u_1v_2 - u_2v_1)(\mathbf{i}\wedge\mathbf{j}).

    It is possible to view the "infinitesimals" dx and dy as basis elements of a vector space, by viewing them as the FUNCTIONS:

    dx(x,y) = x
    dy(x,y) = y

    and then we get:

    \int_a^b\int_c^d 1\ dxdy = \int_{[a,b] \times [c,d]} 1 dx \wedge dy.

    Note that orientation comes into play here by which of a and b, and c and d, we put at the "top" of the integrals.
    Thanks from gralla55
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    Re: Determinants as scalings factors and cross products

    I used your first suggestion, and given that the determinant calculates the "volume" spanned by it's column / row vectors, it is indeed easy to show why the cross product works!

    For instance, in two dimensions, if it is known that:

    |a c|
    |b d|

    is equal to the area spanned by (a,c) and (b,d), I simply rewrite the formula for the determinant as so:

    ad - bc = (a,c) dot (d,-b)

    And since this is the case, one can clearly see that (d,-b) has to be a vector with the same length as (d,b), but perpendicular to it.

    For 3 dimensions, I just applied the same logic:

    |a b c|
    |d e f|
    |g h i|

    = a|e f| - b|d f| + c|d e|
    |h i| |g i| |g h|

    = (a,b,c) dot (|e f| , -|d f| , |d e|)
    |h i| |g i| |g h|


    So again the right vector has to be equal to a vector perpendicular to (d,e,f) and (g,h,i), with the same magnitude as the area between them. Applying the same logic to n dimensions, I can see why the "cross product" always makes sense.

    The only thing missing from this picture, is an explanation of why the recursive formula for the determinant actually calculates the (signed) volume spanned by the row vectors of the matrix. My starting point was that this was true, but I have no "intuitive feel" of why it should be. I've tried to derive the formula myself, but I've had little luck so far. That is to say, I've been able to use only geometry to get the formula for the 2x2 determinant, but I'm struggling go from there to the 3-dimensional case.

    Basically what I'm looking for, is some kind of geometric logic behind how the determinant formula works for higher and higher dimensions. Thanks again for pointing me in the right directions!
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