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I understand how to use determinants but I can't make a mental representation of them

  1. Jul 17, 2012 #1
    I know how to use determinants to solve a system of linear equations, I know I can use them to find the rank of a matrix and find out if a system is linear dependant/independant. However, I still don't really "get" determinants. To me they're some sort of magic box that I can use to calculate all kind of stuff related to matrices.

    Could someone suggest a good website that might help me visualize what determinants really are? If I recall correctly, Gilbert Strang said in one of his first lectures that he doesn't like determinants for that reason (the "magic box" factor).

    I'm not a math major, I'm studying electrical engineering so if the website has too much advanced math, I'll probably end up understanding even less about the topic :)
     
  2. jcsd
  3. Jul 17, 2012 #2
    Re: I understand how to use determinants but I can't make a mental representation of

    There is is a sort-of intuitive understanding of the determinant of a 2x2 matrix, in which the absolute value of the determinant corresponds to the area of the parallelogram with sides given by the column vectors of the matrix. Unfortunately, the simple geometric interpretation of the determinant doesn't really tell you why determinant has all of the properties that it does, or why it's so useful. The determinant, like so many things in freshman/sophomore mathematics, more of less has to be taken at face value without much understanding of the mechanics behind it.
     
  4. Jul 17, 2012 #3

    micromass

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    Re: I understand how to use determinants but I can't make a mental representation of

    A nice geometric interpretation of determinants is that they are the "signed volume".

    For example, consider two points (a,b) and (c,d) in the plane. You can look at the parallellogram with vertices (0,0), (a,b), (c,d) and (a+c,b+d). The area (which is 2-dimensional volume) is given by

    [tex]\det \left(\begin{array}{cc} a & b\\ c & d\end{array}\right)[/tex]

    Why "signed" area? Well, if you switch around (a,b) and (c,d), then the determinant switches signs. So we need to take orientation into account.

    Similarly, in 3D, we can look at (a,b,c), (d,e,f) and (g,h,i). Three points. They span a paralellepiped. The volume (now the actual volume) is given by

    [tex]\det \left(\begin{array}{ccc} a & b & c \\ d & e & f\\ g & h & i\end{array}\right)[/tex]

    Again, this is a signed volume.

    With this interpretation, I hope some theorems about determinants make more sense. For example, the rank of a matrix

    [tex]\det \left(\begin{array}{ccc} a & b & c \\ d & e & f\\ g & h & i\end{array}\right)[/tex]

    is the number of linearly independent rows (or columns) of the matrix. So, if the rank is not three, then there is a vector which is a linear combination of the other vectors. This means that our 3 vectors lie in a plane. But then the paralellepid spanned by those vectors also lies in a plane. This the volume is 0. So we can easily see that if the rank is smaller than 3, then the determinant is 0. The converse is also easily seen.
     
  5. Jul 17, 2012 #4

    chiro

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    Re: I understand how to use determinants but I can't make a mental representation of

    You can also think of the sign of the determinant giving the orientation of a particular linear operator (i.e. a square matrix).

    You may be aware of cross products in three dimensional space, but the idea of being able to multiply vectors and get inverses in general is part of a field known as geometric algebra.

    When we think of multiplication and division we usually only think of real numbers or complex numbers, but there is an analog for vectors in general.

    As a result, what happens is that for vectors, we get orientation information and we can encapsulate vector by vector multiplication by a geometric product which has two elements: the inner product and the outer product (for 3D geometry the inner product is the dot product and the outer product is the cross product).

    So because of this orientation and its relation to the determinant, we can figure out how the determinant describes the 'handedess' of a basis (left or right handed) and this is known as chirality.

    In higher-dimensional geometry, things aren't even necessarily commutative and this becomes important because the order of how things are multiplied becomes essential in understanding what goes on.

    To understand this, you need to understand how rotations and in general matrices are not commutative (i.e. AB != BA in general). The rotation matrices help explain the ideas presented above in specific examples, and if you go deeper you'll find that with more complex objects, it gets even wackier (like spinors).
     
  6. Jul 17, 2012 #5
    Re: I understand how to use determinants but I can't make a mental representation of

    At the risk of going way over your head, let me introduce you to the relationship between linear operators and the "outer product."

    The outer product is denoted by [itex]\wedge[/itex]: i.e. for two vectors [itex]a[/itex] and [itex]b[/itex], the outer product of [itex]a[/itex] and [itex]b[/itex] is [itex]a \wedge b[/itex]. That's great to say, but what is it? Conventionally, we say it denotes the plane spanned by [itex]a[/itex] and [itex]b[/itex]. In this way, it's similar to the cross product, but it can be used in 2d and 4d (and any dimension) spaces, instead of just 3d.

    Why am I talking about the outer product? Because linear operators are often extended to work on multiple vectors through the outer product. Let's say you have the xy-plane, or [itex]\hat x \wedge \hat y[/itex]. One can say that a linear operator [itex]\underline A[/itex] acts on this like so:

    [tex]\underline A(\hat x \wedge \hat y) \equiv \underline A(\hat x) \wedge \underline A(\hat y)[/tex]

    In other words, the operator acts on each of the vectors individually and then you wedge them.

    Now, if you're in 2 dimensions, there is only one real wedge product: [itex]\hat x \wedge \hat y[/itex], and everything else, even if you change basis, is just some multiple of that. There is only one "plane" in a 2d space; this should be obvious. This means that [itex]\underline A(\hat x \wedge \hat y) = \alpha \hat x \wedge \hat y[/itex] for some scalar [itex]\alpha[/itex]. This number [itex]\alpha[/itex] is called the determinant. Geometrically, it tells you how the unit area (or volume in 3d) of the space is dilated or shrunk thanks to the action of the operator. If it is negative, it tells you that the unit area (or volume) has changed its orientation.

    I hope in this way, you can see that determinants are no magic box--they have a direct geometric interpretation, and for this reason they're very important.
     
  7. Jul 18, 2012 #6

    AlephZero

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    Re: I understand how to use determinants but I can't make a mental representation of

    A fairly simple way into the geometrical interpretation is to look at transforming between different coordinate systems. This leads to the "jacobian matrix" where the terms in the matrix are the partial derivatives relating the two systems. The determinant of that matrix (which is often just called the "Jacobian") tells you how areas or volumes are measured in the two different systems. For example transforming between cartesian and polar coordinates on a plane you get expressions like ##\iint \dots dx\,dy## and ##\iint \dots r\,dr\,d\theta##. The ##r## here is actually the value of a determinant!
     
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