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Intutition of the determinant

  1. Jul 9, 2015 #1
    Hey all, the determinant is used in interseting places and I know how to use it but I don't know the intuition of it,Can anyone elucidate it for me? Any help apreciated.
  2. jcsd
  3. Jul 9, 2015 #2
    Let's start with two dimensions. Suppose you have a 2 by 2 matrix. Write it in the form [itex] P = [x_1 \space \space x_2] [/itex] where [itex] x_1, x_2 \in \mathbb{R}^2 [/itex]. Then the vectors [itex] x_1, x_2 [/itex] represent the edges of a 2 dimensional parallelopiped, P, in [itex] \mathbb{R}^2 [/itex]. The determinant actually gives the oriented (or "signed") area of P. Let's check that the determinant is a "good" area function. Intuitively, if we were to multiple one of these vectors by a scalar, say c, then that should affect the area right? It should multiply the area by c as well. Well if you know your determinant properties, then you will know that [itex] det[cx_1 \space \space x_2] = cdet[x_1 \space \space x_2] = det[x_1 \space \space cx_2] [/itex]. Great! Now let's another property that should be intuitive: If the two vectors [itex] x_1, x_2 [/itex] were linearly dependent then they don't really form a parallelopiped because one is a scalar multiple of the other. They form more of a line segment. So their area should be 0. Well, as you should also know, the determinant of any set of linearly dependent vectors is 0. One last thing that may be less intuitive at first: the concept of orientation. Orientation is like defining which direction is positive and which direction is negative. For example, on the number line we say that the direction to the right is positive and the direction to the left is negative. Well if you have your parallelopiped P, then writing P as [itex] x_1 \space \space x_2 [/itex] is designating an orientation. Reversing the order of the vectors should reverse the orientation, i.e. it should multiply the area by -1. Again, by the usual determinant properties, [itex] det[x_1 \space space x_2] = -det[x_2 \space \space x_1] [/itex].

    So the determinant fulfills the properties that we feel a volume function should fulfill. Note that even though we just worked in [itex] \mathbb{R}^2 [/itex], this readily generalizes to [itex] \mathbb{R}^n [/itex]. Now, a question may have arose while reading this: the determinant gives the oriented area of an n-dimensional parallelopiped in [itex] \mathbb{R}^n [/itex]. But what if we had a k-dimensional parallelopiped that happened to be embedded in [itex] \mathbb{R}^n [/itex]? We would often like to find the k-dimensional volume (note that area is 2-dimensional volume), but the determinant is only defined for n by n matrices. Writing our k-dimensional parallelopiped in matrix form we would have [itex] x_1 \space \space x_2 \space \space ... x_k] [/itex] for [itex] x_i \in \mathbb{R}^n [/itex], but this is an n by k matrix. We can't find the determinant of such a matrix.

    BUT, as it turns out, we can still find the volume of such an object using the determinant. Without going into too much detail, one of the most important properties that we want satisfied is that its k-dimensional volume should be the same whether it was embedded in [itex] \mathbb{R}^k [/itex] or [itex] \mathbb{R}^n [/itex] right? Well, if we define the volume of this k-dimensional paralleopiped as [itex] \sqrt{det[P^{t}P]} [/itex] where [itex] P^t [/itex] denotes the transpose of P, then this volume function gives the k-dimensonal volume of P, regardless of whether it is embedded in k-dimensional space or n-dimensional space. Note that this volume function always returns a non-negative number (returns 0 if and only if the edeges of P are linearly dependent), so it doesn't give the oriented volume - it gives the "absolute" volume.

    I hope the concept of orientation wasn't too confusing. I know I may have gone over it a bit too quickly without giving the best details. Let me know if you would like me to elaborate on it.
  4. Jul 12, 2015 #3
    So determinant gives you the area of a parallelogram or parallelopiped in 2d, volume in 3d and so on
  5. Jul 12, 2015 #4
    Why it matter which vector goes first to determine orientation
  6. Jul 12, 2015 #5
    You guna have to show the proof Johnnyg
  7. Jul 13, 2015 #6
    We will do this in R^2 to make visualization easy. R^2 will have two oritentations, lets call them the right handed orientation (RHO) and the left handed orientation (LHO). A parallelogram P = [v_1 v_2] is right handed (belongs to the RHO) if det(P) > 0 and left handed (belongs to the LHO) if det(P) < 0.

    Suppose that P = [v_1 v_2] is right handed. Then you must rotate the vector v_1 counter clockwise through an angle less than pi to make it lie ontop of v_2. So det(P) > 0. Now switch the order of the vectors. That is, write P' = [v_2 v_1]. Visually, the parallelogram looks the same, but instead of rotating v_1 to v_2, you are rotating v_2 to v_1. That is, you are changing your notion of direction (orientation). But det(P') < 0 since you changed the order of the vectors, so this parallelogram that looks the same but has opposite direction (orientation) is now left handed.

    So you see, in short, changing the order of the vectors changes the sign of the determinant, and the sign of the determinant is what determines orientation.
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