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Conceptual Questions/Eigenvectors and Eigenvalues

  1. Nov 2, 2006 #1
    This is how the book introduced eigenvectors:

    I do not get how the normal vector of [tex] x-y = 0 [/tex] is [tex] <1,-1> [/tex]. Isn't that saying that the x-component is 1 and the y-component is -1? Also how did they get the vector equation [tex] <x,y> = t<1,-1> + <a,b> [/tex]? Finally, why does [tex] \vec{OQ} = \vec{OP} +2\vec{PM} [/tex]?

    Thanks
     
  2. jcsd
  3. Nov 2, 2006 #2
    Conceptual Question/Eigenvectors

    This is how the book introduced eigenvectors:

    I do not get how the normal vector of [tex] x-y = 0 [/tex] is [tex] <1,-1> [/tex]. Isn't that saying that the x-component is 1 and the y-component is -1? Also how did they get the vector equation [tex] <x,y> = t<1,-1> + <a,b> [/tex]? Finally, why does [tex] \vec{OQ} = \vec{OP} +2\vec{PM} [/tex]?

    Thanks
     
  4. Nov 3, 2006 #3

    matt grime

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    Draw a picture: the line x-y=0 goes a bit like this /, so its normal goes a bit like \.


    Because that is the equation of a line through (a,b) and parallel to (1,-1): it certainly goes through that point, and can be seen to go in the right direction, again, by drawing a diagram.


    My advice once more is: draw a picture: draw the lines, the points, the line segments and just look at the picture to see what's going on.
     
  5. Nov 3, 2006 #4

    HallsofIvy

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    Yes, that's exactly what it is saying. x-y= 0 is the same as y= x. Any vector in the direction of that line must have y component equal to x component: one such vector is <1, 1>. Any normal vector to that line must have dot product with that vector equal to 0: one such vector is <1, -1> since then the dot product is 1(1)+ (-1)(1)= 0. By the way, note that your quote does not say the normal vector, it says a normal vector. There are an infinite number of normal vectors to a line.

    Presumably, you have already learned that a vector equation for a line in the direction of vector <A, B>, through the point (x0, y0) is of the form t<A,B>+ <x0,y0>. t just measures the "distance" along the line from the point (x0,y0) to the point (x,y).

    Remember that you are trying to find the point Q "symmetric" to P in the line y= x. That is, it lies on the normal line PQ with distance MQ equal to distance PM. To go from P to Q you would travel the vector [itex]\vec{PM}[/itex] and then the vector [itex]\vec{MQ}[/itex]. But, by symmetry, the two vectors are equal! [itex]\vec{PM}= \vec{MQ}[/itex] and so [itex]\vec{PQ}= 2\vec{PM}[/itex]. Of course, to go from O to Q you could go from O to P and then from P to Q: [itex]\vec{OQ}= \vec{OP}+ \vec{PQ}= \vec{OP}+ 2\vec{PM}[/itex].
     
  6. Nov 3, 2006 #5

    HallsofIvy

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    This was posted in both the homework and mathematics sections so I merged the threads.

    Please do not double post!
     
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