Conceptual Questions/Eigenvectors and Eigenvalues

[sherlockjones]

This is how the book introduced eigenvectors:

In the xy-plane let us find a point of reflection [tex] Q [/tex] of a general point [tex] P = (a,b) [/tex] in the line [tex] y = x [/tex]. The line [tex] x-y = 0 [/tex] has a normal vector [tex] <-1, 1> [/tex] and so a vector equation of the straight line through [tex] P [/tex] and perpendicular to [tex] y = x [/tex] is given by [tex] <x,y> = t<1,-1> + <a,b> = <a+t,b-t> [/tex]. The point of intersection [tex] M = (x,y) [/tex] of the two lines is obtained from [tex] a+t = b-t [/tex] or [tex] t = \frac{1}{2}(b-a) [/tex]. Thus [tex] M = (\frac{1}{2}(a+b), \frac{1}{2}(a+b)) [/tex]. The point [tex] Q = (b,a) [/tex] because [tex] \vec{OQ} = \vec{OP} +2\vec{PM} [/tex]. Thus [tex] w = Av [/tex] where [tex] A [/tex] is the identity matrix.

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

 

[sherlockjones]

Conceptual Question/Eigenvectors

This is how the book introduced eigenvectors:

In the xy-plane let us find a point of reflection [tex] Q [/tex] of a general point [tex] P = (a,b) [/tex] in the line [tex] y = x [/tex]. The line [tex] x-y = 0 [/tex] has a normal vector [tex] <-1, 1> [/tex] and so a vector equation of the straight line through [tex] P [/tex] and perpendicular to [tex] y = x [/tex] is given by [tex] <x,y> = t<1,-1> + <a,b> = <a+t,b-t> [/tex]. The point of intersection [tex] M = (x,y) [/tex] of the two lines is obtained from [tex] a+t = b-t [/tex] or [tex] t = \frac{1}{2}(b-a) [/tex]. Thus [tex] M = (\frac{1}{2}(a+b), \frac{1}{2}(a+b)) [/tex]. The point [tex] Q = (b,a) [/tex] because [tex] \vec{OQ} = \vec{OP} +2\vec{PM} [/tex]. Thus [tex] w = Av [/tex] where [tex] A [/tex] is the identity matrix.

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

 

[matt grime]

I do not get how the normal vector of [tex] x-y = 0 [/tex] is [tex] <1,-1> [/tex].

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

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]?

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.

Finally, why does [tex] \vec{OQ} = \vec{OP} +2\vec{PM} [/tex]?

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.

 

[HallsofIvy]

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?

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.

Also how did they get the vector equation [tex] <x,y> = t<1,-1> + <a,b> [/tex]?

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).

Finally, why does [tex] \vec{OQ} = \vec{OP} +2\vec{PM} [/tex]?

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].

 

[HallsofIvy]

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

Please do not double post!