Conceptual Questions/Eigenvectors and Eigenvalues

Thanks everyone. This thread is in the homework section and I believe it is a homework question. I don't see how it could be duplicated there. If someone duplicated it, they will have to fix it. I'm moving it back to the mathematics section.In summary, the book introduced eigenvectors by finding a point of reflection Q in the xy-plane for a general point P = (a, b) in the line y = x. The normal vector of x-y = 0 is <1, -1> and the vector equation of the straight line through P and perpendicular to y = x is <x, y> = t<1, -1> + <a, b> = <a+t
  • #1
sherlockjones
31
0
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
 
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  • #2
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
 
  • #3
sherlockjones said:
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.
 
  • #4
sherlockjones said:
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].
 
  • #5
This was posted in both the homework and mathematics sections so I merged the threads.

Please do not double post!
 

Related to Conceptual Questions/Eigenvectors and Eigenvalues

1. What is the definition of an eigenvector and eigenvalue?

An eigenvector is a vector that does not change direction when multiplied by a given matrix. It only changes in magnitude. An eigenvalue is a scalar that represents the amount by which the eigenvector is stretched or compressed when multiplied by the matrix.

2. How do you calculate eigenvectors and eigenvalues?

To calculate eigenvectors and eigenvalues, you first need to find the determinant of the given matrix. Then, you can solve for the eigenvalues by setting the determinant equal to zero and solving for the variable. Once you have the eigenvalues, you can plug them back into the matrix to find the corresponding eigenvectors.

3. What are the applications of eigenvectors and eigenvalues?

Eigenvectors and eigenvalues have many applications in physics, engineering, and computer science. In physics, they are used to analyze the behavior of oscillating systems and study quantum mechanics. In engineering, they are used to analyze structural systems and vibrations. In computer science, they are used in machine learning algorithms and data compression techniques.

4. Can a matrix have more than one eigenvector and eigenvalue?

Yes, a matrix can have multiple eigenvectors and eigenvalues. In fact, most matrices have multiple eigenvectors and eigenvalues, unless they are special matrices such as identity matrices or zero matrices.

5. What is the relationship between eigenvectors and eigenvalues?

The relationship between eigenvectors and eigenvalues can be described by the equation Av = λv, where A is the given matrix, v is the eigenvector, and λ is the eigenvalue. This equation shows that the eigenvector and eigenvalue are related to each other and cannot exist without each other in the context of a matrix.

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