Conceptual Questions/Eigenvectors and Eigenvalues

[sherlockjones]

This is how the book introduced eigenvectors:

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

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

Thanks

 

[sherlockjones]

Conceptual Question/Eigenvectors

This is how the book introduced eigenvectors:

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

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

Thanks

 

[matt grime]

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

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 $$<x,y> = t<1,-1> + <a,b>$$?

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 $$\vec{OQ} = \vec{OP} +2\vec{PM}$$?

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 $$x-y = 0$$ is $$<1,-1>$$. 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 $$<x,y> = t<1,-1> + <a,b>$$?

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 $$\vec{OQ} = \vec{OP} +2\vec{PM}$$?

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 $\vec{PM}$ and then the vector $\vec{MQ}$. But, by symmetry, the two vectors are equal! $\vec{PM}= \vec{MQ}$ and so $\vec{PQ}= 2\vec{PM}$. Of course, to go from O to Q you could go from O to P and then from P to Q: $\vec{OQ}= \vec{OP}+ \vec{PQ}= \vec{OP}+ 2\vec{PM}$.

 

[HallsofIvy]

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

Please do not double post!