# Homework Help: Conceptual Questions/Eigenvectors and Eigenvalues

1. Nov 2, 2006

### sherlockjones

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? 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

2. Nov 2, 2006

### sherlockjones

Conceptual Question/Eigenvectors

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? 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

3. Nov 3, 2006

### matt grime

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.

4. Nov 3, 2006

### HallsofIvy

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 $\vec{PM}$ 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: $\vec{OQ}= \vec{OP}+ \vec{PQ}= \vec{OP}+ 2\vec{PM}$.

5. Nov 3, 2006

### HallsofIvy

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