Find Basis of R^n for Diagonal B Matrix of Reflection T in R^3

[Inirit]

Homework Statement

Find a basis B of R^n such that the B matrix B of the given linear transformation T is diagonal.

Reflection T about the plane x_1 - 2x_2 + 2x_3 = 0 in R^3.

The Attempt at a Solution

I just don't even know where to begin. I don't know how to interpret problem or how to understand the transformation. I feel completely lost with it. Can someone point me to at least the first step?

 

[micromass]

I would start with writing the matrix of this transform with respect to the basis (1,0,0), (0,1,0) and (0,0,1).

 

[Inirit]

I can't because I don't even understand the transformation. I don't get what kind of matrix would transform something to reflect it about the plane.

Although... I've just had a thought. If you expressed the plane like this: x_1 = 2x_2 - 2x_3... would the matrix that transforms a vector onto the plane be:

0 2 -2
0 1 0
0 0 1

 

[vela]

Reflection T about the plane x_1 - 2x_2 + 2x_3 = 0 in R^2.

You meant in R³, right?

The Attempt at a Solution

I just don't even know where to begin. I don't know how to interpret problem or how to understand the transformation. I feel completely lost with it. Can someone point me to at least the first step?

It often helps to think about simpler cases where you can easily see the answer and then try to generalize the concept. Suppose you wanted to find the reflection about the xy plane. What would the point (x,y,z) transform to?

 

[Inirit]

Yes that's what I meant, sorry.

And the reflection would be (x,y,-z). That's what I was thinking, you just take the height of the vector relative to the plane and reverse it. But I am stuck on how to express it's height relative to the plane.

But just to help me better understand what it is I'm doing, was my claim in my previous post correct?

 

[vela]

That matrix doesn't work because it takes the point (1,0,0) to the origin.

Think in terms of a vector that is normal to the plane.

 

[Inirit]

Yes it does take it to the origin, but I merely asked if that were a matrix that maps a vector in R^3 to the plane, not it's reflection (I know that's not the goal of the problem, but I am slowly trying to understand this). A vector only with a component that isn't within the space of the plane would be mapped to the origin, wouldn't it?

 

[vela]

Oh, sorry, I didn't read what you wrote carefully enough. Yes, you're right that if a vector has no component that lies in the plane, its projection onto the plane will be 0. Your proposed matrix, however, doesn't give you that projection. (By my calculations, the projection of (1, 0, 0) onto the plane is (8/9, 2/9, -2/9).)

 

[Inirit]

*sigh* I am having a hard time comprehending how this works, and now I feel more lost than ever. I just don't see where to go and how it all connects.

 

[vela]

You have the right idea to break the vector x into a component that lies in the plane and a component perpendicular to the plane. You just have to calculate them correctly. From the equation for the plane, you should be able to identify a vector that's normal to the plane. Use it to find the component of x perpendicular to the plane.

 

[Inirit]

<1,-2,2> would be the normal vector, but I don't understand where to go from here.

Edit: Thank you for trying to help me, but it's obvious that I am too far lost with the entire concept. I was hoping that solving this problem would point me in the right direction for understanding, but seeing as how I can't even make the first step I think it's a lost cause.