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Finding the matrix for a reflection about a plane in R^3

  1. Aug 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Let L: R^3 -> R^3 be the linear transformation that is defined by the reflection about the plane P: 2x + y -2z = 0 in R^3. Namely, L(u) = u if u is the vector that lies in the plane P; and L(u) = -u if u is a vector perpendicular to the plane P. Find an orthonormal basis for R^3 and a matrix A such that A is diagonal and A is the matrix representation of L with respect to the orthonormal basis.

    2. Relevant equations

    3. The attempt at a solution
    Basically I tried this problem the only way I knew how, and it was very long and tedious and didn't come out right. I am hoping someone can either (or both) see where I went wrong, or suggest a more streamlined approach.

    First off, I was a little unsure by what was meant towards the end of the question about an orthonormal basis. Specifically, is it referring to an orthonormal basis in terms of the domain of L or the range of L?

    I chose to assume if was referring to the domain since this fit in with the only way I knew how to approach this problem. The standard basis for R^3 is of course orthonormal. My attempt was to use the theorem that a matrix can be found by applying the transformation to each basis vector, then the columns are the coefficients of this transformation written as a linear combination of the basis vectors of the space being transformed to.

    So first off I chose a basis of R^3 in terms of the plane. I choose (1 -2 0) since it satisfies the equation of the plane, the normal vector (2 1 -2), and the cross product of the normal vector and the first vector which is (-4 2 -5). Let's call this set of vectors B

    So then I need to reflect the standard basis (let's call it A) across the plane. Rather than calculate out the reflections of A, I thought I could go straight to writing the standard basis as a linear combination of B and then switch the sign of the coefficient of (2 1 -2) since this is the portion (or the projection of the vector onto the normal vector) that will be reflected direction across. In other words, any vector will be equal to its projection into the plane + it's projection onto the normal of the plane. Then reflecting it is changing the sign of the projection onto the normal vector.

    So if the matrix C is the B as column vectors:

    1 -4 2
    -2 2 1
    0 -5 -2

    I solved Cx = e1, Cx = e2, Cx = e3. (I checked these x's a few times so I am pretty confident about this part). Then I put these x's as the column vectors of a matrix and switched the sign of the last row (again thinking this was writing the reflection of the standard basis across the plane in terms of B). This yielded

    18 -36 0
    -8 -4 -10
    -20 -10 20

    all times (1/90). I thought this was the answer but I check multiplying the normal vector but did not get the negative of the normal vector as output.

    So, where did I go wrong? And or, what is a streamlined approach. This method seemed way to long, not to mention just incorrect.

    PS. I can't figure out how to add matrices officially on this. Is it listed in the Latex Reference button popup?
  2. jcsd
  3. Aug 21, 2012 #2


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    You have the right idea. You're just doing more work than is necessary.

    You seem to have a sign error in your second basis vector. It should also satisfy the equation for the plane, but it doesn't. So you need to fix that. So let's assume you've done that. You will have constructed an orthogonal basis for R3. Now you just need to make it orthonormal.

    If B={b1, b2, b3} is your basis, you found these vectors so that they satisfy

    L(b1) = b1
    L(b2) = b2
    L(b3) = -b3

    Now based on this, what is the matrix representing L relative to your basis B?
  4. Aug 21, 2012 #3
    Hmm, well the second vector was the vector normal to the plane. I presumed that I needed this one to span R^3. That is, there are only two orthogonal vectors in a plane. Is that not the case? I don't see what else I could do. Also, it is not clear to me why they need to be orthonormal, unless you are saying that because that is what the question specified (I was unsure and thought maybe they meant starting from the standard basis that is orthonormal). So are you saying

    L(b1) = b1
    L(b2) = b2
    L(b3) = -b3

    implies that if A (the transformation matrix I'm trying to find) =

    a1 a2 a3
    a4 a5 a6
    a7 a8 a9

    then the first one says

    a1b11 + a2b12 + a3b13 = b11
    a3b21 etc ..

    and I could solve for the ai's since there would be a total of 9 equations and 9 unkowns? Thanks a bunch this is really helpful!
  5. Aug 21, 2012 #4


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    I meant second according to the order you used them to construct matrix C. The vector was the cross product of the first vector and the normal. (-4, 2, -5) doesn't lie in the plane as it should.

    Your idea for how to construct this basis is fine. You just need to get the calculation right. My other point was that you're asked to find an orthonormal basis, but you've only found an orthogonal basis. You still need to normalize the basis vectors you found.

    As far as finding the matrix representing L, you're doing too much work. The n-tuples (1, 2, 0), (2, 1, -2), and whatever the other one works out to be are representations of b1, b3, and b2, relative to the canonical basis {e1, e2, e3}. If you solve all of those linear equations using those n-tuples, you'll find a matrix that represents L, but it will be relative to the canonical basis. Once you have this, you can calculate what the matrix should be relative to B by doing a change of basis. But this would be the long and painful way to do it.

    The problem is asking you to find the matrix relative to basis B. You should be able to simply write down what that matrix is without doing any work. Think about the representation of b1, b2, and b3 relative to the basis B.
    Last edited: Aug 21, 2012
  6. Aug 22, 2012 #5
    Sorry, I missed the sign. It is (-4, -2, -5). So are you saying that it is only asking for the matrix that takes a vector input in terms of the basis B? As in (1, 2, 4) really means 1b1 + 2b2 + 43? Then it would just be:

    1 0 0
    0 1 0
    0 0 -1

    I think I have a hard time understanding how a question specifies both the basis of the vectors being inputed and the basis of the output. So if we wanted the matrix that took as input vectors in basis B and output in that standard basis it would just be:

    1 -4 -2
    2 -2 -1
    0 -5 2

    I had thought it was asking about taking input in terms of the standard basis and output in the standard basis. I am still unclear on how I would find that. What is the efficient approach? Thanks so much Vela. I think I have had a fundamental misunderstanding about transformations and bases but hopefully I am beginning to see the light at the end of the tunnel.
  7. Aug 22, 2012 #6
    I have been thinking about it more. So to find the matrix to take input in the form of the standard basis, reflect about the plane and output in terms of the standard matrix I could find the change of basis matrix for the standard basis to B (call it X), then find the matrix for reflecting a vector given in B about the axis and out putting it in the standard basis (which I think you led me to above? Let's call it Y). Then YX would be the desired matrix?
  8. Aug 22, 2012 #7

    Ray Vickson

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    If [itex] \vec{a} = (a_x,a_y,a_z)[/itex] is a unit normal to the plane, the equation of the plane can be written as [itex] \vec{a}\cdot \vec{r} = 0, [/itex], where [itex] \vec{r} = (x,y,z) \in R^3.[/itex] (Here, [itex] \cdot [/itex] denotes the inner product opertion.) In your case,
    [tex] \vec{a} = \frac{(2,1,-2)}{\sqrt{2^2 + 1^2 + 2^2}} = \frac{(2,1,-2)}{3}.[/tex]

    We can write [itex] \vec{r} = \vec{u} + \vec{v} [/itex] for any vector [itex] \vec{r} [/itex], where [itex] \vec{u} [/itex] is the part of [itex] \vec{r}[/itex] that lies in the plane and [itex] \vec{v}[/itex] is the component perpendicular to the plane. We have
    [tex] \vec{u} =\vec{r} - (\vec{a} \cdot \vec{r}) \vec{a},\; \vec{v} = (\vec{a} \cdot \vec{r}) \vec{a}.[/tex] The vector [itex] \vec{r}^\prime [/itex] obtained by reflecting [itex] \vec{r}[/itex] in the plane is
    [tex] \vec{r}^\prime = \vec{r} - 2 (\vec{a} \cdot \vec{r}) \vec{a}.[/tex]
    You can write this out in terms of x, y and z, so obtain the linear transformation.

  9. Aug 22, 2012 #8
    So that is kind of reminiscent of the Gram-Schmidt process in that you are finding the projection onto the plan and subtracting that from the vector and then just flipping what is left?

    Thanks, it is really helpful to see an efficient way to go straight to the transformation all in terms of the standard basis. Thanks Vela too!
  10. Aug 22, 2012 #9


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    The question asked you to "Find an orthonormal basis for R^3 and a matrix A such that A is diagonal and A is the matrix representation of L with respect to the orthonormal basis." Your first matrix is exactly what's asked for: it's diagonal (because of the way you chose the basis vectors) and it represents L with respect to basis B.

    I think the intent of this question is really a conceptual one. You can find the answer dealing with the standard basis and then transforming to the B basis, but if you understand what the question is asking for and think about it in terms of B, you can simply write down the answer, like you did.

    Still, it's good practice to be able to find the matrix representing L in the standard basis.

    Let's use the unnormalized basis for right now to avoid having to write a lot of fractions, and let's say X is given by
    $$ X = \begin{pmatrix}
    1 & -4 & 2 \\
    -2 & -2 & 1 \\
    0 & -5 & -2
    \end{pmatrix}$$ which takes a coordinate vector relative to B and maps it to the coordinate vector relative to the standard basis. Then ##X^{-1}## goes in the other direction. The diagonal matrix A maps between coordinate vectors relative to B.

    If you want to go from a standard-basis representation to a standard-basis representation with L, you first use ##X^{-1}## to convert from the standard basis to B. If ##\vec{x}## is the coordinate vector relative to the standard basis, ##X^{-1}\vec{x}## is the representation relative to B. Now you apply A to get ##AX^{-1}\vec{x}##, which gives you the reflected vector represented in basis B. Finally, you apply X to convert that result back to the standard basis, which gives you ##XAX^{-1}\vec{x}##. So the matrix C for the linear transformation L relative to the standard basis is given by ##C=XAX^{-1}##.
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