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Math Help - Matrix of linear transformation

  1. #1
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    Matrix of linear transformation

    Hi again everyone!

    I have another question:

    Let T : R2 -> R2 be the transformation that rotates each point in R2 about the origin through an angle σ, with counterclockwise rotation for a positive angle. Find the standard matrix A of this transformation.

    I understand that a standard matrix for a linear transformation T is the matrix relative to a standard basis, but for some reason I feel like I don't have enough information in this exercise to find it?

    Thanks again for taking a look at this, you people on this forum are really the best
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  2. #2
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    Re: Matrix of linear transformation

    Find Ai and Aj in terms of i and j.

    If v is any vector xi+yj, then Av is xAi+yAj.

    Note
    Ai =cosσi +sinσj
    Aj = your turn
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  3. #3
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    Re: Matrix of linear transformation

    As Hartlw suggests, you can find the matrix representation of a linear transformation by determining what the transformation does to each basis vector. In a given basis, the basis vectors themselves are represented by \begin{bmatrix} 1 \\ 0 \end{bmatrix} and \begin{bmatrix}0 \\ 1 \end{bmatrix}. And, of course,
    \begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}a \\ c\end{bmatrix} and \begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}= \begin{bmatrix}b \\ s\end{bmatrix}.

    That is, the columns of the matrix representation of a linear transformation, in a given basis, are the vectors the linear transformation maps those basis vectors into.

    Now, what does a rotation, through \sigma degrees maps (1, 0) into a point (x, y) at distance 1 from (0, 0). If you drop a vertical to the x-axis you have a right triangle with legs of length "x" and "y", hypotenuse of length 1, and with angle \sigma. Then sin(\sigma)= \frac{y}{1}= y and cos(\sigma)= \frac{x}{1}= x. That is, the rotation maps the (1, 0) into (cos(\sigma), sin(\sigma)). Similarly, the point (0, 1) is mapped into (x, y)= (-sin(\sigma), cos(\sigma))

    Therefore the matrix representation of that rotation is ???????
    Last edited by HallsofIvy; May 28th 2013 at 02:12 PM.
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  4. #4
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    Re: Matrix of linear transformation

    Quote Originally Posted by Nora314 View Post
    Hi again everyone!

    I have another question:

    Let T : R2 -> R2 be the transformation that rotates each point in R2 about the origin through an angle σ, with counterclockwise rotation for a positive angle. Find the standard matrix A of this transformation.

    I understand that a standard matrix for a linear transformation T is the matrix relative to a standard basis, but for some reason I feel like I don't have enough information in this exercise to find it?

    Thanks again for taking a look at this, you people on this forum are really the best
    To transform the point \displaystyle \begin{align*} \left[ \begin{matrix} x \\ y \end{matrix} \right] \end{align*} to \displaystyle \begin{align*} \left[ \begin{matrix} x' \\ y' \end{matrix} \right]  \end{align*} by rotating by an angle of \displaystyle \begin{align*} \sigma \end{align*} anticlockwise, first note that we can write \displaystyle \begin{align*} x = r\cos{(\theta)} \end{align*} and \displaystyle \begin{align*} y = r\sin{(\theta)} \end{align*}, and so after the rotation we have \displaystyle \begin{align*} x' = r\cos{(\theta + \sigma)} \end{align*} and \displaystyle \begin{align*} y' = r\sin{(\theta + \sigma)}  \end{align*}. Simplifying these using the angle sum identities we find

    \displaystyle \begin{align*} x' &= r\cos{(\theta + \sigma)} \\ &= r\left[ \cos{(\theta)}\cos{(\sigma)} - \sin{(\theta)}\sin{(\sigma)} \right] \\ &= r\cos{(\theta)}\cos{(\sigma)} - r\sin{(\theta)}\sin{(\sigma)} \\ &= x\cos{(\sigma)} - y\sin{(\sigma)} \end{align*}

    and

    \displaystyle \begin{align*} y' &= r\sin{(\theta + \sigma)} \\ &= r\left[ \sin{(\theta)}\cos{(\sigma)} + \cos{(\theta)}\sin{(\sigma)} \right] \\ &= r\sin{(\theta)}\cos{(\sigma)} + r\cos{(\theta)}\sin{(\sigma)} \\ &= y\cos{(\sigma)} + x\sin{(\sigma)} \end{align*}

    Therefore when we write these in the matrix form we have

    \displaystyle \begin{align*} \left[ \begin{matrix} x' \\ y' \end{matrix} \right] &= \left[ \begin{matrix} x\cos{(\sigma)} - y\sin{(\sigma)} \\ x\sin{(\sigma)} + y\cos{(\sigma)} \end{matrix} \right] \\ \left[ \begin{matrix} x' \\ y' \end{matrix} \right] &= \left[ \begin{matrix} \cos{(\sigma)} & -\sin{(\sigma)} \\ \sin{(\sigma)} & \phantom{-}\cos{(\sigma)} \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] \end{align*}

    So the transformation matrix is \displaystyle \begin{align*} \left[ \begin{matrix} \cos{(\sigma)} & -\sin{(\sigma)} \\ \sin{(\sigma)} & \phantom{-}\cos{(\sigma)} \end{matrix} \right]  \end{align*}.
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  5. #5
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    Re: Matrix of linear transformation

    Wow, thank you everyone I worked through all the answers and now I feel like I finally understand what they mean when they say "find the standard matrix of linear transformation", I finally understand that it is all about seeing what the transformation does to the basis vectors. I think I did not properly understand that when I asked the question.

    Also, I liked the alternative way of solving the exercise that "prove it" showed.

    Thank you all for the help! I really appreciate it!
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