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Math Help - [SOLVED] Practice test

  1. #1
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    [SOLVED] Practice test

    I have my final coming just around the corner and I got this practice test. Any help would be great.

    1. Define T: R^2 \rightarrow R \ by \ T \begin{bmatrix}x_{1}\\x_{2}\end{bmatrix} = det \begin{bmatrix}x_{1}&1\\x_{2}&1\end{bmatrix}.

    Is T a linear transformation? If not, explain why not and provide a counterexample. If so, what is the geometric interpretation of the kernel of T?

    2. Diagonalize \begin{bmatrix}1&1\\-2&4\end{bmatrix}
    (I got Eigenvalues 2 and 3 and Eigenvectors \begin{bmatrix}1\\1\end{bmatrix} and \begin{bmatrix}1\\2\end{bmatrix})

    3. Given T: R^2 \rightarrow R^2 is linear and T(1,2) = (3,4) and T(3,4) = (1,2), then T(4,3) = ?
    (I got (-8, -9))
    Last edited by CaptainBlack; May 12th 2010 at 07:29 PM.
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  2. #2
    Lord of certain Rings
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    1)

    T \begin{bmatrix}x_{1}\\x_{2}\end{bmatrix} = det  \begin{bmatrix}x_{1}&1\\x_{2}&1\end{bmatrix} = x_1 - x_2 = \begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}x_{1}\\x_{2}\end{bmat  rix}
    T is indeed linear!

    Ker T = \{{\bf x} = (x_1,x_2) \in \mathbb{R}^2| T{\bf x} = x_1 - x_2 = 0\}
    Ker T is the line passing through origin with a slope of 45^{\circ} with the x_2 (or x_1) axis on the x_1-x_2 plane.
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  3. #3
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    Quote Originally Posted by Isomorphism View Post
    1)

    T \begin{bmatrix}x_{1}\\x_{2}\end{bmatrix} = det  \begin{bmatrix}x_{1}&1\\x_{2}&1\end{bmatrix} = x_1 - x_2 = \begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}x_{1}\\x_{2}\end{bmat  rix}
    T is indeed linear!

    Ker T = \{{\bf x} = (x_1,x_2) \in \mathbb{R}^2| T{\bf x} = x_1 - x_2 = 0\}
    Ker T is the line passing through origin with a slope of 45^{\circ} with the x_2 (or x_1) axis on the x_1-x_2 plane.
    Thanks. I'm kind of struggling in the class. Could you please explain to me how you got x_1 - x_2 =  \begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}x_{1}\\x_{2}\end{bmat  rix}? I'm confused as to how you got that 1, -1 on the diagonal.
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by chickeneaterguy View Post
    Thanks. I'm kind of struggling in the class. Could you please explain to me how you got x_1 - x_2 =  \begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}x_{1}\\x_{2}\end{bmat  rix}? I'm confused as to how you got that 1, -1 on the diagonal.
    Just use matrix multiplication and show right hand side of the equation is equal to left hand side
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  5. #5
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    Quote Originally Posted by Isomorphism View Post
    Just use matrix multiplication and show right hand side of the equation is equal to left hand side
    Oh, okay. I see now. Thanks again
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