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Math Help - Eigenvector, linear transformation problem

  1. #1
    MHF Contributor arbolis's Avatar
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    Eigenvector, linear transformation problem

    Let T: \mathbb{R}^2 \to \mathbb{R}^2 defined by T(x,y)=(3x+y, 2x+2y).
    Show that there exists a basis \bold {B} =\{ \alpha _1 , \alpha _2 \} of \mathbb{R}^2 formed by eigenvectors.
    Find the matrix of T with respect of the canonical basis and the matrix of T with respect of the basis of eigenvectors of \bold {B}.

    My attempt: I don't know how to show the first part.
    I think I've found the matrix of T with respect to the canonical basis :
    T(1,0)=(3,2)
    T(0,1)=(1,2). Hence the matrix is \begin{bmatrix} 3 & 1 \\ 2 & 2   \end{bmatrix}. Let's call it T'.
    I don't know how to get it with respect to another basis, even if I realize I must know it since it's very basic.
    From memory, I think that eigenvalues of T are the \lambda in \det [I \lambda - T']=0 \Leftrightarrow \lambda=1 \text{ or } 4. I guess that by finding the eigenvalues or T it would help to find its eigenvectors... I'm lost! I'd appreciate a bit of help if you can. Thanks.
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    Quote Originally Posted by arbolis View Post
    Hence the matrix is \begin{bmatrix} 3 & 1 \\ 2 & 2   \end{bmatrix}. Let's call it T'.
    I don't know how to get it with respect to another basis, even if I realize I must know it since it's very basic.
    From memory, I think that eigenvalues of T are the \lambda in \det [I k - T']=0 \Leftrightarrow k=1 \text{ or } 4. I guess that by finding the eigenvalues or T it would help to find its eigenvectors... I'm lost! I'd appreciate a bit of help if you can. Thanks.
    You established that k=1\text{ or }4. Therefore, there is a non-trivial solution to (working with k=1 first):
    \begin{bmatrix} 3 & 1 \\ 2 & 2   \end{bmatrix} \begin{bmatrix}x_1\\x_2 \end{bmatrix} = 1\cdot \begin{bmatrix}x_1\\x_2\end{bmatrix}.
    This corresponds to:
    \left\{ \begin{array}{c}3x_1 + 1x_2 = x_1 \\ 2x_1 + 2x_2 = x_2 \end{array} \right. \implies \left\{ \begin{array}{c} 2x_1 + x_2 = 0 \\ 2x_1 + x_2 = 0 \end{array} \right.
    The solution to this system is: x_1 = t\text{ and }x_2 = -2t, t\in \mathbb{R}.
    The eigenvectors corresponding to the eigenvalue one are therefore: \left\{ k \begin{bmatrix}1\\-2\end{bmatrix} : k \in \mathbb{R}^{\times} \right\}
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    MHF Contributor arbolis's Avatar
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    Thank you TPH,
    by the way, what does \mathbb{R}^{\times} mean?
    Last edited by arbolis; February 13th 2009 at 05:54 AM.
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    Quote Originally Posted by arbolis View Post
    Thank you TPH,
    by the way, what does \mathbb{R}^{\times} mean?
    No, turns out ThePerfectHacker didn't mean what I thought at all!
    Last edited by HallsofIvy; February 13th 2009 at 09:09 AM.
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    Quote Originally Posted by arbolis View Post
    Thank you TPH,
    by the way, what does \mathbb{R}^{\times} mean?
    We define \mathbb{R}^{\times} = \{ x\in \mathbb{R} | x\not = 0\}.
    The reason I did that because I wanted to avoid the zero vector, usually when we look for eigenvectors we ignore the trivial ones.
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  6. #6
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    We define \mathbb{R}^{\times} = \{ x\in \mathbb{R} | x\not = 0\}.
    The reason I did that because I wanted to avoid the zero vector, usually when we look for eigenvectors we ignore the trivial ones.
    Ah ok. I knew it as R^{*}.
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