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Thread: Eigenvector, linear transformation problem

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    MHF Contributor arbolis's Avatar
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    Eigenvector, linear transformation problem

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

    My attempt: I don't know how to show the first part.
    I think I've found the matrix of $\displaystyle T$ with respect to the canonical basis :
    $\displaystyle T(1,0)=(3,2)$
    $\displaystyle T(0,1)=(1,2)$. Hence the matrix is $\displaystyle \begin{bmatrix} 3 & 1 \\ 2 & 2 \end{bmatrix}$. Let's call it $\displaystyle T'$.
    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 $\displaystyle T$ are the $\displaystyle \lambda$ in $\displaystyle \det [I \lambda - T']=0 \Leftrightarrow \lambda=1 \text{ or } 4$. I guess that by finding the eigenvalues or $\displaystyle T$ 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 $\displaystyle \begin{bmatrix} 3 & 1 \\ 2 & 2 \end{bmatrix}$. Let's call it $\displaystyle T'$.
    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 $\displaystyle T$ are the $\displaystyle \lambda$ in $\displaystyle \det [I k - T']=0 \Leftrightarrow k=1 \text{ or } 4$. I guess that by finding the eigenvalues or $\displaystyle T$ it would help to find its eigenvectors... I'm lost! I'd appreciate a bit of help if you can. Thanks.
    You established that $\displaystyle k=1\text{ or }4$. Therefore, there is a non-trivial solution to (working with $\displaystyle k=1$ first):
    $\displaystyle \begin{bmatrix} 3 & 1 \\ 2 & 2 \end{bmatrix} \begin{bmatrix}x_1\\x_2 \end{bmatrix} = 1\cdot \begin{bmatrix}x_1\\x_2\end{bmatrix}$.
    This corresponds to:
    $\displaystyle \left\{ \begin{array}{c}3x_1 + 1x_2 = x_1 \\ 2x_1 + 2x_2 = x_2 \end{array} \right. \implies \left\{ \begin{array}{c} 2x_1 + x_2 = 0 \\ 2x_1 + x_2 = 0 \end{array} \right.$
    The solution to this system is: $\displaystyle x_1 = t\text{ and }x_2 = -2t, t\in \mathbb{R}$.
    The eigenvectors corresponding to the eigenvalue one are therefore: $\displaystyle \left\{ k \begin{bmatrix}1\\-2\end{bmatrix} : k \in \mathbb{R}^{\times} \right\}$
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    MHF Contributor arbolis's Avatar
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    Thank you TPH,
    by the way, what does $\displaystyle \mathbb{R}^{\times}$ mean?
    Last edited by arbolis; Feb 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 $\displaystyle \mathbb{R}^{\times}$ mean?
    No, turns out ThePerfectHacker didn't mean what I thought at all!
    Last edited by HallsofIvy; Feb 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 $\displaystyle \mathbb{R}^{\times}$ mean?
    We define $\displaystyle \mathbb{R}^{\times} = \{ x\in \mathbb{R} | x\not = 0\}$.
    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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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    We define $\displaystyle \mathbb{R}^{\times} = \{ x\in \mathbb{R} | x\not = 0\}$.
    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 $\displaystyle R^{*}$.
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