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Let $L:\Bbb R^3\rightarrow \Bbb R^3$ be defined by $L[a_1,a_2,a_3]=[-a_2, a_1+a_2, a_1-a_3]$. Using natural basis for $\Bbb R^3$, find the eigenvalues and eigenvectors of $L$.

How do I represent with a characteristic equation?

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  • $\begingroup$ As you wrote in your question: use the natural basis of $R^3$ to write down the square matrix associated to $L$. Then compute its characteristic polynomial ($\det( \dots)$) and solve. $\endgroup$ – Crostul May 12 '15 at 21:28
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We are asked to find all eigenpairs of the linear map \begin{array}{ccc} \Bbb R^3 & \xrightarrow{L} & \Bbb R^3 \\ \begin{bmatrix}x\\y\\z \end{bmatrix} & \mapsto & \begin{bmatrix} -y\\x+y\\x-z\end{bmatrix} \end{array} The standard basis for $\Bbb R^3$ is $\beta=\{e_1,e_2,e_3\}$ where \begin{align*} e_1 &= \begin{bmatrix}1\\0\\0\end{bmatrix} & e_2 &= \begin{bmatrix}0\\1\\0\end{bmatrix} & e_3 &= \begin{bmatrix}0\\0\\1\end{bmatrix} \end{align*} To compute the matrix of $L$ relative to $\beta$, note that \begin{array}{rcrcrcrc} L(e_1) & = & \color{red}{0}\,e_1 & + & \color{green}{1}\,e_2 & + & \color{blue}{1}\,e_3 \\ L(e_2) & = & \color{red}{-1}\,e_1 & + & \color{green}{1}\,e_2 & + & \color{blue}{0}\,e_3 \\ L(e_3) & = & \color{red}{0}\,e_1 & + & \color{green}{0}\,e_2 & + & \color{blue}{-1}\,e_3 \end{array} This implies $$ [L]_\beta^\beta= \begin{bmatrix} \color{red}{0} & \color{red}{-1} & \color{red}{0} \\ \color{green}{1} & \color{green}{1} & \color{green}{0} \\ \color{blue}{1} & \color{blue}{0} & \color{blue}{-1} \end{bmatrix} $$ Our problem is now equivalent to finding the eigenvalues and corresponding eigenvectors of $[L]_\beta^\beta$. Do you know how to do this?

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The matrix for $L$ is

$$\left[\begin{array}{ccc} 0 & 1 & 1\\ -1 & 1 &0\\ 0 & 0 & -1\\ \end{array}\right] $$ So you need the determinant of $$\left[\begin{array}{ccc} -\lambda & 1 & 1\\ -1 & 1-\lambda &0\\ 0 & 0 & -1-\lambda\\ \end{array}\right] $$ Set it equal to zero and solve for the three roots. $\lambda=-1$ clearly makes the matrix singular, so that is a root, which should make finding the other two easy.

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  • $\begingroup$ where did you get the matrix for L? $\endgroup$ – randev May 12 '15 at 22:00
  • $\begingroup$ Sorry I had it wrong (typo), I think it's right now $\endgroup$ – Gregory Grant May 12 '15 at 22:05

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