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I am given the $2\times2$ matrix $$A = \begin {bmatrix} -2&-1 \\\\ 15&6 \ \end{bmatrix}$$

I calculated the Eigenvalues to be 3 and 1. How do I find the vectors? If I plug the value back into the character matrix, I get $$B = \begin {bmatrix} -5&1 \\\\ 15&3 \ \end{bmatrix}$$

Am I doing this right? What would the eigenvector be?

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Almost right, only the $1$ in the upper right hand corner of $B$ should be a $-1$. Can you find the eigenvectors now?

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Solve the equation $Bx = 0$. If $Bx=0$ then $(A-\lambda I)x = 0$, so $Ax = \lambda I x = \lambda x$, so $x$ is an eigenvector for $\lambda$.

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Remember what the word "eigenvector" means. If $3$ is an eigenvalue, then you're looking for a vector satisfying this: $$A = \begin {bmatrix} -2&-1 \\\\ 15&6 \ \end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix} = 3\begin{bmatrix} x \\ y\end{bmatrix}$$

Solve that. You'll get infinitely many solutions since every scalar multiple of a solution is also a solution.

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