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→Eigenbasis: clean up; most of what was said here already had been above, or is evident |
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For the transformation matrix
:<math>A = \begin{bmatrix} 3 & -1\\-1 & 3 \end{bmatrix},</math>
the vector
:<math>v = \begin{bmatrix} 4 \\ -4 \end{bmatrix}</math>
is an eigenvector with eigenvalue 2. Indeed,
:<math>A v = \begin{bmatrix} 3 & -1\\-1 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ -4 \end{bmatrix} = \begin{bmatrix} 3 \cdot 4 + -1 \cdot (
::<math> = \begin{bmatrix} 8 \\ -8 \end{bmatrix} = 2 \cdot \begin{bmatrix} 4 \\ -4 \end{bmatrix}.</math>
On the other hand the vector
:<math>v = \begin{bmatrix} 0 \\ 1 \end{bmatrix}</math>
is ''not'' an eigenvector, since
:<math>\begin{bmatrix} 3 & -1\\-1 & 3 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \cdot 0 + 1 \cdot 1 \\ 1 \cdot 0 + 3 \cdot 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \end{bmatrix},</math>
and this vector is not a multiple of the original vector <math>v</math>.
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