If say, a matrix A is s.t. A(0,3,5) = (0,6,10), then obviously, the eigenvalue is 2.
But how do you explain why (5,3,0) is also an eigenvector with eigenvalue of 2?
uggh, sorry, I changed the question slightly so I could try work out the question myself for a different set of vectors. The original question was:
Suppose that for some 3x3 matrix A, we have A(1,0,-1) = (2,0,-2), and A(1,2,0) = (2,4,0).
a) Find one eigenvalue of A.
b) Explain why (0,2,1) is an eigenvector for this eigenvalue and hence find A(0,2,1).
You changed the question...""slightly""??! I suppose "slightly" must have another meaning in the southern hemisphere...
Anyway, since $\displaystyle (0,2,1)=(1,2,0)-(1,0,-1)$ and $\displaystyle A$ is linear then a linear combination of eigenvectors corresponding to one single eigenvalue is
again an eigenvector corresponding to the same eigenvalue, and we check:
$\displaystyle A(0,2,1)=A(1,2,0)-A(1,0,-1)=(2,4,0)-(2,0,-2)=(0,4,2)=2\cdot (0,2,1)$
Tonio
No - he's right. It is a change of basis (a simple column swap).
If $\displaystyle A =\left( \begin{array}{ccc}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3 \end{array} \right)$ then define $\displaystyle A^{\prime} = \left( \begin{array}{ccc}
c_1 & b_1 & a_1 \\
c_2 & b_2 & a_2 \\
c_3 & b_3 & a_3 \end{array} \right)
$.
Then $\displaystyle A^{\prime}\left( \begin{array}{c}5\\ 3\\ 0\end{array}\right) = \left(\begin{array}{c}10\\ 6\\ 0\end{array}\right)$ as required...
Unless I am missing something?
Yes, that mixing the matrix'x columns is mixing the basis of the vector space which was chosen to represent the operator, and then you also must change the representation of each vector according to the new basis...or else you must check whether any given equality is true wrt the same basis.
For example, $\displaystyle \begin{pmatrix}1&\!\!\!-5&3\\1&7&\!\!\!-3\\3&0&2\end{pmatrix}\begin{pmatrix}0\\3\\5\end{pm atrix}=\begin{pmatrix}0\\6\\10\end{pmatrix}=2\cdot \begin{pmatrix}0\\3\\5\end{pmatrix}$ , but $\displaystyle \begin{pmatrix}1&\!\!\!-5&3\\1&7&\!\!\!-3\\5&3&0\end{pmatrix}\begin{pmatrix}5\\3\\0\end{pm atrix}\neq 2\cdot \begin{pmatrix}5\\3\\0\end{pmatrix}$ .
Tonio