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I'm trying to find the Eigenvalue and Eigenvector for the Linear transformation:
$T:\mathbb{Z}_2^4 \to \mathbb{Z}_2^4: (x_1,x_2,x_3,x_4)=(x_1+x_3,-2x_1-x_3,x_2+x_4,x_2-x_4)$
My problem is with $\mathbb{Z}_2^4$ cause I've never worked with this before.
So what I did:
I've found the representing matrix of the transformation using the standard basis of $\mathbb{R}^4$ and got $$ [T]_B^B= \left( \begin{array}{ccc} 1 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & -1 \end{array} \right)$$ Then did $det([T]_B^B-\lambda I)$ and received $\lambda ^2(\lambda +1)(\lambda -1)$ as characteristic polynomial, so I've said that there are two Eigenvalue: $$\lambda_1=0, \lambda_2=1, \lambda_3=-1 \equiv 1 \pmod 2$$ So for zero I got $$ sp \left( \begin{array}{ccc} 0 \\ 0 \\ 0 \\ 0 \end{array} \right)$$ and for 1 I got $$ sp \left( \begin{array}{ccc} 0 \\ 0 \\ 0 \\ 1 \end{array} \right)$$
Is this right?

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  • $\begingroup$ The zero vector is never an eigenvector. 0101 is an eigenvector for the eigenvalue zero. And 0001 is not an eigenvector for the eigenvalue 1, but 1000 is. $\endgroup$ – Gerry Myerson Nov 21 '14 at 11:52
  • $\begingroup$ I fixed my eigenvector of 1, didn't quite understand how you got the answer for zero and will thank you if you explain. $\endgroup$ – user114138 Nov 21 '14 at 12:20
  • $\begingroup$ The 2nd & 4th columns are equal, so 0101 is an eigenvector for the eigenvalue zero. $\endgroup$ – Gerry Myerson Nov 21 '14 at 22:21

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