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December 16th, 2017, 02:45 AM   #1
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Finding eigenvectors without knowing the matrix?

Let $\displaystyle B=A^{12}-8A^{7}+5A^{5}+4I$,$\displaystyle A=\begin{pmatrix}
2 & 1 & 2\\
0 & 4 & 1\\
0 & -2 & 1
\end{pmatrix}$

Find two independent eigenvectors for B.

How do I do this?
I found that the eigenvalues of A are 3 , 2 (double root) and that the eigenspaces are $\displaystyle V(2)={x_{1}\begin{pmatrix}
1 & 0 & 0
\end{pmatrix}^{t},x_{1}\epsilon\mathbb{R}}$
and $\displaystyle V(3)={x_{1}\begin{pmatrix}
1 & -1 & 1
\end{pmatrix}^{t},x_{1}\epsilon\mathbb{R}}$
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December 16th, 2017, 03:26 AM   #2
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Recall that (subject to certain conditions)$$A=SPS^{-1}$$ where $S$ is formed from the eigenvectors of $A$ and $P$ is the diagonal matrix having the eigenvalues of $A$ as elements in the same order as the corresponding eigenvectors.
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December 16th, 2017, 04:30 AM   #3
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Quote:
Originally Posted by v8archie View Post
Recall that (subject to certain conditions)$$A=SPS^{-1}$$ where $S$ is formed from the eigenvectors of $A$ and $P$ is the diagonal matrix having the eigenvalues of $A$ as elements in the same order as the corresponding eigenvectors.
But neither A nor B here satisfy these conditions (they both have an eigenvalue whose geometric multiplicity is less than its algebraic multiplicity), so they aren't diagonalisable.
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December 20th, 2017, 04:27 AM   #4
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If v is an eigenvector of A, with eigenvalue $\lambda$ then $Bv= A^{12}v- 8A^7v+ 5A^5v+ 4v= \lambda^{12}v- 8\lambda^7v+ 5\lambda^5v+ 4v= (\lambda^{12}- 8\lambda^7+ 5\lambda^5+ 4)v$.

That is, any eigenvector of A, with eigenvalue $\lambda$, is an eigenvector of B with eigenvalue $\lambda^{12}- 8\lambda^7+ 5\lambda^5+ 4$
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