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 December 16th, 2017, 03:45 AM #1 Member   Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 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 $\displaystyle V(3)={x_{1}\begin{pmatrix} 1 & -1 & 1 \end{pmatrix}^{t},x_{1}\epsilon\mathbb{R}}$
 December 16th, 2017, 04:26 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,502 Thanks: 2511 Math Focus: Mainly analysis and algebra 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.
December 16th, 2017, 05:30 AM   #3
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Quote:
 Originally Posted by v8archie 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.

 December 20th, 2017, 05:27 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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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