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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: 7,599 Thanks: 2587 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 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 307 Thanks: 101 Math Focus: Number Theory, Algebraic Geometry  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: 896 
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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eigenvectors, finding, knowing, matrix 
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