The eigenvalues of the matrix $$ \begin{pmatrix} 0 & 3 & 0 \\ 3 & 0 & 4 \\ 0 & 4 & 0 \\ \end{pmatrix} $$ are $5$, $0$, and $‐5$.
Question: Decompose the vector $(50, 0, 0)^T$ as a linear combination of eigenvectors.
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$\begingroup$ from professor. I don't what does it mean? I found material abot matrix decomposition but I couldn't see anything about vector. $\endgroup$– Murat CabukJan 30, 2016 at 10:27
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$\begingroup$ Have you computed the eigenvectors corresponding to your eigenvalues? That would be the first step. $\endgroup$– Matthew CassellJan 30, 2016 at 10:28
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$\begingroup$ yes I have eigenvectors : for -5 = [1; -5/3; 4/3] , for 5 = [1; 5/3 ; 4/3] and for 0 = [1;0;-3/4] $\endgroup$– Murat CabukJan 30, 2016 at 10:33
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$\begingroup$ So now you have a simple linear equation system.... $\endgroup$– adjanJan 30, 2016 at 10:35
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$\begingroup$ So the question you are now being asked is to find scalars $a, b$ and $c$ such that $$av_{1} + bv_{2} + cv_{3} = (50, 0, 0)^{T}$$ where the $v_{i}$ are the eigenvectors you have found. $\endgroup$– Matthew CassellJan 30, 2016 at 10:38
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1 Answer
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First, determine the eigenvectors $u_i$ for each eigenvalue $\lambda_i$ ($i \in \{1, 2, 3\}$).
Then, solve the linear equation system:
$$\begin{pmatrix} -- u_1 -- \\ -- u_2 -- \\ -- u_3 -- \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \\ \gamma \end{pmatrix} = \begin{pmatrix} 50\\ 0 \\ 0 \end{pmatrix} $$
For $\alpha, \beta, \gamma \in \mathbb{R}$
Then your decomposition is:
$$\alpha u_1 + \beta u_2 + \gamma u_3 = (50, 0, 0)^T$$
