# vector decomposition using eigenvectors

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.

• from professor. I don't what does it mean? I found material abot matrix decomposition but I couldn't see anything about vector. Jan 30, 2016 at 10:27
• Have you computed the eigenvectors corresponding to your eigenvalues? That would be the first step. Jan 30, 2016 at 10:28
• 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] Jan 30, 2016 at 10:33
• So now you have a simple linear equation system.... Jan 30, 2016 at 10:35
• 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. Jan 30, 2016 at 10:38

First, determine the eigenvectors $u_i$ for each eigenvalue $\lambda_i$ ($i \in \{1, 2, 3\}$).
$$\begin{pmatrix} -- u_1 -- \\ -- u_2 -- \\ -- u_3 -- \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \\ \gamma \end{pmatrix} = \begin{pmatrix} 50\\ 0 \\ 0 \end{pmatrix}$$
For $\alpha, \beta, \gamma \in \mathbb{R}$
$$\alpha u_1 + \beta u_2 + \gamma u_3 = (50, 0, 0)^T$$