MHB Eigenspace (E's question at Yahoo Answers)

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The eigenvalues of the matrix A are identified as (-9, 0, 9), with the largest being 9. The eigenspace corresponding to the eigenvalue λ=9 is determined by the kernel of (A - 9I). The general form of the eigenvectors for this eigenvalue is expressed as (x1, x2, x3) = α(-1, 0, 1) for any real number α. This indicates that the eigenvectors are scalar multiples of the vector (-1, 0, 1). The discussion provides a clear mathematical derivation of the eigenspace associated with the largest eigenvalue.
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Hello E,

The eigenspace associated to $\lambda=9$ is $\ker (A-9I)$, that is: $$\ker (A-9 I)\equiv{}\begin{bmatrix}{-9}&{\;\;0}&{-9}\\{\;\;0}&{-9}&{\;\;0}\\{-9}&{\;\;0}&{-9}\end{bmatrix} \begin{bmatrix}{x_1}\\{x_2}\\{x_3}\end{bmatrix}= \begin{bmatrix}{0}\\{0}\\{0}\end{bmatrix} \Leftrightarrow \left \{ \begin{matrix} x_1+x_3=0\\x_2=0\end{matrix}\right.\Leftrightarrow \left \{ \begin{matrix} x_1=-\alpha\\x_2=0\\x_3=\alpha\end{matrix}\right. \;(\alpha \in\mathbb{R})$$ so, the general form of the eigenvectors corresponding to $\lambda=9$ is $(x_1,x_2,x_3)^T=\alpha(-1,0,1)^T$ with $\alpha \in\mathbb{R}$.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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