Let
\[ A = \left[ \begin{array}{c c c} 5 & 0 & 0 \\ 1 & 5 & 0 \\ 0 & 1 & 5 \end{array} \right]. \]
For which \(X\) does there exist a scalar \(c\) such that \(AX = cX\) ?
Solution
Let \(X=(x_1,x_2,x_3)^t\). If \(X=0\), then \(AX=cX\) for every scalar \(c\). Assume \(X\ne0\). The equation \(AX=cX\) is \[ (5-c)x_1=0, \] \[ x_1+(5-c)x_2=0, \] \[ x_2+(5-c)x_3=0. \] If \(c\ne5\), the first equation gives \(x_1=0\), then the second gives \(x_2=0\), and the third gives \(x_3=0\), contrary to \(X\ne0\). Thus \(c=5\).
For \(c=5\), the second equation gives \(x_1=0\) and the third gives \(x_2=0\), while \(x_3\) is arbitrary. Hence the nonzero solutions are precisely \[ X=t\begin{bmatrix}0\\0\\1\end{bmatrix},\qquad t\ne0, \] with \(c=5\). Including the zero vector, the vectors for which some scalar \(c\) satisfies \(AX=cX\) are exactly the multiples of \((0,0,1)^t\).