Solution:

$A^{2}=A A=\left[\begin{array}{lll}{2} & {4} & {4} \\ {4} & {2} & {4} \\ {4} & {4} & {2}\end{array}\right]\left[\begin{array}{lll}{2} & {4} & {4} \\ {4} & {2} & {4} \\ {4} & {4} & {2}\end{array}\right]$

$A^{2}=\left[\begin{array}{ccc}{4+16+16} & {8+8+16} & {8+16+8} \\ {8+8+16} & {16+4+16} & {16+8+8} \\ {8+16+8} & {16+8+8} & {16+16+4}\end{array}\right]$

$A^{2}=\left[\begin{array}{lll}{36} & {32} & {32} \\ {32} & {36} & {32} \\ {32} & {32} & {36}\end{array}\right]$

$8 A=8\left[\begin{array}{lll}{2} & {4} & {4} \\ {4} & {2} & {4} \\ {4} & {4} & {2}\end{array}\right]=\left[\begin{array}{ccc}{16} & {32} & {32} \\ {32} & {16} & {32} \\ {32} & {32} & {16}\end{array}\right]$

$\therefore A^{2}-8 A=\left[\begin{array}{ccc}{36} & {32} & {32} \\ {32} & {36} & {32} \\ {32} & {32} & {36}\end{array}\right]-\left[\begin{array}{ccc}{16} & {32} & {32} \\ {32} & {16} & {32} \\ {32} & {32} & {16}\end{array}\right]=\left[\begin{array}{ccc}{20} & {0} & {0} \\ {0} & {20} & {0} \\ {0} & {0} & {20}\end{array}\right]$