A square matrix $A$ is called half-magic if the sum of the numbers in each row and column is the same. The common sum in each row and column is denoted by $s(A)$ and is called the magic sum of the matrix $A$. Let $V$ be the vector space of $2\times 2$ half-magic squares.
(a) Find
$s(\left[\begin{array}{cc} -6 &1\cr 1 &-6 \end{array}\right]) =$ .
Consider the linear operator $L:V\to V$ that subtracts the magic sum from every entry in the matrix. More preciselly, $L(A) := A-s(A) \left[\begin{array}{cc} 1 &1\cr 1 &1 \end{array}\right]$.
(b) Find
$L(\left[\begin{array}{cc} -6 &1\cr 1 &-6 \end{array}\right]) =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$.
(c) Find an ordered basis $B$ for $V$.
$B = ($ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$, $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$ $)$.
(d) Find the matrix $[L]_B^B$ of $L$ in your chosen basis $B$.
$[L]_B^B =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$.

You can earn partial credit on this problem.