The problem
Let $$ B = \left( \begin{array}{c c c} 1 & 0.5 & 0.2 \\ 0.5 & 1 & 0.5 \\ 0.2 & 0.5 & 1 \\ \end{array} \right) $$ Let $I = \left\{ 1,3 \right\}.$ We define $Q = (q_{i,j})$ s.t $q_{i,j} = 1$ if $i = j \in I$ and $q_{i,j} = 0,$ otherwise. Let $c = (1,1,1)$ and $x$ a vector.
Show $QBQx = Qc$, with $Qx = x$, has one, and only one, solution.
My works
I found $$ Q = \left( \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ Then $$ QBQ = \left( \begin{array}{c c c} 1 & 0 & 0.2 \\ 0 & 0 & 0 \\ 0.2 & 0 & 1 \\ \end{array} \right) $$ But $QBQ$ is not inversible. I don't understand how we can show the equation $QBQx = Qc$ has one solution...
Thank you
According to your question, we have to satisfy 2 conditions:$QBQx=Qc$ and $Qx=x$.
Let $x=(x_1,x_2,x_3)'$, then from $Qx=x$, we have $x_2 =0$.
Now consider, $QBQx=Qc$, then $\begin{pmatrix} 1&0&0.2\\0&0&0\\0.2&0&1 \end{pmatrix}\begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix}=\begin{pmatrix} 1\\0\\1 \end{pmatrix} = \begin{pmatrix} x_1+0.2x_3\\0\\0.2x_1+x_3 \end{pmatrix}$
This gives $x_1=\frac{80}{96}, x_3=\frac{160}{192}$.