Show that $DLD^{-1} = dL$ for $D$ a diagonal matrix, $L$ a sub-diagonal matrix, and $U$ a super-diagonal matrix

Question

Let $D$ be a diagonal matrix with $D_{i, i} = d^{i-1},$ where $d$ is some real constant. Let $L$ be a matrix with entries only on its first sub-diagonal, and let $U$ be a matrix with entries only on its first super-diagonal. Then $DLD^{-1} = dL, DUD^{-1}=\frac{1}{d}U.$

Part of some derivations regarding the SOR optimization method. Is there any nice way to check this claim?

Answer 1

Since $D$ is diagonal, $(DLD^{-1})_{i+1,i} = (D)_{i+1,i+1} (L)_{i+1,i} (D^{-1})_{i,i}$ = $d^{i} (L)_{i+1,i} d^{1-i} = d (L)_{i+1,i}$. Similarly $(DUD^{-1})_{i,i+1} = d^{i-1} (U)_{i,i+1} d^{-i} = (U)_{i,i+1} /d$.