"Basic" properties of linear/dual operators Let $L: X \to Y$ be a linear operator between real vector spaces $X$ and $Y$. The dual
operator $L' : Y \to X$ is defined by
$$(L' y)(x) = y (Lx)$$
for all $x \in X$ and $y \in Y$. My book leaves several "basic" properties as exercises,
but I was wondering how to prove two of them in detail as practice. I tried searching for these properties online but could not find them, so I apologize if this is a duplicate. Let $X,Y,Z$ be real vector spaces,
1) If $A: X \to Y$ and $B: Y \to Z$ are linear operators
then $(AB)' = B'A'$.
2) If $C: X \to X$ is a linear operator and $C^{-1} : X \to X$ 
exists then $(C')^{-1} :X \to X$ exists and $(C') ^{-1} = (C^{-1})'$.
Some of my thoughts on $(1)$: shouldn't it also be assumed that $AB$ exists, or is that 
implied? I also
struggled with this since I don't understand why we have $B'A'$ instead of $A'B'$
since everything we are working with has a linearity property. 
 A: First, as noted by Jochen, the dual operator goes from the dual space $Y'$ to the dual space $X'$.
1) Let $A: X \rightarrow Y$ and $B: Y \rightarrow Z$ be linear operators. Then $BA: X \rightarrow Z$ is well defined linear operator. (note that here we have $BA$, not $AB$ which does not make sense). Now $(BA)': Z' \rightarrow X'$ and for $z' \in Z'$ and $x \in X$ we have
$$((BA)'z')(x) = z'(BAx) = (B'z')(Ax) = (A'B'z')(x).$$
This holds for any $x \in X$ so the operators $(BA)'z'$ and $A'B'z'$ are equal elements of $X'$. Now this holds for any $z' \in Z'$, hence, $(BA)'$ and $A'B'$ are equal operators from $Z'$ to $X'$.
2) Let $C: X \rightarrow X$ be an invertible operator with inverse $D := C^{-1}$. We want to show that $(C')^{-1}$ exists and is equal to $D'$. First note that the dual operator to identity is also identity, ie. $(I_X)' = I_{X'}$, as for any $x \in X$ and $x' \in X'$ we have
$$((I_X)'x')(x) = x'(I_X x) = x'(x) = (I_{X'}x')(x).$$
Now from 1) adn the fact that $D = C^{-1}$ we have
\begin{align}I_{X'} = (I_X)' &=  (CD)' = D'C' \\ &= (DC)' = C'D'.\end{align}
Hence $C'D' = I_{X'} = D'C'$ and $(C')^{-1}$ is invertible with inverse $D' = (C^{-1})'$.
