# Prove that $(R+S)' \simeq R' + S'$ for any two representations R and S of a group G.

The Contragradient or Dual Representation is defined as:

My trail: $$((R+S)'(g)f)(x) = f((R +S)(g)^{-1} x)$$, $$(g \in G, f \in V', x \in V)$$

But then I am stucked, how to distribute the inverse, Is it allowed to write it as $$f ((R^{-1} + S^{-1})(g)x)$$ ...... if I can do so then the solution will be very easy ..... so if anyone give me a hint about the solution I would appreciate this very much?

• Transpose is linear. – David Hill Feb 13 at 5:16
• could you please say a little more details?@DavidHill – hopefully Feb 15 at 0:05
• @DavidHill but my definition for duality does not contain transpose ...... from where the transpose will come? – hopefully Feb 16 at 2:14