Given an orthonormal basis $\beta$, transformation $T$ and $T'$, show that $M_\beta(T')$ is the transpose of $M_\beta(T)$

Question

Let $\beta=\{f_1,f_2,...,f_n\}$ be an orthonormal basis of an inner product space $V$. Given a linear operator $T:V \rightarrow V$, define $T':V \rightarrow V$ by $$T'(v)=\sum_{i=1}^n \langle v,T(f_i) \rangle f_i$$ Show that $M_\beta(T')$ is the transpose of $M_\beta(T)$.

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We know that for any $v \in V$, we can write $v= \sum_{i=1}^n \langle v,f_i \rangle f_i$. So in this case $T′(v)=∑_{i=1}^n⟨T′(v),f_i⟩f_i=∑_{i=1}^n⟨v,T(f_i)⟩f_i$. Sadly I do not really know how we can use this to our advantage. I have thought about using the fact that $⟨T′(v),f_i⟩ = ⟨v,T(f_i)⟩$, but I do not see how I can continue. Could anyone provide me with a hint?

Answer 1

But, isn't it true that $\langle Tv,w\rangle=\langle v,T^tw\rangle$? This appears to make it rather trivial, since then $T'$ and $T^t$ agree on any basis.