Given 2 different orthogonal bases, prove that projection does not depend on the choice of basis. That is (working over $\mathbb{R}^n$), if we have 2 different orthogonal bases, $\mathcal{B} = \{b_1, ...b_n\}$ and $\mathcal{C} = \{c_1, ...c_n\}$, such that $$T_1(v) = \frac{\langle v, b_1\rangle}{\langle b_1, b_1\rangle}b_1 + ... + \frac{\langle v, b_n\rangle}{\langle b_n, b_n\rangle}b_n$$ and $$T_2(v) = \frac{\langle v, c_1\rangle}{\langle c_1, c_1\rangle}c_1 + ... + \frac{\langle v, c_n\rangle}{\langle c_n, c_n\rangle}c_n$$
show that $T_1 = T_2$.
My idea is to prove that both $T_1$ and $T_2$ perform the same transformation in terms of a standard basis, if we use the coordinates for the basis elements in $\mathcal{C}$ and $\mathcal{B}$. Does this seems like a valid way to go?
 A: This is a weird question, since it would be more natural to show that $T_1$ is the identity (then of course so is $T_2$, since the information given about it is identical). The key to this is to first compute the scalar productof some $b_i$ with an arbitrary linear combination $v=\lambda_1b_1+\cdots+\lambda_nb_n$ of $b_1,\ldots,b_n$, that is (since that is a basis) with an arbitrary vector:
$$\def\scal#1#2{\left<{#1},{#2}\right>}
  \scal{b_i}v
  =\scal{b_i}{\lambda_1b_1+\cdots+\lambda_nb_n}
  =\lambda_1\scal{b_i}{b_1}+\cdots+\lambda_n\scal{b_i}{b_n}
  =\lambda_i\scal{b_i}{b_i},
$$
since all dropped scalar products are zero by orthogonality. Then
$$
  T_1(v)=\frac{\scal{b_1}v}{\scal{b_1}{b_1}}b_1 + \cdots 
       + \frac{\scal{b_n}v}{\scal{b_n}{b_n}}b_n
       =\frac{\lambda_1\scal{b_1}{b_1}}{\scal{b_1}{b_1}}b_1 + \cdots 
       + \frac{\lambda_n\scal{b_n}{b_n}}{\scal{b_n}{b_n}}b_n
$$
which simplifies to
$$ T_1(v)=\lambda_1b_1 + \cdots \lambda_nb_n=v
$$
as promised.
A: Let's call our vector space $V$. One approach is to start with the fact that $\mathcal{B}$ and $\mathcal{C}$ are both bases of $V$. In particular, they span $V$. That means that for each $v \in V$, there exist scalars $\beta_1, \dots, \beta_n$, and $\gamma_1, \dots, \gamma_n$, such that 
\begin{align}
v &= \beta_1 b_1 + \cdots + \beta_n b_n, \\
v &= \gamma_1 c_1 + \cdots + \gamma_n c_n.
\end{align}
By orthogonality of $\mathcal{B}$ and $\mathcal{C}$ (consider the inner product of $v$ with some element of $\mathcal{B}$, and do the same for $\mathcal{C}$), we have that for each $1 \leq j \leq n$ $$\beta_j = \frac{\langle v, b_j \rangle}{\langle b_j, b_j \rangle}, \ \text{and} \ \gamma_j = \frac{\langle v, c_j \rangle}{\langle c_j, c_j \rangle}.$$ So for each $v \in V$, we have $T_1(v) = v = T_2(v)$, which implies that $T_1 = T_2 = \text{id}_V$.
