Background
I have been self-studying the Discrete Fourier Transform. Suppose we have a discrete, real data vector $\vec{f}=(f_0,f_1,...,f_{N-1})^T$. The discrete Fourier transform and inverse transform can be written
\begin{align*} \hat{f}_k &= \frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}f_je^{-i2\pi k \frac{j}{N}}\\ f_j&=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1} \hat{f_k} e^{i2\pi k \frac{j}{N}} \end{align*}
Question
I'm trying to understand this in terms of basis vectors. Denote the normalized basis vector corresponding to frequency $k$ as
$$ \vec{\phi}_k =\frac{1}{\sqrt{N}}(1,e^{i2\pi k \frac{1}{N}},e^{i2\pi k \frac{2}{N}},...,e^{i2\pi k \frac{N-1}{N}}) $$
Then the coefficient $\hat{f}_k$ should correspond to the projection of the data onto basis vector $\vec{\phi}_k$. However, I'm not super familiar with complex vectors and complex projections. When I naively "project" the data, I have something like this:
$$ \hat{f}_k=\vec{f}\cdot\vec{\phi}_k= \vec{f}^{\dagger}\vec{\phi}_k=\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}f_je^{i2\pi k \frac{j}{N}} $$
This expression is missing the negative sign. It seems that instead my projection should be $\vec{\phi}_k^{\dagger}\vec{f}$. What am I misunderstanding here?
If this projection is correct (albeit unconventional), how do I transform back to the data? Thinking of vectors and projections again, I think we should write
$$ \vec{f}=\hat{f}_0\vec{\phi}_0 + \hat{f}_1\vec{\phi}_1 + ... + \hat{f}_{N-1}\vec{\phi}_{N-1} $$
which is equivalent to
$$ f_j =\sum_{k=0}^{N-1}\hat{f}_k e^{i2\pi k \frac{j}{N}} $$
Since my convention takes away the negative sign in the projection step, I was thinking by symmetry I would have a negative in the exponential for the inverse step, but there isn't a negative.
Edit - Update 1
Perhaps one reason that we should write the projection as $\vec{\phi}_k^{\dagger}\vec{f}$ is that it conforms with writing the DFT as a matrix-vector operation.
The discrete Fourier transform can be written as follows. Let $\omega_N=e^{-i2\pi \frac{1}{N}}$.
$$ \hat{f}= \begin{bmatrix} \hat{f_0}\\ \hat{f_1}\\ \vdots \\ \hat{f}_{N-1} \end{bmatrix} = F\vec{f} = \begin{bmatrix} \vec{\phi}^\dagger_0 \\ \vec{\phi}^\dagger_1 \\ \vdots \\ \vec{\phi}^\dagger_{N-1} \end{bmatrix} \begin{bmatrix} f_0\\ f_1\\ \vdots \\ f_{N-1} \end{bmatrix} = \frac{1}{\sqrt{N}} \begin{bmatrix} 1 & 1 & 1 &\cdots & 1 \\ 1 & \omega_N & \omega_N^2 & \cdots & \omega_N^{N-1} \\ \vdots \\ 1 & \omega_N^{N-1} & \omega_N^{2(N-1)} & \cdots & \omega_N^{(N-1)^2} \end{bmatrix} \begin{bmatrix} f_0\\ f_1\\ \vdots \\ f_{N-1} \end{bmatrix} $$
Also, by writing my projection as
$$ \tilde{f}_k=\vec{f}\cdot\vec{\phi}_k= \vec{f}^{\dagger}\vec{\phi}_k=\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}f_je^{i2\pi k \frac{j}{N}} $$
I've essentially defined my projection coefficients as the complex conjugate of the convention. Denote my coefficents as $\tilde{f}_k$ and the conventional coefficients as $\hat{f}_k$. Then
$$ \overline{\hat{f}_k} = \overline{\vec{\phi}^\dagger_k\vec{f}} = \vec{f}^\dagger\vec{\phi}_k = \tilde{f}_k $$
Essentially, my definition of $\tilde{f}_k$ implies that my basis vectors are the complex conjugates of the basis vectors used in the conventional definition. This question shows that if a set of complex vectors form a basis for some space, then so does the set of conjugate vectors.
The moral of the story is that deciding between projection coefficients $$\tilde{f}_k = \vec{f}^T\vec{\phi}_k$$ or $$ \hat{f}_k=\vec{\phi}_k^\dagger \vec{f} $$ is equivalent to choosing basis vectors $\left\{\overline{\vec{\phi}_k}\right\}_{k=0}^{N-1}$ or $\left\{\vec{\phi}_k\right\}_{k=0}^{N-1}$, respectively. In either case, the DFT matrix provides an invertible transformation from real space to frequency space.
