**Proposition 2**. *Fractional-order matrix is used as the basis function, so the M-WFRFT has unitarity.*

**Proof.** The calculation of the fractional power of the matrix is applied to the eigenvalues, so eigenvalue decomposition of the matrix is required. Therefore, the eigendecomposition of the matrix can be expressed as

$$F = VDV^H,\tag{32}$$

where *F* is the DFT matrix, *V* is the eigenvector, and *D* is the eigenvalue matrix.

In [26,27], the eigenvalues of the DFT can be expressed as *λ<sup>n</sup>* = *enπi*/2. Then, the possible values of the eigenvalue are *λ<sup>r</sup>* = {1, −1, *i*, −*i*};*r* = 1, 2, ··· , *n*. In this way, the eigenvalue matrix *D* can be expressed as

$$D = \begin{pmatrix} \lambda\_1 & 0 & \cdots & 0 \\ 0 & \lambda\_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda\_n \end{pmatrix} . \tag{33}$$

Then, the fractional power operation of matrix *F* can be expressed as

$$F^{4l/M} = VD^{4l/M}V^H.\tag{34}$$

where *l* = 0, 1, ··· , *M* − 1. For Equation (10), *Yk* can be expressed as

$$\begin{split} Y\_k &= u^{0 \times k} I + u^{1 \times k} \times F^{\frac{4}{M}} + \cdots + u^{(M-1) \times k} \times F^{\frac{4(M-1)}{M}} \\ &= u^{0 \times k} V D^0 V^H + u^{1 \times k} V D^{4/M} V^H + \cdots + u^{(M-1) \times k} V D^{4(M-1)/M} V^H. \end{split} \tag{35}$$
