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Block Diagonalization of Block Circulant Quaternion Matrices and the Fast Calculation for T-Product of Quaternion Tensors

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Abstract

With quaternion matrices and quaternion tensors being gradually used in the color image and color video processing, the block diagonalization of block circulant quaternion matrices has become a key issue in the establishment of T-product based methods for quaternion tensors. Out of this consideration, we aim at establishing a fast calculation approach for the block diagonalization of block circulant quaternion matrices with the help of the fast Fourier transform (FFT). We first show that the discrete Fourier matrix \(\mathbf {F_p}\) cannot diagonalize \(p\times p\) circulant quaternion matrices, nor can the unitary quaternion matrices \(\mathbf {F_p}\textbf{j}\) and \(\mathbf {F_p}(1+\textbf{j})/\sqrt{2}\) with \(\textbf{j}\) being an imaginary unit of quaternion algebra. Then we prove that the unitary octonion matrix \(\mathbf {F_p}\textbf{p}\) with \(\textbf{p}=\textbf{l},\textbf{il}\) or \((\textbf{l}+\textbf{il})/\sqrt{2}\) (\(\textbf{l}, \textbf{il}\) being imaginary units of octonion algebra) can diagonalize a circulant quaternion matrix of size \(p\times p\), which further means that a block circulant quaternion matrix of size \(mp\times np\) can be block diagonalized at the cost of \(O(mnp\log p)\) via the FFT. As one of applications, we give a fast algorithm to speed up the calculation of the T-product between \(m\times n\times p\) and \(n\times s\times p\) third-order quaternion tensors via FFTs, whose computational magnitude is almost 1/p of the original one. As another application, we propose an effective compression strategy for third-order quaternion tensors with a certain low-rankness. Simulations on the color image and color video compression demonstrate that our compression strategy with no QSVD involved, can achieve higher quality compression in terms of PSNR values at much less time costs, compared with the QSVD-based methods.

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Data Availability

We do not analyse or generate any data sets, because our work proceeds within a theoretical and mathematical approach.

Notes

  1. In [17], the tubal-scalars of \(\mathcal {S}\) are called eigentuples of tensor \(\mathcal {A}\). For more relevant studies on the eigentuples and the corresponding eigenmatrices, the linear combinations with tubal-scalars, tensor polynomials and the computation of tensor eigendecomposition for third-order tensors, based on the T-product algebra and the diagonalization theory of circulant matrices, please see [17] for reference.

  2. The Matlab command \(svd(\hat{\mathcal {A}}(:,:,r),k)\) means to compute the first k largest singular values for each matrix \(\hat{\mathcal {A}}(:,:,r)\).

  3. It can be seen that when the quaternion tensor truncation problem for tensor \(\mathcal {A}\) degenerates to the real case, the term “\((\mathbf {P_p}\otimes \mathbf {I_m})unfold(\overline{\hat{\mathcal {A}}_k^{(1,\textbf{i})}})\)” in the 5-th line of Algorithm 5.2 becomes “\(unfold({\hat{\mathcal {A}}_k^{(1,\textbf{i})}})\)” as a result of the conjugate symmetry of the fourier transform performing on real numbers (please see the reference [17] by Kilmer, Braman, Hao and Hoover in 2013), which implies that our method provided by Algorithm 5.2 reduces to the classic truncated approximation strategy for third-order real tensors given in reference [18], i.e. Algorithm 2.2.

  4. The MATLAB code can be downloaded from https://hkumath.hku.hk/\(\sim \)mng/mng_files/LANQSVDToolbox.zip.

  5. It can be seen that in the compression for quaternion tensor \(\overrightarrow{\mathcal {M}}\), there are mainly Fourier transform and vector norm computations involved in Algorithm 5.2. So, we refer our method here to as Fourier transform based quaternion vector norm (FT-QVN) method.

  6. It is observed that this way of real tensor construction achieves the best performance for color image processing [22].

  7. http://trace.eas.asu.edu/yuv/.

  8. It is easy to see that the FT-QSVD method is equivalent to the FT-QTF method given in [26] without theoretical proof. So, the proof in Subsection 5.2 provides theoretical analysis for the FT-QTF method.

References

  1. Braman, K.: Third-order tensors as linear operators on a space of matrices. Linear Algebra Appl. 433, 1241–1253 (2010)

    Article  MathSciNet  Google Scholar 

  2. Bader, B.W., Kolda, T.G., others.: Matlab tensor toolbox version 3.6, available online, (2023) http://www.tensortoolbox.org/

  3. Cayley, A.: On Jacobi’s elliptic functions, in reply to the Rev. B. Bronwin; and on quaternions. Philos. Mag. 26, 208–211 (1845)

    Google Scholar 

  4. Chan, R.H., Ng, M.K.: Conjugate gradient methods for Toeplitz systems. SIAM Rev. 38, 427–482 (1996)

    Article  MathSciNet  Google Scholar 

  5. Cheung, W., Ng, M.K.: Block-circulant preconditioners for systems arising from discretization of the three-dimensional convection-diffusion equation. J. Comput. Appl. Math. 140, 143–158 (2002)

    Article  MathSciNet  Google Scholar 

  6. Chen, Y., Xiao, X., Zhou, Y.: Low-rank quaternion approximation for color image processing. IEEE Trans. Image Process. 29, 1426–1439 (2020)

    Article  MathSciNet  Google Scholar 

  7. Dixon, G.M.: Division Algebras: Octonions. Quaternions, Complex Numbers and the Algebraic Design of Physics, Kluwer, Dordrecht (1994)

    Book  Google Scholar 

  8. Ell, T.A., Le Bihan, N., Sangwine, S.J.: Quaternion Fourier transforms for signal and image processing. John Wiley & Sons, New Jersey (2014)

    Book  Google Scholar 

  9. Golub, G.H., Van Loan, C.F.: Matrix Computations, 4th edn. Johns Hopkins University Press, Baltimore (2013)

    Book  Google Scholar 

  10. Gai, S., Yang, G., Wan, M., Wang, L.: Denoising color images by reduced quaternion matrix singular value decomposition. Multidim. Syst. Sign. Process. 26(1), 307–320 (2015)

    Article  MathSciNet  Google Scholar 

  11. Hamilton, W.R.: Elements of Quaternions. Longmans Green, London, U.K. (1866)

    Google Scholar 

  12. Huckle, T.: Circulant and Skew Circulant Matrices for Solving Toeplitz Matrix Problems. SIAM J. Matrix Anal. Appl. 13, 767–777 (1992)

    Article  MathSciNet  Google Scholar 

  13. Hosny, K.M., Darwish, M.M.: New set of multi-channel orthogonal moments for color image representation and recognition. Pattern Recogn. 88, 153–173 (2019)

    Article  Google Scholar 

  14. He, Z.H., Wang, X.X., Zhao, Y.F.: Eigenvalues of quaternion tensors with applications to color video processing. J. Sci. Comput. (2023). https://doi.org/10.1007/s10915-022-02058-5

    Article  MathSciNet  Google Scholar 

  15. Jia, Z.G., Ng, M.K., Song, G.J.: Robust quaternion matrix completion with applications to image inpainting. Numer. Linear Algebra Appl. 26, e2245 (2019)

    Article  MathSciNet  Google Scholar 

  16. Jia, Z.G., Ng, M.K., Song, G.J.: Lanczos method for large-scale quaternion singular value decomposition. Numer. Algorithm 82, 699–717 (2019)

    Article  MathSciNet  Google Scholar 

  17. Kilmer, M.E., Braman, K., Hao, N., Hoover, R.C.: Third-order tensors as operators on matrices: a theoretical and computational framework with applications in imaging. SIAM J. Matrix Anal. Appl. 34, 148–172 (2013)

    Article  MathSciNet  Google Scholar 

  18. Kilmer, M.E., Martin, C.D.: Factorization strategies for third-order tensors. Linear Algebra Appl. 435, 641–658 (2011)

    Article  MathSciNet  Google Scholar 

  19. Kilmer, M.E., Martin, C.D., Perrone, L.: A third-order generalization of the matrix svd as a product of third-order tensors, Tech. Report TR-2008-4, Tufts University, Computer Science Department, (2008)

  20. Lund, K.: The tensor t-function: a definition for functions of third-order tensors. Numer. Linear Algebra Appl. 27, e2288 (2020)

    Article  MathSciNet  Google Scholar 

  21. Lu, C., Feng, J., Chen, Y., Liu, W., Lin, Z., Yan, S.: Tensor robust principal component analysis: Exact recovery of corrupted low-rank tensors via convex optimization, CVPR, pp. 5249–5257 (2016)

  22. Lu, C., Peng, X., Wei, Y.: Low-rank tensor completion with a new tensor nuclear norm induced by invertible linear transforms, CVPR, pp. 5996–6004 (2019)

  23. Martin, D., Fowlkes, C., Tal, D., Malik, J.: A database of human segmented natural images and its application to evaluating segmentation algorithms and measuring ecological statistics. In Proc. IEEE Int’l Conf. Computer Vision, vol. 2, pp. 416–423 (2001)

  24. Miao, J.F., Kou, K.I., Liu, W.K.: Low-rank quaternion tensor completion for recovering color videos and images. Pattern Recogn. 107, 107505 (2020)

    Article  Google Scholar 

  25. Miao, J.F., Kou, K.I.: Quaternion tensor singular value decomposition using a flexible transform-based approach. Signal Process. 206, 108910 (2023)

    Article  Google Scholar 

  26. Pan, J., Ng, M.K.: Block diagonalization of quaternion circulant matrices with applications to quaternion tensor singular value decomposition, (2023), arXiv preprint arXiv:2302.04086

  27. Qi, L., Luo, Z., Wang, Q.W., Zhang, X.: Quaternion matrix optimization: motivation and analysis. J. Optim. Theory Appl. 193, 621–648 (2022)

    Article  MathSciNet  Google Scholar 

  28. Qin, Z.Z., Ming, Z.Y., Zhang, L.P.: Singular value decomposition of third order quaternion tensors. Appl. Math. Lett. 123, 107597 (2022)

    Article  MathSciNet  Google Scholar 

  29. Rezghi, M., Eldèn, L.: Diagonalization of tensors with circulant structure. Linear Algebra Appl. 435, 422–447 (2011)

    Article  MathSciNet  Google Scholar 

  30. Song, G.J., Ng, M.K., Zhang, X.J.: Robust tensor completion using transformed tensor singular value decomposition. Numer. Linear Algebra Appl. 27, e2299 (2020)

    Article  MathSciNet  Google Scholar 

  31. Shi, J., Zheng, X., Wu, J., Gong, B., Zhang, Q., Ying, S.: Quaternion Grassmann average network for learning representation of histopathological image. Pattern Recogn. 89, 67–76 (2019)

    Article  Google Scholar 

  32. Zhang, F.: Quaternions and matrices of quaternions. Linear Algebra Appl. 251, 21–57 (1997)

    Article  MathSciNet  Google Scholar 

  33. Zhang, Z., Aeron, S.: Exact tensor completion using t-SVD. IEEE Trans. Signal Process. 65, 1511–1526 (2017)

    Article  MathSciNet  Google Scholar 

  34. Zhao, X., Bai, M., Ng, M.K.: Nonconvex Optimization for Robust Tensor Completion from Grossly Sparse Observations. J. Sci. Comput. 85(46), 1–32 (2020)

    MathSciNet  Google Scholar 

  35. Zhang, Z., Ely, G., Aeron, S., Hao, N., Kilmer, M.: Novel methods for multilinear data completion and de-noising based on tensor-svd. In: Proceedings of the 2014 IEEE Conference on Computer Vision and Pattern Recognition, ser. CVPR, pp. 3842–3849 (2014)

  36. Zhang, J., Saibaba, A.K., Kilmer, M.E., Aeron, S.: A randomized tensor singular value decomposition based on the t-product. Numer. Linear Algebra Appl. 25(5), e2179 (2018)

    Article  MathSciNet  Google Scholar 

  37. Zheng, M.M., Ni, G.Y.: Approximation strategy based on the T-product for third-order quaternion tensors with application to color video compression. Appl. Math. Lett. 140, 108587 (2023)

    Article  MathSciNet  Google Scholar 

Download references

Funding

The authors would like to thank anonymous referees for their valuable comments and suggestions that helped us to improve the quality of this paper. This work is supported by National Natural Science Foundations of China (Grant No.12371315), the Natural Science Foundation of Hunan Province, China (Grant No.2022JJ40543) and China Postdoctoral Science Foundation (Grant No.2021MD703978).

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Correspondence to Guyan Ni.

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Appendix

Appendix

Here, we provide the proofs of Proposition 4.1, Proposition 4.2, Theorem 5.1 and Theorem 5.3.

A.1. Proof of Proposition 4.1:

Proof. (i) From the octonion multiplication rules in Table 2, it can be seen that if \(u\in \{\textbf{l},\textbf{il},\textbf{jl},\textbf{kl}\}\), then \((p+q\textbf{j})u=u\overline{(p+q\textbf{j})}=u(\bar{p}-q\textbf{j}).\) Thus, it can be deduced

$$\begin{aligned} (p+q\textbf{j})\cdot [(r+s\textbf{j})\textbf{l}] =[(r+s\textbf{j})\textbf{l}]\cdot \overline{(p+q\textbf{j})}. \end{aligned}$$

(ii) From the octonion multiplication rules in Table 2, it follows that

$$\begin{aligned} \begin{array}{rcl} (p\textbf{l})\cdot q\cdot (r\textbf{l}) &{}=&{}(a\textbf{l}+b\textbf{il})(c+d\textbf{i})(e\textbf{l}+f\textbf{il})\\ &{}=&{}-ace-bce\textbf{i}+ade\textbf{i}-bde +acf\textbf{i}-bcf+adf+bdf\textbf{i}\\ &{}=&{}-(a+b\textbf{i})(c-d\textbf{i})(e-f\textbf{i})=-p\bar{q}\bar{r}, \end{array} \\ \begin{array}{rcl} [p(\textbf{jl})]\cdot q\cdot [r(\textbf{jl})]&{}=&{}(a\textbf{jl}+b\textbf{kl})(c+d\textbf{i})(e\textbf{jl}+f\textbf{kl})\\ &{}=&{}-ace-bce\textbf{i}+ade\textbf{i}-bde +acf\textbf{i}-bcf+adf+bdf\textbf{i}\\ &{}=&{}-(a+b\textbf{i})(c-d\textbf{i})(e-f\textbf{i})=-p\bar{q}\bar{r}. \end{array} \end{aligned}$$

Thus, it can be deduced \((p\textbf{l})\cdot q\cdot (r\textbf{l})=-p\bar{q}\bar{r}=[p(\textbf{jl})]\cdot q\cdot [r(\textbf{jl})].\)

(iii) Similar to the proof of (ii), from Table 2, we can obtain that

$$\begin{aligned} \begin{array}{rcl} (p\textbf{l})\cdot (q\textbf{j})\cdot (r\textbf{l}) &{}=&{}(a\textbf{l}+b\textbf{il})(c\textbf{j}+d\textbf{k})(e\textbf{l}+f\textbf{il})\\ &{}=&{}ace\textbf{j}+bce\textbf{k}+ ade\textbf{k}-bde\textbf{j}-acf\textbf{k}+bcf\textbf{j}+ adf\textbf{j}+bdf\textbf{k}\\ &{}=&{}(a+b\textbf{i})(c+d\textbf{i})(e-f\textbf{i})\textbf{j}=pq\bar{r}\textbf{j}, \end{array} \\ \begin{array}{rcl} [p(\textbf{jl})]\cdot (q\textbf{j})\cdot [r(\textbf{jl})]&{}=&{} (a\textbf{jl}+b\textbf{kl})(c\textbf{j}+d\textbf{k})(e\textbf{jl}+f\textbf{kl})\\ &{}=&{}ace\textbf{j}+bce\textbf{k}+ ade\textbf{k}-bde\textbf{j}-acf\textbf{k}+bcf\textbf{j}+ adf\textbf{j}+bdf\textbf{k}\\ &{}=&{}(a+b\textbf{i})(c+d\textbf{i})(e-f\textbf{i})\textbf{j}=pq\bar{r}\textbf{j}, \end{array} \end{aligned}$$

thus, it follows that \((p\textbf{l})\cdot (q\textbf{j})\cdot (r\textbf{l})=pq\bar{r}\textbf{j}=[p(\textbf{jl})]\cdot (q\textbf{j})\cdot [r(\textbf{jl})].\)

(iv) Similarly, from Table 2, we can obtain that

$$\begin{aligned} \begin{array}{rcl} (p\textbf{l})\cdot q\cdot [r(\textbf{jl})] &{}=&{}(a\textbf{l}+b\textbf{il})(c+d\textbf{i})(e\textbf{jl}+f\textbf{kl})\\ &{}=&{}(ace\textbf{j}-bce\textbf{k}+ade\textbf{k}+bde\textbf{j} -acf\textbf{k}-bcf\textbf{j}+adf\textbf{j}-bdf\textbf{k}\\ &{}=&{}(a-b\textbf{i})(c+d\textbf{i})(e-f\textbf{i})\textbf{j}=\bar{p}{q}\bar{r}\textbf{j}, \end{array} \\ \begin{array}{rcl} [p(\textbf{jl})]\cdot q\cdot (r\textbf{l}) &{}=&{}(a\textbf{jl}+b\textbf{kl})(c+d\textbf{i})(e\textbf{l}+f\textbf{il})\\ &{}=&{}(ace\textbf{j}-bce\textbf{k}+ade\textbf{k}+bde\textbf{j} -acf\textbf{k}-bcf\textbf{j}+adf\textbf{j}-bdf\textbf{k}\\ &{}=&{}(a-b\textbf{i})(c+d\textbf{i})(e-f\textbf{i})\textbf{j}=\bar{p}{q}\bar{r}\textbf{j}, \end{array} \end{aligned}$$

thus, it follows that \((p\textbf{l})\cdot q\cdot [r(\textbf{jl})]=\bar{p}q\bar{r}\textbf{j}=-[p(\textbf{jl})]\cdot q\cdot (r\textbf{l}).\)

(v) Similarly, from Table 2, we can obtain that

$$\begin{aligned} \begin{array}{rcl} (p\textbf{l})\cdot (q\textbf{j})\cdot [r(\textbf{jl})] &{}=&{}(a\textbf{l}+b\textbf{il})(c\textbf{j}+d\textbf{k})(e\textbf{jl}+f\textbf{kl})\\ &{}=&{}ace-bce\textbf{i}-ade\textbf{i}-bde -acf\textbf{i}-bcf-adf+bdf\textbf{i}\\ &{}=&{}(a-b\textbf{i})(c-d\textbf{i})(e-f\textbf{i})=\bar{p}\bar{q}\bar{r}, \end{array} \\ \begin{array}{rcl} [p(\textbf{jl})]\cdot (q\textbf{j})\cdot (r\textbf{l}) &{}=&{}(a\textbf{jl}+b\textbf{kl})(c\textbf{j}+d\textbf{k})(e\textbf{l}+f\textbf{il})\\ &{}=&{}ace-bce\textbf{i}-ade\textbf{i}-bde -acf\textbf{i}-bcf-adf+bdf\textbf{i}\\ &{}=&{}(a-b\textbf{i})(c-d\textbf{i})(e-f\textbf{i})=\bar{p}\bar{q}\bar{r}, \end{array} \end{aligned}$$

thus, it follows that \((p\textbf{l})\cdot (q\textbf{j})\cdot [r(\textbf{jl})]=\bar{p}\bar{q}\bar{r}=-[p(\textbf{jl})]\cdot (q\textbf{j})\cdot (r\textbf{l}).\) \(\square \)

A.2. Proof of Proposition 4.2:

Proof. (i) From Proposition 4.1, it follows that

$$\begin{aligned}{} & {} p[(q\textbf{l})r(s\textbf{l})]=p(-q\bar{r}\bar{s}),\ p(q\textbf{l})r(s\textbf{l})=(q\textbf{l})\bar{p}r(s\textbf{l})=-qp\bar{r}\bar{s},\\{} & {} p[(q\textbf{l})(r\textbf{j})(s\textbf{l})]=p(qr\bar{s}\textbf{j}), \ p(q\textbf{l})(r\textbf{j})(s\textbf{l})=(q\textbf{l})\bar{p}(r\textbf{j})(s\textbf{l}) =qpr\bar{s}\textbf{j}. \end{aligned}$$

Noting that \(p,q,r,s\in \mathbb {C}\), thus we have that

$$\begin{aligned} p[(q\textbf{l})r(s\textbf{l})]=p(q\textbf{l})r(s\textbf{l})\quad \text{ and }\quad p[(q\textbf{l})(r\textbf{j})(s\textbf{l})]=p(q\textbf{l})(r\textbf{j})(s\textbf{l}). \end{aligned}$$

(ii) From Proposition 4.1 (i) and (ii), it follows that

$$\begin{aligned} (p\textbf{j})[(q\textbf{l})r(s\textbf{l})]=-p\bar{q}rs\textbf{j}. \end{aligned}$$

By the octonion multiplication rules shown in Table 2, we can obtain that

$$\begin{aligned} \begin{array}{rcl} &{} &{}(p\textbf{j})(q\textbf{l})(r\textbf{j})(s\textbf{l})\\ &{}=&{}(a\textbf{j}+b\textbf{k})(c\textbf{l}+d\textbf{il})(e\textbf{j}+f\textbf{k})(g\textbf{l}+h\textbf{il})\\ &{}=&{}-aceg\textbf{j}-bceg\textbf{k}-adeg\textbf{k}+bdeg\textbf{j} -acfg\textbf{k}+bcfg\textbf{j}+adfg\textbf{j}+bdfg\textbf{k}\\ &{} &{}+aceh\textbf{k}-bceh\textbf{j}-adeh\textbf{j}-bdeh\textbf{k} -acfh\textbf{j}-bcfh\textbf{k}-adfh\textbf{k}+bdfh\textbf{j}\\ &{}=&{}-pqr\bar{s}\textbf{j}\\ \end{array} \end{aligned}$$

From Proposition 4.1 (i) and (iii), it follows that

$$\begin{aligned} (p\textbf{j})[(q\textbf{l})(r\textbf{j})(s\textbf{l})]=(p\textbf{j})(qr\bar{s}\textbf{j})=-p\bar{q}\bar{r}s. \end{aligned}$$

By the octonion multiplication rules shown in Table 2, we can obtain that

$$\begin{aligned} \begin{array}{rcl} &{} &{}(p\textbf{j})(q\textbf{l})(r\textbf{j})(s\textbf{l})\\ &{}=&{}(a\textbf{j}+b\textbf{k})(c\textbf{l}+d\textbf{il})(e\textbf{j}+f\textbf{k})(g\textbf{l}+h\textbf{il})\\ &{}=&{}-aceg-bceg\textbf{i}-adeg\textbf{i}+bdeg +acfg\textbf{i}-bcfg-adfg-bdfg\textbf{i}\\ &{} &{}+aceh\textbf{i}-bceh-adeh-bdeh\textbf{i} +acfh+bcfh\textbf{i}+adfh\textbf{i}-bdfh\\ &{}=&{}-pq\bar{r}\bar{s}. \end{array} \end{aligned}$$

Hence, it follows that

$$\begin{aligned} (p\textbf{j})[(q\textbf{l})r(s\textbf{l})]=(p\textbf{j})(\bar{q}\textbf{l})r(\bar{s}\textbf{l})\quad \text{ and }\quad (p\textbf{j})[(q\textbf{l})(r\textbf{j})(s\textbf{l})]=(p\textbf{j})(\bar{q}\textbf{l})(r\textbf{j})(\bar{s}\textbf{l}). \end{aligned}$$

In addition, by the octonion multiplication rules shown in Table 2, it is similar to obtain that (iii), (iv), (v) and (vi) hold. \(\square \)

A.3. Proof of Theorem 5.1:

Proof. To show that (5.2) holds, we only need to prove that

$$\begin{aligned} bcirc(\mathcal {C})=(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})[{Diag}({\hat{\mathcal {C}}^{(1,\textbf{i})}})+{Diag}({\hat{\mathcal {C}}^{(\textbf{j},\textbf{k})}})\textbf{j}](-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}}). \end{aligned}$$
(7.1)

From Theorem 4.4 and (5.1), it follows that

$$\begin{aligned} {Diag}(\hat{\mathcal {A}})={Diag}(\hat{\mathcal {A}}^{(1,\textbf{i})})+{Diag}(\hat{\mathcal {A}}^{(\textbf{j},\textbf{k})})\textbf{j}=(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{m}})bcirc(\mathcal {A})(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}}), \end{aligned}$$
(7.2)
$$\begin{aligned} {Diag}(\hat{\mathcal {B}})={Diag}(\hat{\mathcal {B}}^{(1,\textbf{i})})+{Diag}(\hat{\mathcal {B}}^{(\textbf{j},\textbf{k})})\textbf{j}=(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{m}})bcirc(\mathcal {B})(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}}). \end{aligned}$$
(7.3)

Then by Proposition 4.2 (iii), (iv) and (7.2), it follows that

$$\begin{aligned} bcirc(\mathcal {A})=(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{m}}){Diag}(\hat{\mathcal {A}}^{(1,\textbf{i})})(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}})+(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})[{Diag}(\hat{\mathcal {A}}^{(\textbf{j},\textbf{k})})\textbf{j}](\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}}).\nonumber \\ \end{aligned}$$
(7.4)

By Proposition 4.2 (iii), (iv) and (7.3), it follows that

$$\begin{aligned} bcirc(\mathcal {B})=(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}}){Diag}(\hat{\mathcal {B}}^{(1,\textbf{i})})(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{s}})+(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})[{Diag}(\hat{\mathcal {B}}^{(\textbf{j},\textbf{k})})\textbf{j}](\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}}).\nonumber \\ \end{aligned}$$
(7.5)

By a similar proof as [20, Lemma 2.3], we can obtain that the following property holds for quaternion tensors \(\mathcal {A}\) and \(\mathcal {B}\), that is:

$$\begin{aligned} bcirc(\mathcal {A}*\mathcal {B})=bcirc(\mathcal {A})\cdot bcirc(\mathcal {B}). \end{aligned}$$
(7.6)

Thereby, from (7.4), (7.5) and (7.6), it is deduced that

$$\begin{aligned} \begin{array}{rl} &{}bcirc(\mathcal {C})=bcirc(\mathcal {A}*\mathcal {B})=bcirc(\mathcal {A})\cdot bcirc(\mathcal {B})\\ =&{}[(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{m}}){Diag}(\hat{\mathcal {A}}^{(1,\textbf{i})})(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}})]\cdot [(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}}){Diag}(\hat{\mathcal {B}}^{(1,\textbf{i})})(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{s}})]\\ &{}+\;[(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})({Diag}(\hat{\mathcal {A}}^{(\textbf{j},\textbf{k})})\textbf{j})(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})]\cdot [(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})({Diag}(\hat{\mathcal {B}}^{(\textbf{j},\textbf{k})})\textbf{j})(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}})]\\ &{}+\;[(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{m}}){Diag}(\hat{\mathcal {A}}^{(1,\textbf{i})})(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}})]\cdot [(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})({Diag}(\hat{\mathcal {B}}^{(\textbf{j},\textbf{k})})\textbf{j})(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}})]\\ &{}+\;[(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})({Diag}(\hat{\mathcal {A}}^{(\textbf{j},\textbf{k})})\textbf{j})(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})]\cdot [(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}}){Diag}(\hat{\mathcal {B}}^{(1,\textbf{i})})(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{s}})]. \end{array}\nonumber \\ \end{aligned}$$
(7.7)

By (4.1) and (2.4), it is easy to see \(\mathbf {{P}_{p}}=\mathbf {F_{p}}\mathbf {F_{p}}=\mathbf {F^*_{p}}\mathbf {F^*_{p}} =-(\mathbf {F_{p}}\textbf{l})(\mathbf {F^*_{p}}\textbf{l}) =-(\mathbf {F^*_{p}}\textbf{l})(\mathbf {F_{p}}\textbf{l})\in \mathbb {R}^{p\times p}\) and

$$\begin{aligned} \mathbf {{P}_{p}}\otimes \mathbf {I_m} =-(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_m})(\mathbf {F^*_{p}}{\textbf{l}}\otimes \mathbf {I_m})=-(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_m})(\mathbf {F_{p}}{\textbf{l}}\otimes \mathbf {I_m}). \end{aligned}$$
(7.8)

Denote \(\hat{\textbf{A}_1}={Diag}(\hat{\mathcal {A}}^{(1,\textbf{i})}),\hat{\textbf{A}_2}={Diag}(\hat{\mathcal {A}}^{(\textbf{j},\textbf{k})}), \hat{\textbf{B}_1}={Diag}(\hat{\mathcal {B}}^{(1,\textbf{i})}),\;\text{ and }\;\hat{\textbf{B}_2}={Diag}(\hat{\mathcal {B}}^{(\textbf{j},\textbf{k})}).\) Then, from Proposition 4.2 (i), (v), (vi) and (7.8), it follows

$$\begin{aligned} \begin{array}{lcl} &{}&{}[(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{m}})\;{Diag}(\hat{\mathcal {A}}^{(1,\textbf{i})})\;(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}})]\cdot [(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}})\;{Diag}(\hat{\mathcal {B}}^{(1,\textbf{i})})\;(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{s}})]\\ &{}=&{}(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{m}})\;\hat{\textbf{A}_1}\;(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}}) (-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}})\;\hat{\textbf{B}_1}\;(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{s}})\\ &{}=&{}(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{m}})\;\hat{\textbf{A}_1}\;[(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}})\; (-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}})]\;\hat{\textbf{B}_1}\;(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{s}})\\ &{}=&{}(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{m}})\;\hat{\textbf{A}_1}\;\hat{\textbf{B}_1}\;(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{s}})\\ &{}=&{}[(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\;(\mathbf {{P}_{p}}\otimes \mathbf {I_{m}})]\cdot \hat{\textbf{A}_1}\;\hat{\textbf{B}_1} \cdot [(\mathbf {{P}_{p}}\otimes \mathbf {I_{s}})\;(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}})]\\ &{}=&{}(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\cdot [(\mathbf {{P}_{p}}\otimes \mathbf {I_{m}})\;\hat{\textbf{A}_1}\;\hat{\textbf{B}_1}\;(\mathbf {{P}_{p}}\otimes \mathbf {I_{s}})]\cdot (\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}})\\ &{}=&{}(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\cdot [(\mathbf {{P}_{p}}\otimes \mathbf {I_{m}})\;{Diag}({\hat{\mathcal {A}}^{(1,\textbf{i})}})\;{Diag}({\hat{\mathcal {B}}^{(1,\textbf{i})}})\;(\mathbf {{P}_{p}}\otimes \mathbf {I_{s}})]\cdot (\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}}). \end{array} \end{aligned}$$
(7.9)

From Proposition 4.2 (ii), (v), (vi) and (7.8), it follows

$$\begin{aligned} \begin{array}{ll} &{}[(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\;({Diag}(\hat{\mathcal {A}}^{(\textbf{j},\textbf{k})})\textbf{j})\;(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})]\cdot [(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})\;({Diag}(\hat{\mathcal {B}}^{(\textbf{j},\textbf{k})})\textbf{j})\;(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}})]\;\\ =&{}[(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\;(\mathbf {\hat{\textbf{A}_2}}\textbf{j})\;(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})]\cdot \{(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}})\;[(\mathbf {{P}_{p}}\otimes \mathbf {I_{n}})\;(\hat{\textbf{B}_2}\textbf{j})\;(\mathbf {{P}_{p}}\otimes \mathbf {I_{s}})]\;(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{s}})\}\;\\ =&{}[(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\;(\hat{\textbf{A}_2}\textbf{j})\;(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})]\cdot (-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})\cdot [(\mathbf {{P}_{p}}\otimes \mathbf {I_{n}})\;(\hat{\textbf{B}_2}\textbf{j})\;(\mathbf {{P}_{p}}\otimes \mathbf {I_{s}})]\cdot (\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}})\\ =&{}(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\;(\mathbf {\hat{\textbf{A}_2}}\textbf{j})\cdot [(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}}) \;(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}})]\cdot [(\mathbf {{P}_{p}}\otimes \mathbf {I_{n}})\;(\hat{\textbf{B}_2}\textbf{j})\;(\mathbf {{P}_{p}}\otimes \mathbf {I_{s}})]\cdot (\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}})\\ =&{}(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\cdot [-\overline{\hat{\textbf{A}_2}}\;(\mathbf {{P}_{p}}\otimes \mathbf {I_{n}})\;\hat{\textbf{B}}_2\;(\mathbf {{P}_{p}}\otimes \mathbf {I_{s}})]\cdot (\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}})\\ =&{}(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\cdot [-{Diag}(\overline{{\hat{\mathcal {A}}^{(\textbf{j},\textbf{k})}}})\;(\mathbf {{P}_{p}}\otimes \mathbf {I_{n}})\;{Diag}(\hat{\mathcal {B}}^{(\textbf{j},\textbf{k})})\;(\mathbf {{P}_{p}}\otimes \mathbf {I_{s}})]\cdot (\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}}). \end{array}\nonumber \\ \end{aligned}$$
(7.10)

From Proposition 4.2 (i), (v), (vi) and (7.8), it follows

$$\begin{aligned} \begin{array}{lcl} &{}&{}[(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{m}})\;{Diag}(\hat{\mathcal {A}}^{(1,\textbf{i})})\;(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}})]\cdot [(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})\;({Diag}(\hat{\mathcal {B}}^{(\textbf{j},\textbf{k})})\textbf{j})\;(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}})]\\ &{}=&{}\{(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\cdot [(\mathbf {{P}_{p}}\otimes \mathbf {I_{m}})\;\hat{\textbf{A}_1}\;(\mathbf {{P}_{p}}\otimes \mathbf {I_{n}})]\cdot (\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})\}\cdot [(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})\;(\hat{\textbf{B}_2}\textbf{j})\;(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}})]\;\\ &{}=&{}(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\cdot [(\mathbf {{P}_{p}}\otimes \mathbf {I_{m}})\;\hat{\textbf{A}_1}\;(\mathbf {{P}_{p}}\otimes \mathbf {I_{n}})]\cdot [(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})\; (-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})]\cdot (\hat{\textbf{B}_2}\textbf{j})\cdot (\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}})\\ &{}=&{}(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\cdot [(\mathbf {{P}_{p}}\otimes \mathbf {I_{m}})\;\overline{\hat{\textbf{A}_1}}\;(\mathbf {{P}_{p}}\otimes \mathbf {I_{n}})\;(\hat{\textbf{B}_2}\textbf{j})]\cdot (\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}})\\ &{}=&{}(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\cdot [(\mathbf {{P}_{p}}\otimes \mathbf {I_{m}})\;{Diag}(\overline{\hat{\mathcal {A}}^{(1,\textbf{i})}})\;(\mathbf {{P}_{p}}\otimes \mathbf {I_{n}})\;({Diag}(\hat{\mathcal {B}}^{(\textbf{j},\textbf{k})})\textbf{j})]\cdot (\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}}). \end{array}\nonumber \\ \end{aligned}$$
(7.11)

From Proposition 4.2 (ii), (v) and (vi), it follows

$$\begin{aligned} \begin{array}{lcl} &{}&{}[(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\;({Diag}(\hat{\mathcal {A}}^{(\textbf{j},\textbf{k})})\textbf{j})\;(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})]\cdot [(-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}})\;{Diag}(\hat{\mathcal {B}}^{(1,\textbf{i})})\;(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{s}})]\\ &{}=&{}[(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\;({Diag}(\hat{\mathcal {A}}^{(\textbf{j},\textbf{k})})\textbf{j})\;(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})]\cdot (-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{n}})\cdot {Diag}(\hat{\mathcal {B}}^{(1,\textbf{i})})\cdot (\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}})\\ &{}=&{}(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\cdot ({Diag}(\hat{\mathcal {A}}^{(\textbf{j},\textbf{k})})\textbf{j})\cdot [(\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}})\; (-\mathbf {F_{p}}\textbf{l}\otimes \mathbf {I_{n}})]\cdot {Diag}(\hat{\mathcal {B}}^{(1,\textbf{i})})\cdot (\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}})\\ &{}=&{}(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\cdot [{Diag}(\hat{\mathcal {A}}^{(\textbf{j},\textbf{k})}){Diag}(\hat{\mathcal {B}}^{(1,\textbf{i})})\textbf{j}]\cdot (\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}}). \end{array}\nonumber \\ \end{aligned}$$
(7.12)

Thus, by (7.7), (7.9), (7.10), (7.11), (7.12), (7.2) and (7.3), we obtain that

$$\begin{aligned} \begin{array}{c} bcirc(\mathcal {C})=(\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{m}})\;[{Diag}({\hat{\mathcal {C}}^{(1,\textbf{i})}})+{Diag}({\hat{\mathcal {C}}^{(\textbf{j},\textbf{k})}})\textbf{j}]\;(-\mathbf {F^*_{p}}\textbf{l}\otimes \mathbf {I_{s}}), \end{array} \end{aligned}$$

i.e., (7.1) holds. \(\square \)

A.4. Proof of Theorem 5.3:

Proof. From \(\hat{\mathcal {A}}=\textrm{fft}(\overline{\mathcal {A}},[\;],3)\), it follows that

$$\begin{aligned} \begin{array}{rcl} &{}&{}{Diag}(\hat{A}^{(1)},\hat{A}^{(2)},\ldots ,\hat{A}^{(p)}) =(\mathbf {F_p}\textbf{l}\otimes \mathbf {I_m})bcirc(\mathcal {A})(-\mathbf {F_p}\textbf{l}\otimes \mathbf {I_m})\\ &{}=&{}(\mathbf {F_p}\otimes \mathbf {I_m})bcirc(\overline{\mathcal {A}^{(1,\textbf{i})}})(\mathbf {F^*_p}\otimes \mathbf {I_m}) -(\mathbf {F_p}\otimes \mathbf {I_m})bcirc(\mathcal {A}^{(\textbf{j},\textbf{k})}) (\mathbf {F^*_p}\otimes \mathbf {I_m})\textbf{j}. \end{array} \end{aligned}$$

By Lemma 5.2, it follows from (5.10) that \(\hat{\mathcal {A}_k}=\textrm{fft}(\overline{\mathcal {A}_k},[\;],3)\), thus

$$\begin{aligned} \begin{array}{rcl} &{}&{}{Diag}(\hat{A}_k^{(1)},\hat{A}_k^{(2)},\ldots ,\hat{A}_k^{(p)})= (\mathbf {F_p}\textbf{l}\otimes \mathbf {I_m})bcirc(\mathcal {A}_k)(-\mathbf {F_p}\textbf{l}\otimes \mathbf {I_m})\\ &{}=&{}(\mathbf {F_p}\otimes \mathbf {I_m})bcirc(\overline{\mathcal {A}_k^{(1,\textbf{i})}}) (\mathbf {F^*_p}\otimes \mathbf {I_m}) -(\mathbf {F_p}\otimes \mathbf {I_m})bcirc(\mathcal {A}_k^{(\textbf{j},\textbf{k})}) (\mathbf {F^*_p}\otimes \mathbf {I_m})\textbf{j}. \end{array} \end{aligned}$$

Furthermore, Since \(\mathbf {F_p}\otimes \mathbf {I_m}\) is unitary, it follows that

$$\begin{aligned} \begin{array}{rcl} ||\mathcal {A}-\mathcal {A}_k||_F^2 &{}=&{}\frac{1}{p}||\;bcirc(\mathcal {A})-bcirc(\mathcal {A}_k)\;||_F^2\\ &{}=&{}\frac{1}{p}||\;bcirc(\overline{\mathcal {A}^{(1,\textbf{i})}})- bcirc(\overline{\mathcal {A}_k^{(1,\textbf{i})}})\;||_F^2 \\ &{}&{} \quad +\frac{1}{p}||\;bcirc(\mathcal {A}^{(\textbf{j},\textbf{k})})- bcirc(\mathcal {A}_k^{(\textbf{j},\textbf{k})})\;||_F^2\\ &{}=&{}\frac{1}{p}||\;{Diag}(\hat{A}^{(1)},\hat{A}^{(2)},\ldots ,\hat{A}^{(p)}) -{Diag}(\hat{A}_k^{(1)},\hat{A}_k^{(2)},\ldots ,\hat{A}_k^{(p)})\;||_F^2\\ &{}=&{}\frac{1}{p}\sum _{i=1}^p||\;\hat{\mathcal {S}}(k+1:m,k+1:n,i)\;||_F^2. \end{array} \end{aligned}$$
(7.13)

Assume that \(\widetilde{\mathcal {A}}\in M\), then there exist \(\mathcal {X}\in \mathbb {Q}^{m\times k\times p}\), \(\mathcal {Y}\in \mathbb {Q}^{k\times n\times p}\) such that

$$\begin{aligned} \begin{array}{rcl} &{}&{}{Diag}(X^{(1)}Y^{(1)},X^{(2)}Y^{(2)},\ldots ,X^{(p)}Y^{(p)}) =(\mathbf {F_p}\textbf{l}\otimes \mathbf {I_m})bcirc(\widetilde{\mathcal {A}})(-\mathbf {F_p}\textbf{l}\otimes \mathbf {I_m})\\ &{}=&{}(\mathbf {F_p}\otimes \mathbf {I_m})bcirc(\overline{\widetilde{\mathcal {A}}_1})(\mathbf {F^*_p}\otimes \mathbf {I_m}) -(\mathbf {F_p}\otimes \mathbf {I_m})bcirc(\widetilde{\mathcal {A}}_2)(\mathbf {F^*_p}\otimes \mathbf {I_m})\textbf{j}, \end{array} \end{aligned}$$

where \(\widetilde{\mathcal {A}}=\widetilde{\mathcal {A}}_1+\widetilde{\mathcal {A}}_2\textbf{j}\) with \(\widetilde{\mathcal {A}}_1\), \(\widetilde{\mathcal {A}}_2\in \mathbb {C}^{m\times n\times p}\). Then similar to (7.13), we can obtain that

$$\begin{aligned} \begin{array}{rcl} ||\mathcal {A}-\widetilde{\mathcal {A}}||_F^2 &{}=&{}\frac{1}{p}||\;bcirc(\mathcal {A})-bcirc(\widetilde{\mathcal {A}})\;||_F^2\\ &{}=&{}\frac{1}{p}||\;bcirc(\overline{\mathcal {A}^{(1,\textbf{i})}})-bcirc(\overline{\widetilde{\mathcal {A}}_1})\;||_F^2 +\frac{1}{p}||\;bcirc(\mathcal {A}^{(\textbf{j},\textbf{k})})-bcirc(\widetilde{\mathcal {A}}_2)\;||_F^2\\ &{}=&{}\frac{1}{p}||\;{Diag}(\hat{A}^{(1)},\hat{A}^{(2)},\ldots ,\hat{A}^{(p)}) -{Diag}(X^{(1)}Y^{(1)},X^{(2)}Y^{(2)},\ldots ,X^{(p)}Y^{(p)})\;||_F^2\\ &{}=&{}\frac{1}{p}\sum _{i=1}^p||\;\hat{\mathcal {S}}(k+1:m,k+1:n,i)\;||_F^2\\ &{}=&{}\frac{1}{p}\sum _{i=1}^p||\;\hat{\mathcal {A}}(:,:,i)-\mathcal {X}(:,:,i) \mathcal {Y}(:,:,i)\;||_F^2. \end{array} \end{aligned}$$

For every \(i\in [p]\), denote the QSVD of \(\mathcal {X}(:,:,i) \mathcal {Y}(:,:,i)\) as

$$\begin{aligned} \mathcal {X}(:,:,i) \mathcal {Y}(:,:,i)= \widetilde{\mathcal {U}}(:,:,i)\widetilde{\mathcal {S}}(:,:,i)\widetilde{\mathcal {V}}(:,:,i), \end{aligned}$$

then it follows from \(\mathcal {X}\in \mathbb {Q}^{m\times k\times p}\) and \(\mathcal {Y}\in \mathbb {Q}^{k\times n\times p}\) that \(\widetilde{\mathcal {S}}(k+1:m,k+1:n,i)=0\). Hence, it is deduced that

$$\begin{aligned} \begin{array}{rcl} ||\mathcal {A}-\widetilde{\mathcal {A}}||_F^2 &{}=&{}\frac{1}{p}\sum _{i=1}^p||\;\hat{\mathcal {A}}(:,:,i)-\mathcal {X}(:,:,i) \mathcal {Y}(:,:,i)\;||_F^2\\ &{}=&{}\frac{1}{p}\sum _{i=1}^p||\;\hat{\mathcal {S}}(:,:,i)-\widetilde{\mathcal {S}}(:,:,i)\;||_F^2\\ &{}\ge &{} \frac{1}{p}\sum _{i=1}^p||\;\hat{\mathcal {S}}(k+1:m,k+1:n,i)||, \end{array} \end{aligned}$$

which, together with (7.13), implies that \(\mathcal {A}_k=argmin_{\widetilde{\mathcal {A}}\in M}||\mathcal {A} -\widetilde{\mathcal {A}}||_F\). \(\square \)

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Zheng, MM., Ni, G. Block Diagonalization of Block Circulant Quaternion Matrices and the Fast Calculation for T-Product of Quaternion Tensors. J Sci Comput 100, 69 (2024). https://doi.org/10.1007/s10915-024-02623-0

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