User talk:Oyz

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[[User:ClockworkSoul|User:ClockworkSoul/sig]] 05:37, 1 Dec 2004 (UTC)

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Oleg Alexandrov (talk) 17:08, 7 April 2006 (UTC)Reply


ejπ=1

some notes

Complex-conjugate multiplications with complex-swap


Efficient Implementation for Complex-Conjugate Multiplications with Complex-Swap

Coexistence of complex and complex-conjugate multiplications

abstract

background and summary

descriptions

embodiments

claims



Householder transformation


  • H=I2 v v*
  • v=ejθ e1w
  • Hw=e1
  • He1=w
  • ||v||=1
  • ||w||=1
  • H*H=I
  • R=ΘW
  • W*W=I,Θ*Θ=I.
  • Γ*R*RΓ=I
  • R+2RΓΓ*R*W=0
  • RΓ+2RΓΓ*R*(ΘR)Γ=0
  • RΓ+2RΓΓ*R*ΘΓ2RΓΓ*R*RΓ=0
  • 2RΓΓ*R*ΘΓ=RΓ
  • R2RΓΓ*R*Θ=0
  • RΓΓ*R*(W+Θ)=0
  • (ΘW)ΓΓ*(ΘW)*(Θ+W)=0
  • Γ*(Θ*WW*Θ)=0
  • It implies Γ is not full-rank. It contradicts with Γ*R*RΓ=I.
  • Therefore, Θ*W=W*Θ
  • Since Θ or W can not be Hermitian matrices, the failure of the generalization is proved.

The case of one-rank modification is the only possible one for the reflection with any desired hyperplane.

  • But multiple-rank reflection transform can be used for finding the basis of the null space!

Order-recursive calculation of SVD via column-wise augmentation


Low-latency SVD

applications to mimo detector, steering matrix gain ...

introduction

* motivation
  * real-time or massive data application: small processing resource or high data volume.
  * column- or row-wise data insertion: cache structure or memory limitation.
  * need to update inovative column information...
* enabling ideas
  * rank-one update formula: adding new column
  * solving secular equation
  * bi-digonalization for numerical stability

approach

  • order-recursive formula:
𝐀n+1=(𝐀n𝐜n+1)
  • consider the SVD of A,,n,,is available: {{{#!latex
$$
\mathbf A_n = \mathbf U_n \mathbf \Sigma_n \mathbf V_n^*
$$

}}}

* uninary matrices can be used to obtain an almost diagonalized matrix: {{{#!latex
$$
\mathbf U_n^* \ \mathbf A_{n+1} 
\begin{pmatrix}
\mathbf V_n & \mathbf 0 \\
\mathbf 0^* & 1
\end{pmatrix}
=
\begin{pmatrix}
\mathbf \Sigma_n[1:n,1:n] &  \mathbf d_{n+1}[1:n] \\
\mathbf 0_{(m-n)\times n} &  \mathbf d_{n+1}[n+1:m] \\
\end{pmatrix}
$$
where $\mathbf d_{n+1} = \mathbf U_n^* \ \mathbf  c_{n+1}$.

}}}

* Using Householder transformation, the upper-triangular form can be obtained (tall matrix assumed.): {{{#!latex
$$
\begin{pmatrix}
\mathbf I_n & \mathbf O \\
\mathbf O   & \mathbf H_{m-n}
\end{pmatrix}
\begin{pmatrix}
\mathbf \Sigma_n[1:n,1:n] &  \mathbf d_{n+1}[1:n] \\
\mathbf 0_{(m-n)\times n} &  \mathbf d_{n+1}[n+1:m] \\
\end{pmatrix}
=
\begin{pmatrix}
\mathbf \Sigma_n[1:n,1:n] &  \mathbf d_{n+1}[1:n] \\
\mathbf 0_{n}^* &  f_{n+1} \\
\mathbf 0_{(m-n-1)\times n} &  \mathbf 0_{m-n-1} \\
\end{pmatrix}
$$
where $\mathbf d_{n+1} = \mathbf U_n^* \ \mathbf  c_{n+1}$.

}}}

* The almost diagonal matrix can be diagonalized by means of the previous approaches.
* Among them, the secular equation solving is the best for rank-one update: 
  * it leads to finding simple zeros of polynomials. 
  * linear interplation/iterations are enough.

solving secular equation

  • Summary:
  1. move zero sigmas right-most: column-swap
  2. move up zero d's: column-and-row swap
  3. make a square part by householder transforming residual d.
  4. apply secular equation for the square part of dimension r-q+1.
  5. merge diagonal parts and unitary matrices: singular values are not ordered for calculation speed.
  6. sort the diagonal
  • re-visit formula:
𝐔n* 𝐀n+1(𝐕n𝟎𝟎*1)=(Σn[1:r,1:r]𝐎r×(nr)𝐝n+1[1:r]𝐎(nr)×r𝐎nr𝐝n+1[r+1:n]𝐎(mn)×r𝐎(mn)×(nr)𝐝n+1[n+1:m])

where r is rank of Σn. Note that 𝐝n+1[1:r] may include zeros.

  • more swapping rows and columns for zero singular values and diagonal parts.
  1. move zero sigmas right-most: column-swap {{{#!latex
$$
\mathbf U_n^* \ \mathbf A_{n+1}
\begin{pmatrix}
\mathbf V_n & \mathbf 0 \\
\mathbf 0^* & 1
\end{pmatrix}
\mathbf P_{\mathbf\Sigma_n}
=
\begin{pmatrix}
\mathbf \Sigma_n[1:r,1:r]    &  \mathbf d_{n+1}[1:r]   & \mathbf O_{r\times(n-r)}\\
\mathbf O_{(n-r)\times r}   &  \mathbf d_{n+1}[r+1:n] & \mathbf O_{n-r}        \\
\mathbf O_{(m-n)\times r}    & \mathbf d_{n+1}[n+1:m]  & \mathbf O_{(m-n)\times(n-r)} \\
\end{pmatrix}
$$
where $\mathbf P_{\mathbf \Sigma_n}$ is a proper permutation matrix.

}}}

  1. move up zero d's: column-and-row swap
𝐏𝐝n+1*𝐔n* 𝐀n+1(𝐕n𝟎𝟎*1)𝐏Σn𝐏𝐝n+1=(Σn,0𝐎q×(rq)𝟎q𝐎q×(nr)𝐎(rq)×qΣn,1𝐟n+1[q+1:r]𝐎(rq)×(nr)𝐎(mr)×q𝐎(mr)×(rq)𝐝n+1[r+1:m]𝐎(mr)×(nr))

where 𝐏𝐝n+1 is a proper permutation matrix s.t. the non-zero elements of 𝐝n+1[1:r] form a new vector 𝐟n+1[q+1:r].

  1. make a square part by householder transforming residual d. {{{#!latex
\begin{*align}
\ &
\begin{pmatrix}
\mathbf I_r & \mathbf O \\
\mathbf O   & \mathbf H_{m-r}
\end{pmatrix}
\mathbf P_{\mathbf d_{n+1}}^*
\mathbf U_n^* \ \mathbf A_{n+1}
\begin{pmatrix}
\mathbf V_n & \mathbf 0 \\
\mathbf 0^* & 1
\end{pmatrix}
\mathbf P_{\mathbf \Sigma_n}
\mathbf P_{\mathbf d_{n+1}}
\\
&=
\begin{pmatrix}
\mathbf \Sigma_{n,0}      & \mathbf O_{q\times(r-q)}   &  \mathbf 0_q                & \mathbf O_{q\times(n-r)}\\
\mathbf O_{(r-q)\times q}  & \mathbf \Sigma_{n,1}    &  \mathbf f_{n+1}[q+1:r]     & \mathbf O_{(r-q)\times(n-r)}\\
\mathbf 0_{q}^*           &  \mathbf 0_{r-q}^*}   &  -\mathbf f_{n+1}[r+1] e^{j\angle \mathbf d_{n+1}[r+1]} & \mathbf 0_{n-r}^*        \\
\mathbf O_{(m-r-1)\times q} &  \mathbf O_{(m-r-1)\times(r-q)}    & \mathbf 0_{m-r-1}  & \mathbf O_{(m-r-1)\times(n-r)} \\
\end{pmatrix}
\end{*align}
\\
where $\mathbf f_{n+1}[r+1]=||\mathbf d_{n+1}[r+1:m]||$.

}}}

  1. apply secular equation for the square part of dimension r-q+1. {{{#!latex
$$
\begin{pmatrix}
\mathbf \Sigma_{n,1}    &  \mathbf f_{n+1}[q+1:r] \\
\mathbf 0_{r-q}^*}      &  \mathbf f_{n+1}[r+1]   \\
\end{pmatrix}
=
\mathbf U_{n+\tfrac{1}{2}} \mathbf \Sigma_{n+\tfrac{1}{2}} \mathbf V_{n+\tfrac{1}{2}}^*
$$

}}}

     Note that the coefficients of the secular equation will be non-zero. It leads to easy non-generic soluation.
  1. merge diagonal parts and unitary matrices: singular values are not ordered for calculation speed.
𝐀n+1=𝐔n+1(Σn+1𝐎(mn1)×(n+1))𝐕n+1*

where the unordered diagonal matrix is

Σn+1=(Σn,0𝐎q×(rq+1)𝐎(rq+1)×qΣn+12|𝐎(n+1)×(nr))

and the unitary matrices are calculated by multiplying the intermediate unitary matrices:

𝐔n+1=𝐔n(𝐈r𝐎𝐎𝐇mr)𝐏𝐝n+1(𝐈r𝟎𝐎𝟎*ej𝐝n+1[n+1]𝟎*𝐎𝟎𝐈mr1)(𝐈q𝐎𝐎𝐎𝐔n+12𝐎(rq+1)×(mr1)𝐎𝐎(mr1)×(rq+1)𝐈mr1)

and

𝐕n+1=(𝐕n𝟎𝟎*1)𝐏Σn𝐏𝐝n+1(𝐈q𝐎𝐎𝐎𝐕n+12𝐎𝐎𝐎𝐈nr1).
  1. sort the diagonal

example

* mx2 case
  * formula: {{{#!latex
$$
\mathbf A_{2} =
\begin{pmatrix}
\mathbf a_1 & \mathbf  c_{2} \\
\end{pmatrix}
$$

}}}

  * trivial SVD of a,,1,,: {{{#!latex
$$
\mathbf a_1 = \mathbf u_1 \cdot \sigma_1 \cdot 1
$$

}}}

  * almost digonalization: {{{#!latex
$$
\mathbf U_n^* \ \mathbf A_{2} 
\begin{pmatrix}
          1 & 0 \\
          0 & 1
\end{pmatrix}
=
\begin{pmatrix}
\mathbf \sigma_1 &  d_{1} \\
\mathbf 0_{m-1}        &  \mathbf d[2:m] 
\end{pmatrix}
$$
where $d_{1} = \mathbf u_1^* \ \mathbf  c_{2}$ and 

$\mathbf d[2:m] = \mathbf U_1[2:m]^* \ \mathbf c_{2}$. }}}

  * upper-triangular form is good for numerical stability and compact calculation as well:
* mx3 case
* mx4 case

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