User talk:Oyz

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– Oleg Alexandrov (talk) 17:08, 7 April 2006 (UTC)[reply]

$e^{j\pi }=1$

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

$H=I-2\ v\ v^{*}$
$v=e^{j\theta }\ e_{1}-w$
$Hw=e_{1}$
$He_{1}=w$
$||v||=1$
$||w||=1$
$H^{*}H=I$

$R=\Theta -W$
$W^{*}W=I,\qquad \Theta ^{*}\Theta =I.$
$\Gamma ^{*}R^{*}R\Gamma =I$
$R+2R\Gamma \Gamma ^{*}R^{*}W=0$

$R\Gamma +2R\Gamma \Gamma ^{*}R^{*}(\Theta -R)\Gamma =0$
$R\Gamma +2R\Gamma \Gamma ^{*}R^{*}\Theta \Gamma -2R\Gamma \Gamma ^{*}R^{*}R\Gamma =0$
$2R\Gamma \Gamma ^{*}R^{*}\Theta \Gamma =R\Gamma$

$R-2R\Gamma \Gamma ^{*}R^{*}\Theta =0$

$R\Gamma \Gamma ^{*}R^{*}(W+\Theta )=0$

$(\Theta -W)\Gamma \Gamma ^{*}(\Theta -W)^{*}(\Theta +W)=0$

$\Gamma ^{*}(\Theta ^{*}W-W^{*}\Theta )=0$

It implies $\Gamma$ is not full-rank. It contradicts with $\Gamma ^{*}R^{*}R\Gamma =I$ .

Therefore, $\Theta ^{*}W=W^{*}\Theta$

Since $\Theta$ 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:

\mathbf {A} _{n+1}={\begin{pmatrix}\mathbf {A} _{n}&\mathbf {c} _{n+1}\\\end{pmatrix}}

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:

move zero sigmas right-most: column-swap
move up zero d's: column-and-row swap
make a square part by householder transforming residual d.
apply secular equation for the square part of dimension r-q+1.
merge diagonal parts and unitary matrices: singular values are not ordered for calculation speed.
sort the diagonal

re-visit formula:

\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:r,1:r]&\mathbf {O} _{r\times (n-r)}&\mathbf {d} _{n+1}[1:r]\\\mathbf {O} _{(n-r)\times r}&\mathbf {O} _{n-r}&\mathbf {d} _{n+1}[r+1:n]\\\mathbf {O} _{(m-n)\times r}&\mathbf {O} _{(m-n)\times (n-r)}&\mathbf {d} _{n+1}[n+1:m]\\\end{pmatrix}}

where $r$ is rank of $\Sigma _{n}$ . Note that $\mathbf {d} _{n+1}[1:r]$ may include zeros.

more swapping rows and columns for zero singular values and diagonal parts.

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.

}}}

move up zero d's: column-and-row swap

\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 {O} _{(m-r)\times q}&\mathbf {O} _{(m-r)\times (r-q)}&\mathbf {d} _{n+1}[r+1:m]&\mathbf {O} _{(m-r)\times (n-r)}\\\end{pmatrix}}

where $\mathbf {P} _{\mathbf {d} _{n+1}}$ is a proper permutation matrix s.t. the non-zero elements of $\mathbf {d} _{n+1}[1:r]$ form a new vector $\mathbf {f} _{n+1}[q+1:r]$ .

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.

merge diagonal parts and unitary matrices: singular values are not ordered for calculation speed.

\mathbf {A} _{n+1}=\mathbf {U} _{n+1}{\begin{pmatrix}\mathbf {\Sigma } _{n+1}\\\mathbf {O} _{(m-n-1)\times (n+1)}\end{pmatrix}}\mathbf {V} _{n+1}^{*}

where the unordered diagonal matrix is

\mathbf {\Sigma } _{n+1}={\begin{pmatrix}\left.{\begin{matrix}\mathbf {\Sigma } _{n,0}&\mathbf {O} _{q\times (r-q+1)}\\\mathbf {O} _{(r-q+1)\times q}&\mathbf {\Sigma } _{n+{\tfrac {1}{2}}}\\\end{matrix}}\right|&\mathbf {O} _{(n+1)\times (n-r)}\end{pmatrix}}

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

\mathbf {U} _{n+1}=\mathbf {U} _{n}{\begin{pmatrix}\mathbf {I} _{r}&\mathbf {O} \\\mathbf {O} &\mathbf {H} _{m-r}\end{pmatrix}}\mathbf {P} _{\mathbf {d} _{n+1}}{\begin{pmatrix}\mathbf {I} _{r}&\mathbf {0} &\mathbf {O} \\\mathbf {0} ^{*}&-e^{j\angle \mathbf {d} _{n+1}[n+1]}&\mathbf {0} ^{*}\\\mathbf {O} &\mathbf {0} &\mathbf {I} _{m-r-1}\\\end{pmatrix}}{\begin{pmatrix}\mathbf {I} _{q}&\mathbf {O} &\mathbf {O} \\\mathbf {O} &\mathbf {U} _{n+{\tfrac {1}{2}}}&\mathbf {O} _{(r-q+1)\times (m-r-1)}\\\mathbf {O} &\mathbf {O} _{(m-r-1)\times (r-q+1)}&\mathbf {I} _{m-r-1}\end{pmatrix}}

and

\mathbf {V} _{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 {I} _{q}&\mathbf {O} &\mathbf {O} \\\mathbf {O} &\mathbf {V} _{n+{\tfrac {1}{2}}}&\mathbf {O} \\\mathbf {O} &\mathbf {O} &\mathbf {I} _{n-r-1}\\\end{pmatrix}}.

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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