Help to find a factorization of a matrix into a product and sum of matrices
$begingroup$
if $hspace{0.2cm}$$Zin$ R$^{ntimes n} $ $hspace{0.2cm}$is the down-shift matrix with ones on the first subdiagonal and zeros elsewhere, and $Lin$ R$^{ntimes n} $ $hspace{0.2cm}$ is the lower triangular matrix with $1s$ at the non-zero entries, then the matrix
$$A=left(
begin{array}{ccccc}
w_1 & w_1 &cdots & w_1 \
w_1 & w_2 & cdots& w_2 \
vdots & vdots & ddots& vdots \
w_1 & w_2 & cdots & w_n\
end{array}
right)$$
can be written as $hspace{0.2cm}$ $A=L(D_w-ZD_wZ^T)L^T$$hspace{0.2cm}$ where $hspace{0.2cm}$$D_w=Diag[w_1,cdots,w_n]$
If I have the matrix
$$B=left(
begin{array}{ccccc}
w_1 & w_2 &cdots & w_n \
w_2 & w_2 & cdots& w_n \
vdots & vdots & ddots& vdots \
w_n & w_n & cdots & w_n\
end{array}
right)$$
How can I to factorize it in a similar way to the previous matrix in terms of $Z$ and $L$ ?
linear-algebra matrices factoring
asked Jan 18 at 17:02
mrRmrR
111
$endgroup$
add a comment |
$begingroup$
if $hspace{0.2cm}$$Zin$ R$^{ntimes n} $ $hspace{0.2cm}$is the down-shift matrix with ones on the first subdiagonal and zeros elsewhere, and $Lin$ R$^{ntimes n} $ $hspace{0.2cm}$ is the lower triangular matrix with $1s$ at the non-zero entries, then the matrix
$$A=left(
begin{array}{ccccc}
w_1 & w_1 &cdots & w_1 \
w_1 & w_2 & cdots& w_2 \
vdots & vdots & ddots& vdots \
w_1 & w_2 & cdots & w_n\
end{array}
right)$$
can be written as $hspace{0.2cm}$ $A=L(D_w-ZD_wZ^T)L^T$$hspace{0.2cm}$ where $hspace{0.2cm}$$D_w=Diag[w_1,cdots,w_n]$
If I have the matrix
$$B=left(
begin{array}{ccccc}
w_1 & w_2 &cdots & w_n \
w_2 & w_2 & cdots& w_n \
vdots & vdots & ddots& vdots \
w_n & w_n & cdots & w_n\
end{array}
right)$$
How can I to factorize it in a similar way to the previous matrix in terms of $Z$ and $L$ ?
linear-algebra matrices factoring
asked Jan 18 at 17:02
mrRmrR
111
$endgroup$
add a comment |
$begingroup$
if $hspace{0.2cm}$$Zin$ R$^{ntimes n} $ $hspace{0.2cm}$is the down-shift matrix with ones on the first subdiagonal and zeros elsewhere, and $Lin$ R$^{ntimes n} $ $hspace{0.2cm}$ is the lower triangular matrix with $1s$ at the non-zero entries, then the matrix
$$A=left(
begin{array}{ccccc}
w_1 & w_1 &cdots & w_1 \
w_1 & w_2 & cdots& w_2 \
vdots & vdots & ddots& vdots \
w_1 & w_2 & cdots & w_n\
end{array}
right)$$
can be written as $hspace{0.2cm}$ $A=L(D_w-ZD_wZ^T)L^T$$hspace{0.2cm}$ where $hspace{0.2cm}$$D_w=Diag[w_1,cdots,w_n]$
If I have the matrix
$$B=left(
begin{array}{ccccc}
w_1 & w_2 &cdots & w_n \
w_2 & w_2 & cdots& w_n \
vdots & vdots & ddots& vdots \
w_n & w_n & cdots & w_n\
end{array}
right)$$
How can I to factorize it in a similar way to the previous matrix in terms of $Z$ and $L$ ?
linear-algebra matrices factoring
asked Jan 18 at 17:02
mrRmrR
111
$endgroup$
if $hspace{0.2cm}$$Zin$ R$^{ntimes n} $ $hspace{0.2cm}$is the down-shift matrix with ones on the first subdiagonal and zeros elsewhere, and $Lin$ R$^{ntimes n} $ $hspace{0.2cm}$ is the lower triangular matrix with $1s$ at the non-zero entries, then the matrix
$$A=left(
begin{array}{ccccc}
w_1 & w_1 &cdots & w_1 \
w_1 & w_2 & cdots& w_2 \
vdots & vdots & ddots& vdots \
w_1 & w_2 & cdots & w_n\
end{array}
right)$$
can be written as $hspace{0.2cm}$ $A=L(D_w-ZD_wZ^T)L^T$$hspace{0.2cm}$ where $hspace{0.2cm}$$D_w=Diag[w_1,cdots,w_n]$
If I have the matrix
$$B=left(
begin{array}{ccccc}
w_1 & w_2 &cdots & w_n \
w_2 & w_2 & cdots& w_n \
vdots & vdots & ddots& vdots \
w_n & w_n & cdots & w_n\
end{array}
right)$$
How can I to factorize it in a similar way to the previous matrix in terms of $Z$ and $L$ ?
linear-algebra matrices factoring
linear-algebra matrices factoring
asked Jan 18 at 17:02
mrRmrR
111
asked Jan 18 at 17:02
mrRmrR
111
asked Jan 18 at 17:02
mrRmrR
111
asked Jan 18 at 17:02
mrRmrR
111
asked Jan 18 at 17:02
mrRmrR
111
111
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $K$ denote the matrix with $1$s on the "antidiagonal", i.e.
$$
K = pmatrix{&&1\&cdots\1}
$$
We note that "flipping" the matrix $B$ gives us
$$
KBK = pmatrix{
w_n&w_n&cdots&w_n\
w_n&w_{n-1}&cdots&w_{n-1}\
vdots & vdots & ddots & vdots\
w_n & w_{n-1} & cdots & w_1}
$$
Let $D_hat w$ denote the matrix $operatorname{Diag}[w_n,w_{n-1},dots,w_1]$ (if you prefer, you could write $D_{hat w} = KD_wK$). Using your decomposition, we have
$$
KBK = L(D_{hat w}-ZD_{hat w}Z^T)L^T implies\
B = KL(D_{hat w}-ZD_{hat w}Z^T)L^TK
$$
Or if you prefer,
$$
B = (KL)(D_{hat w}-ZD_{hat w}Z^T)(KL)^T
$$
answered Jan 18 at 17:49
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $K$ denote the matrix with $1$s on the "antidiagonal", i.e.
$$
K = pmatrix{&&1\&cdots\1}
$$
We note that "flipping" the matrix $B$ gives us
$$
KBK = pmatrix{
w_n&w_n&cdots&w_n\
w_n&w_{n-1}&cdots&w_{n-1}\
vdots & vdots & ddots & vdots\
w_n & w_{n-1} & cdots & w_1}
$$
Let $D_hat w$ denote the matrix $operatorname{Diag}[w_n,w_{n-1},dots,w_1]$ (if you prefer, you could write $D_{hat w} = KD_wK$). Using your decomposition, we have
$$
KBK = L(D_{hat w}-ZD_{hat w}Z^T)L^T implies\
B = KL(D_{hat w}-ZD_{hat w}Z^T)L^TK
$$
Or if you prefer,
$$
B = (KL)(D_{hat w}-ZD_{hat w}Z^T)(KL)^T
$$
answered Jan 18 at 17:49
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
add a comment |
$begingroup$
Let $K$ denote the matrix with $1$s on the "antidiagonal", i.e.
$$
K = pmatrix{&&1\&cdots\1}
$$
We note that "flipping" the matrix $B$ gives us
$$
KBK = pmatrix{
w_n&w_n&cdots&w_n\
w_n&w_{n-1}&cdots&w_{n-1}\
vdots & vdots & ddots & vdots\
w_n & w_{n-1} & cdots & w_1}
$$
Let $D_hat w$ denote the matrix $operatorname{Diag}[w_n,w_{n-1},dots,w_1]$ (if you prefer, you could write $D_{hat w} = KD_wK$). Using your decomposition, we have
$$
KBK = L(D_{hat w}-ZD_{hat w}Z^T)L^T implies\
B = KL(D_{hat w}-ZD_{hat w}Z^T)L^TK
$$
Or if you prefer,
$$
B = (KL)(D_{hat w}-ZD_{hat w}Z^T)(KL)^T
$$
answered Jan 18 at 17:49
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
add a comment |
$begingroup$
Let $K$ denote the matrix with $1$s on the "antidiagonal", i.e.
$$
K = pmatrix{&&1\&cdots\1}
$$
We note that "flipping" the matrix $B$ gives us
$$
KBK = pmatrix{
w_n&w_n&cdots&w_n\
w_n&w_{n-1}&cdots&w_{n-1}\
vdots & vdots & ddots & vdots\
w_n & w_{n-1} & cdots & w_1}
$$
Let $D_hat w$ denote the matrix $operatorname{Diag}[w_n,w_{n-1},dots,w_1]$ (if you prefer, you could write $D_{hat w} = KD_wK$). Using your decomposition, we have
$$
KBK = L(D_{hat w}-ZD_{hat w}Z^T)L^T implies\
B = KL(D_{hat w}-ZD_{hat w}Z^T)L^TK
$$
Or if you prefer,
$$
B = (KL)(D_{hat w}-ZD_{hat w}Z^T)(KL)^T
$$
answered Jan 18 at 17:49
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
Let $K$ denote the matrix with $1$s on the "antidiagonal", i.e.
$$
K = pmatrix{&&1\&cdots\1}
$$
We note that "flipping" the matrix $B$ gives us
$$
KBK = pmatrix{
w_n&w_n&cdots&w_n\
w_n&w_{n-1}&cdots&w_{n-1}\
vdots & vdots & ddots & vdots\
w_n & w_{n-1} & cdots & w_1}
$$
Let $D_hat w$ denote the matrix $operatorname{Diag}[w_n,w_{n-1},dots,w_1]$ (if you prefer, you could write $D_{hat w} = KD_wK$). Using your decomposition, we have
$$
KBK = L(D_{hat w}-ZD_{hat w}Z^T)L^T implies\
B = KL(D_{hat w}-ZD_{hat w}Z^T)L^TK
$$
Or if you prefer,
$$
B = (KL)(D_{hat w}-ZD_{hat w}Z^T)(KL)^T
$$
answered Jan 18 at 17:49
OmnomnomnomOmnomnomnom
129k794188
answered Jan 18 at 17:49
OmnomnomnomOmnomnomnom
129k794188
answered Jan 18 at 17:49
OmnomnomnomOmnomnomnom
129k794188
answered Jan 18 at 17:49
OmnomnomnomOmnomnomnom
129k794188
129k794188
add a comment |
add a comment |
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