Mixing
matrix: solve for square matrix with 0 diagonal
$begingroup$
I have an equation of $XI = Y$, where I know $I$ (vector of 1s) and $Y$ (vector of positive integers). I want to find positive square matrix $X$, I also know that the diagonal entries of $X$ are all 0.
So example is $X$ is 3x3, $I$ is 3x1 and $Y$ is 3x1.
How can I solve this? There could be multiple solutions.
matrix-calculus
asked Jan 24 at 11:41
Dirk NachbarDirk Nachbar
1012
$endgroup$
add a comment |
$begingroup$
I have an equation of $XI = Y$, where I know $I$ (vector of 1s) and $Y$ (vector of positive integers). I want to find positive square matrix $X$, I also know that the diagonal entries of $X$ are all 0.
So example is $X$ is 3x3, $I$ is 3x1 and $Y$ is 3x1.
How can I solve this? There could be multiple solutions.
matrix-calculus
asked Jan 24 at 11:41
Dirk NachbarDirk Nachbar
1012
$endgroup$
add a comment |
$begingroup$
I have an equation of $XI = Y$, where I know $I$ (vector of 1s) and $Y$ (vector of positive integers). I want to find positive square matrix $X$, I also know that the diagonal entries of $X$ are all 0.
So example is $X$ is 3x3, $I$ is 3x1 and $Y$ is 3x1.
How can I solve this? There could be multiple solutions.
matrix-calculus
asked Jan 24 at 11:41
Dirk NachbarDirk Nachbar
1012
$endgroup$
I have an equation of $XI = Y$, where I know $I$ (vector of 1s) and $Y$ (vector of positive integers). I want to find positive square matrix $X$, I also know that the diagonal entries of $X$ are all 0.
So example is $X$ is 3x3, $I$ is 3x1 and $Y$ is 3x1.
How can I solve this? There could be multiple solutions.
matrix-calculus
matrix-calculus
asked Jan 24 at 11:41
Dirk NachbarDirk Nachbar
1012
asked Jan 24 at 11:41
Dirk NachbarDirk Nachbar
1012
asked Jan 24 at 11:41
Dirk NachbarDirk Nachbar
1012
asked Jan 24 at 11:41
Dirk NachbarDirk Nachbar
1012
asked Jan 24 at 11:41
Dirk NachbarDirk Nachbar
1012
1012
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $X=(x_{jk})_{j,k=1}^3$ and $Y=(l,m,n)^T$ with $l,m,n in mathbb N.$ Since $x_{11}=x_{22}=x_{33}=0$, we get from $XI=Y$ the equations
$x_{12}+x_{13}=l, x_{21}+x_{23}=m$ and $x_{31}+x_{32}=n.$
Can you proceed ?
answered Jan 24 at 12:08
FredFred
48.3k1849
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add a comment |
$begingroup$
The Moore-Penrose inverse can be used to write the general solution to any linear equation, i.e.
$$(XA = Y) implies X = YA^+ + C - CAA^+$$ where $C$ is an arbitrary matrix.
In the current problem the matrices $(A,Y)$ are replaced by the vectors $(1,y)$, yielding
$$eqalign{
X &= y1^+ + C - C11^+ cr
}$$
For a real vector, the MP-inverse can be written in terms of the transpose
$$a^+ = frac{a^T}{a^Ta}$$
so we can write
$$1^+ = tfrac{1}{3}1^T$$
Extract the $k^{th}$ diagonal element of the matrix by multiplying from the left with $e_k^T$ and from the right with $e_k$, and set it to zero.
$$eqalign{
X_{kk} &= e_k^TXe_k cr
0 &= tfrac{1}{3}y_k + C_{kk} - tfrac{1}{3}e_k^TC1 cr
0 &= y_k + 3C_{kk} - e_k^TC1 cr
}$$
There are more unknowns than equations, so let's constrain the matrix to $C={rm Diag}(c).$
$$eqalign{
0 &= y_k + 3c_{k} - e_k^Tc cr
c_k &= -tfrac{1}{2}y_k &implies
C &= -tfrac{1}{2}{rm Diag}(y) cr
}$$
Putting it all together, and generalizing the dimensions from $(3to n)$ yields
$$eqalign{
X &= tfrac{1}{n}y1^T - tfrac{1}{n-1}{rm Diag}(y) + tfrac{1}{n(n-1)}{rm Diag}(y)11^T cr
}$$
answered Jan 26 at 19:39
greggreg
8,8401824
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $X=(x_{jk})_{j,k=1}^3$ and $Y=(l,m,n)^T$ with $l,m,n in mathbb N.$ Since $x_{11}=x_{22}=x_{33}=0$, we get from $XI=Y$ the equations
$x_{12}+x_{13}=l, x_{21}+x_{23}=m$ and $x_{31}+x_{32}=n.$
Can you proceed ?
answered Jan 24 at 12:08
FredFred
48.3k1849
$endgroup$
add a comment |
$begingroup$
Let $X=(x_{jk})_{j,k=1}^3$ and $Y=(l,m,n)^T$ with $l,m,n in mathbb N.$ Since $x_{11}=x_{22}=x_{33}=0$, we get from $XI=Y$ the equations
$x_{12}+x_{13}=l, x_{21}+x_{23}=m$ and $x_{31}+x_{32}=n.$
Can you proceed ?
answered Jan 24 at 12:08
FredFred
48.3k1849
$endgroup$
add a comment |
$begingroup$
Let $X=(x_{jk})_{j,k=1}^3$ and $Y=(l,m,n)^T$ with $l,m,n in mathbb N.$ Since $x_{11}=x_{22}=x_{33}=0$, we get from $XI=Y$ the equations
$x_{12}+x_{13}=l, x_{21}+x_{23}=m$ and $x_{31}+x_{32}=n.$
Can you proceed ?
answered Jan 24 at 12:08
FredFred
48.3k1849
$endgroup$
Let $X=(x_{jk})_{j,k=1}^3$ and $Y=(l,m,n)^T$ with $l,m,n in mathbb N.$ Since $x_{11}=x_{22}=x_{33}=0$, we get from $XI=Y$ the equations
$x_{12}+x_{13}=l, x_{21}+x_{23}=m$ and $x_{31}+x_{32}=n.$
Can you proceed ?
answered Jan 24 at 12:08
FredFred
48.3k1849
answered Jan 24 at 12:08
FredFred
48.3k1849
answered Jan 24 at 12:08
FredFred
48.3k1849
answered Jan 24 at 12:08
FredFred
48.3k1849
48.3k1849
add a comment |
add a comment |
$begingroup$
The Moore-Penrose inverse can be used to write the general solution to any linear equation, i.e.
$$(XA = Y) implies X = YA^+ + C - CAA^+$$ where $C$ is an arbitrary matrix.
In the current problem the matrices $(A,Y)$ are replaced by the vectors $(1,y)$, yielding
$$eqalign{
X &= y1^+ + C - C11^+ cr
}$$
For a real vector, the MP-inverse can be written in terms of the transpose
$$a^+ = frac{a^T}{a^Ta}$$
so we can write
$$1^+ = tfrac{1}{3}1^T$$
Extract the $k^{th}$ diagonal element of the matrix by multiplying from the left with $e_k^T$ and from the right with $e_k$, and set it to zero.
$$eqalign{
X_{kk} &= e_k^TXe_k cr
0 &= tfrac{1}{3}y_k + C_{kk} - tfrac{1}{3}e_k^TC1 cr
0 &= y_k + 3C_{kk} - e_k^TC1 cr
}$$
There are more unknowns than equations, so let's constrain the matrix to $C={rm Diag}(c).$
$$eqalign{
0 &= y_k + 3c_{k} - e_k^Tc cr
c_k &= -tfrac{1}{2}y_k &implies
C &= -tfrac{1}{2}{rm Diag}(y) cr
}$$
Putting it all together, and generalizing the dimensions from $(3to n)$ yields
$$eqalign{
X &= tfrac{1}{n}y1^T - tfrac{1}{n-1}{rm Diag}(y) + tfrac{1}{n(n-1)}{rm Diag}(y)11^T cr
}$$
answered Jan 26 at 19:39
greggreg
8,8401824
$endgroup$
add a comment |
$begingroup$
The Moore-Penrose inverse can be used to write the general solution to any linear equation, i.e.
$$(XA = Y) implies X = YA^+ + C - CAA^+$$ where $C$ is an arbitrary matrix.
In the current problem the matrices $(A,Y)$ are replaced by the vectors $(1,y)$, yielding
$$eqalign{
X &= y1^+ + C - C11^+ cr
}$$
For a real vector, the MP-inverse can be written in terms of the transpose
$$a^+ = frac{a^T}{a^Ta}$$
so we can write
$$1^+ = tfrac{1}{3}1^T$$
Extract the $k^{th}$ diagonal element of the matrix by multiplying from the left with $e_k^T$ and from the right with $e_k$, and set it to zero.
$$eqalign{
X_{kk} &= e_k^TXe_k cr
0 &= tfrac{1}{3}y_k + C_{kk} - tfrac{1}{3}e_k^TC1 cr
0 &= y_k + 3C_{kk} - e_k^TC1 cr
}$$
There are more unknowns than equations, so let's constrain the matrix to $C={rm Diag}(c).$
$$eqalign{
0 &= y_k + 3c_{k} - e_k^Tc cr
c_k &= -tfrac{1}{2}y_k &implies
C &= -tfrac{1}{2}{rm Diag}(y) cr
}$$
Putting it all together, and generalizing the dimensions from $(3to n)$ yields
$$eqalign{
X &= tfrac{1}{n}y1^T - tfrac{1}{n-1}{rm Diag}(y) + tfrac{1}{n(n-1)}{rm Diag}(y)11^T cr
}$$
answered Jan 26 at 19:39
greggreg
8,8401824
$endgroup$
add a comment |
$begingroup$
The Moore-Penrose inverse can be used to write the general solution to any linear equation, i.e.
$$(XA = Y) implies X = YA^+ + C - CAA^+$$ where $C$ is an arbitrary matrix.
In the current problem the matrices $(A,Y)$ are replaced by the vectors $(1,y)$, yielding
$$eqalign{
X &= y1^+ + C - C11^+ cr
}$$
For a real vector, the MP-inverse can be written in terms of the transpose
$$a^+ = frac{a^T}{a^Ta}$$
so we can write
$$1^+ = tfrac{1}{3}1^T$$
Extract the $k^{th}$ diagonal element of the matrix by multiplying from the left with $e_k^T$ and from the right with $e_k$, and set it to zero.
$$eqalign{
X_{kk} &= e_k^TXe_k cr
0 &= tfrac{1}{3}y_k + C_{kk} - tfrac{1}{3}e_k^TC1 cr
0 &= y_k + 3C_{kk} - e_k^TC1 cr
}$$
There are more unknowns than equations, so let's constrain the matrix to $C={rm Diag}(c).$
$$eqalign{
0 &= y_k + 3c_{k} - e_k^Tc cr
c_k &= -tfrac{1}{2}y_k &implies
C &= -tfrac{1}{2}{rm Diag}(y) cr
}$$
Putting it all together, and generalizing the dimensions from $(3to n)$ yields
$$eqalign{
X &= tfrac{1}{n}y1^T - tfrac{1}{n-1}{rm Diag}(y) + tfrac{1}{n(n-1)}{rm Diag}(y)11^T cr
}$$
answered Jan 26 at 19:39
greggreg
8,8401824
$endgroup$
The Moore-Penrose inverse can be used to write the general solution to any linear equation, i.e.
$$(XA = Y) implies X = YA^+ + C - CAA^+$$ where $C$ is an arbitrary matrix.
In the current problem the matrices $(A,Y)$ are replaced by the vectors $(1,y)$, yielding
$$eqalign{
X &= y1^+ + C - C11^+ cr
}$$
For a real vector, the MP-inverse can be written in terms of the transpose
$$a^+ = frac{a^T}{a^Ta}$$
so we can write
$$1^+ = tfrac{1}{3}1^T$$
Extract the $k^{th}$ diagonal element of the matrix by multiplying from the left with $e_k^T$ and from the right with $e_k$, and set it to zero.
$$eqalign{
X_{kk} &= e_k^TXe_k cr
0 &= tfrac{1}{3}y_k + C_{kk} - tfrac{1}{3}e_k^TC1 cr
0 &= y_k + 3C_{kk} - e_k^TC1 cr
}$$
There are more unknowns than equations, so let's constrain the matrix to $C={rm Diag}(c).$
$$eqalign{
0 &= y_k + 3c_{k} - e_k^Tc cr
c_k &= -tfrac{1}{2}y_k &implies
C &= -tfrac{1}{2}{rm Diag}(y) cr
}$$
Putting it all together, and generalizing the dimensions from $(3to n)$ yields
$$eqalign{
X &= tfrac{1}{n}y1^T - tfrac{1}{n-1}{rm Diag}(y) + tfrac{1}{n(n-1)}{rm Diag}(y)11^T cr
}$$
answered Jan 26 at 19:39
greggreg
8,8401824
edited Jan 26 at 20:53
answered Jan 26 at 19:39
greggreg
8,8401824
answered Jan 26 at 19:39
greggreg
8,8401824
answered Jan 26 at 19:39
greggreg
8,8401824
8,8401824
add a comment |
add a comment |
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