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
47.7k1849
$endgroup$
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
8,6851823
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3085773%2fmatrix-solve-for-square-matrix-with-0-diagonal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
47.7k1849
$endgroup$
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
47.7k1849
$endgroup$
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
47.7k1849
$endgroup$
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
47.7k1849
answered Jan 24 at 12:08
FredFred
47.7k1849
answered Jan 24 at 12:08
FredFred
47.7k1849
answered Jan 24 at 12:08
FredFred
47.7k1849
47.7k1849
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
8,6851823
$endgroup$
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
8,6851823
$endgroup$
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
8,6851823
$endgroup$
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
8,6851823
edited Jan 26 at 20:53
answered Jan 26 at 19:39
greggreg
8,6851823
answered Jan 26 at 19:39
greggreg
8,6851823
answered Jan 26 at 19:39
greggreg
8,6851823
8,6851823
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3085773%2fmatrix-solve-for-square-matrix-with-0-diagonal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
