Matrix Commutation, is it valid here?
$begingroup$
I'm trying to obtain the Ridge Regression solution from the mean of predictive distribution of a Gaussian Process with a linear kernel.
The mean of the predictive distribution of a GP is $$boldsymbol k^T(boldsymbol K+sigma^2I)^{-1}y$$
where $boldsymbol K=k(boldsymbol X^T, boldsymbol X)$ is the Kernel Gram Matrix (Symetric Positive definite) and $boldsymbol k = k(x_{n+1},boldsymbol X)$. (Note that $x_{n+1}$ is a vector while $boldsymbol X$ is a matrix
Now I'm assuming a linear kernel, thus $k(boldsymbol X^T, boldsymbol X) = boldsymbol X^T boldsymbol X$ and $boldsymbol k = k(x_{n+1},boldsymbol X) = x_{n+1}^T* boldsymbol X$.
So I can write
$$boldsymbol k^T(boldsymbol K+sigma^2*I)^{-1}y = (x_{n+1} boldsymbol X^T) (boldsymbol X^T boldsymbol X+sigma^2*I)^{-1}y $$
Now the issue:
The Ridge Regression should be $$x_{n_1}(X^TX+sigma ^2 I)^{-1}boldsymbol X^T y$$ while I have
$$(x_{n+1} boldsymbol X^T) (boldsymbol X^T boldsymbol X+sigma^2*I)^{-1}y $$
So I need to commute $boldsymbol X^T$ and $(boldsymbol K+sigma^2I)^{-1}$ but I know matrix cannot commute, so I'm stucked.
What am I missing? Can you help me?
Note that $(boldsymbol K+sigma^2I)^{-1}$ is a symmetric matrix.
linear-algebra matrices
asked Jan 12 at 10:12
Tommaso BendinelliTommaso Bendinelli
13610
$endgroup$
add a comment |
$begingroup$
I'm trying to obtain the Ridge Regression solution from the mean of predictive distribution of a Gaussian Process with a linear kernel.
The mean of the predictive distribution of a GP is $$boldsymbol k^T(boldsymbol K+sigma^2I)^{-1}y$$
where $boldsymbol K=k(boldsymbol X^T, boldsymbol X)$ is the Kernel Gram Matrix (Symetric Positive definite) and $boldsymbol k = k(x_{n+1},boldsymbol X)$. (Note that $x_{n+1}$ is a vector while $boldsymbol X$ is a matrix
Now I'm assuming a linear kernel, thus $k(boldsymbol X^T, boldsymbol X) = boldsymbol X^T boldsymbol X$ and $boldsymbol k = k(x_{n+1},boldsymbol X) = x_{n+1}^T* boldsymbol X$.
So I can write
$$boldsymbol k^T(boldsymbol K+sigma^2*I)^{-1}y = (x_{n+1} boldsymbol X^T) (boldsymbol X^T boldsymbol X+sigma^2*I)^{-1}y $$
Now the issue:
The Ridge Regression should be $$x_{n_1}(X^TX+sigma ^2 I)^{-1}boldsymbol X^T y$$ while I have
$$(x_{n+1} boldsymbol X^T) (boldsymbol X^T boldsymbol X+sigma^2*I)^{-1}y $$
So I need to commute $boldsymbol X^T$ and $(boldsymbol K+sigma^2I)^{-1}$ but I know matrix cannot commute, so I'm stucked.
What am I missing? Can you help me?
Note that $(boldsymbol K+sigma^2I)^{-1}$ is a symmetric matrix.
linear-algebra matrices
asked Jan 12 at 10:12
Tommaso BendinelliTommaso Bendinelli
13610
$endgroup$
add a comment |
$begingroup$
I'm trying to obtain the Ridge Regression solution from the mean of predictive distribution of a Gaussian Process with a linear kernel.
The mean of the predictive distribution of a GP is $$boldsymbol k^T(boldsymbol K+sigma^2I)^{-1}y$$
where $boldsymbol K=k(boldsymbol X^T, boldsymbol X)$ is the Kernel Gram Matrix (Symetric Positive definite) and $boldsymbol k = k(x_{n+1},boldsymbol X)$. (Note that $x_{n+1}$ is a vector while $boldsymbol X$ is a matrix
Now I'm assuming a linear kernel, thus $k(boldsymbol X^T, boldsymbol X) = boldsymbol X^T boldsymbol X$ and $boldsymbol k = k(x_{n+1},boldsymbol X) = x_{n+1}^T* boldsymbol X$.
So I can write
$$boldsymbol k^T(boldsymbol K+sigma^2*I)^{-1}y = (x_{n+1} boldsymbol X^T) (boldsymbol X^T boldsymbol X+sigma^2*I)^{-1}y $$
Now the issue:
The Ridge Regression should be $$x_{n_1}(X^TX+sigma ^2 I)^{-1}boldsymbol X^T y$$ while I have
$$(x_{n+1} boldsymbol X^T) (boldsymbol X^T boldsymbol X+sigma^2*I)^{-1}y $$
So I need to commute $boldsymbol X^T$ and $(boldsymbol K+sigma^2I)^{-1}$ but I know matrix cannot commute, so I'm stucked.
What am I missing? Can you help me?
Note that $(boldsymbol K+sigma^2I)^{-1}$ is a symmetric matrix.
linear-algebra matrices
asked Jan 12 at 10:12
Tommaso BendinelliTommaso Bendinelli
13610
$endgroup$
I'm trying to obtain the Ridge Regression solution from the mean of predictive distribution of a Gaussian Process with a linear kernel.
The mean of the predictive distribution of a GP is $$boldsymbol k^T(boldsymbol K+sigma^2I)^{-1}y$$
where $boldsymbol K=k(boldsymbol X^T, boldsymbol X)$ is the Kernel Gram Matrix (Symetric Positive definite) and $boldsymbol k = k(x_{n+1},boldsymbol X)$. (Note that $x_{n+1}$ is a vector while $boldsymbol X$ is a matrix
Now I'm assuming a linear kernel, thus $k(boldsymbol X^T, boldsymbol X) = boldsymbol X^T boldsymbol X$ and $boldsymbol k = k(x_{n+1},boldsymbol X) = x_{n+1}^T* boldsymbol X$.
So I can write
$$boldsymbol k^T(boldsymbol K+sigma^2*I)^{-1}y = (x_{n+1} boldsymbol X^T) (boldsymbol X^T boldsymbol X+sigma^2*I)^{-1}y $$
Now the issue:
The Ridge Regression should be $$x_{n_1}(X^TX+sigma ^2 I)^{-1}boldsymbol X^T y$$ while I have
$$(x_{n+1} boldsymbol X^T) (boldsymbol X^T boldsymbol X+sigma^2*I)^{-1}y $$
So I need to commute $boldsymbol X^T$ and $(boldsymbol K+sigma^2I)^{-1}$ but I know matrix cannot commute, so I'm stucked.
What am I missing? Can you help me?
Note that $(boldsymbol K+sigma^2I)^{-1}$ is a symmetric matrix.
linear-algebra matrices
linear-algebra matrices
asked Jan 12 at 10:12
Tommaso BendinelliTommaso Bendinelli
13610
asked Jan 12 at 10:12
Tommaso BendinelliTommaso Bendinelli
13610
edited Jan 12 at 10:49
Tommaso Bendinelli
asked Jan 12 at 10:12
Tommaso BendinelliTommaso Bendinelli
13610
asked Jan 12 at 10:12
Tommaso BendinelliTommaso Bendinelli
13610
asked Jan 12 at 10:12
Tommaso BendinelliTommaso Bendinelli
13610
13610
add a comment |
add a comment |
0
active
oldest
votes
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%2f3070757%2fmatrix-commutation-is-it-valid-here%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3070757%2fmatrix-commutation-is-it-valid-here%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
