Eigen Values of a product of two matrices
$begingroup$
Suppose we have a matrix $C = DP$ where D is a real diagonal matrix with all the entries positive, and P is a real matrix whose eigen values are positive and all the elements are positive. Moreover, each row of P sums to the same value. Can we conclude that C has all the eigen values positive?
linear-algebra
asked Nov 28 '18 at 13:07
Suhan ShettySuhan Shetty
1077
$endgroup$
add a comment |
$begingroup$
Suppose we have a matrix $C = DP$ where D is a real diagonal matrix with all the entries positive, and P is a real matrix whose eigen values are positive and all the elements are positive. Moreover, each row of P sums to the same value. Can we conclude that C has all the eigen values positive?
linear-algebra
asked Nov 28 '18 at 13:07
Suhan ShettySuhan Shetty
1077
$endgroup$
$begingroup$
If the matrix $P$ is symmetric (in addition to the above conditions), I suspect that $C$ can be shown to have all positive eigenvalues. But I'm not 100% sure.
$endgroup$
– Michael Seifert
Nov 28 '18 at 14:01
$begingroup$
@MichaelSeifert Indeed. If $P$ is symmetric, then $C=DP$ is similar to $D^{-1/2}CD^{1/2}=D^{1/2}PD^{1/2}$, which is positive definite.
$endgroup$
– user1551
Nov 29 '18 at 11:41
$begingroup$
Why is $D^{1/2}PD{1/2}$ positive definite if P is symmetric? @user1551
$endgroup$
– Suhan Shetty
Nov 29 '18 at 12:26
$begingroup$
@SuhanShetty In your question, $P$ is assumed to be a real matrix with a positive spectrum. If it is also symmetric, it is positive definite. Hence $D^{1/2}PD^{1/2}$ (which is congruent to $P$) is positive definite too.
$endgroup$
– user1551
Nov 29 '18 at 12:42
add a comment |
$begingroup$
Suppose we have a matrix $C = DP$ where D is a real diagonal matrix with all the entries positive, and P is a real matrix whose eigen values are positive and all the elements are positive. Moreover, each row of P sums to the same value. Can we conclude that C has all the eigen values positive?
linear-algebra
asked Nov 28 '18 at 13:07
Suhan ShettySuhan Shetty
1077
$endgroup$
Suppose we have a matrix $C = DP$ where D is a real diagonal matrix with all the entries positive, and P is a real matrix whose eigen values are positive and all the elements are positive. Moreover, each row of P sums to the same value. Can we conclude that C has all the eigen values positive?
linear-algebra
linear-algebra
asked Nov 28 '18 at 13:07
Suhan ShettySuhan Shetty
1077
asked Nov 28 '18 at 13:07
Suhan ShettySuhan Shetty
1077
asked Nov 28 '18 at 13:07
Suhan ShettySuhan Shetty
1077
asked Nov 28 '18 at 13:07
Suhan ShettySuhan Shetty
1077
asked Nov 28 '18 at 13:07
Suhan ShettySuhan Shetty
1077
1077
$begingroup$
If the matrix $P$ is symmetric (in addition to the above conditions), I suspect that $C$ can be shown to have all positive eigenvalues. But I'm not 100% sure.
$endgroup$
– Michael Seifert
Nov 28 '18 at 14:01
$begingroup$
@MichaelSeifert Indeed. If $P$ is symmetric, then $C=DP$ is similar to $D^{-1/2}CD^{1/2}=D^{1/2}PD^{1/2}$, which is positive definite.
$endgroup$
– user1551
Nov 29 '18 at 11:41
$begingroup$
Why is $D^{1/2}PD{1/2}$ positive definite if P is symmetric? @user1551
$endgroup$
– Suhan Shetty
Nov 29 '18 at 12:26
$begingroup$
@SuhanShetty In your question, $P$ is assumed to be a real matrix with a positive spectrum. If it is also symmetric, it is positive definite. Hence $D^{1/2}PD^{1/2}$ (which is congruent to $P$) is positive definite too.
$endgroup$
– user1551
Nov 29 '18 at 12:42
add a comment |
$begingroup$
If the matrix $P$ is symmetric (in addition to the above conditions), I suspect that $C$ can be shown to have all positive eigenvalues. But I'm not 100% sure.
$endgroup$
– Michael Seifert
Nov 28 '18 at 14:01
$begingroup$
@MichaelSeifert Indeed. If $P$ is symmetric, then $C=DP$ is similar to $D^{-1/2}CD^{1/2}=D^{1/2}PD^{1/2}$, which is positive definite.
$endgroup$
– user1551
Nov 29 '18 at 11:41
$begingroup$
Why is $D^{1/2}PD{1/2}$ positive definite if P is symmetric? @user1551
$endgroup$
– Suhan Shetty
Nov 29 '18 at 12:26
$begingroup$
@SuhanShetty In your question, $P$ is assumed to be a real matrix with a positive spectrum. If it is also symmetric, it is positive definite. Hence $D^{1/2}PD^{1/2}$ (which is congruent to $P$) is positive definite too.
$endgroup$
– user1551
Nov 29 '18 at 12:42
$begingroup$
If the matrix $P$ is symmetric (in addition to the above conditions), I suspect that $C$ can be shown to have all positive eigenvalues. But I'm not 100% sure.
$endgroup$
– Michael Seifert
Nov 28 '18 at 14:01
$begingroup$
If the matrix $P$ is symmetric (in addition to the above conditions), I suspect that $C$ can be shown to have all positive eigenvalues. But I'm not 100% sure.
$endgroup$
– Michael Seifert
Nov 28 '18 at 14:01
$begingroup$
@MichaelSeifert Indeed. If $P$ is symmetric, then $C=DP$ is similar to $D^{-1/2}CD^{1/2}=D^{1/2}PD^{1/2}$, which is positive definite.
$endgroup$
– user1551
Nov 29 '18 at 11:41
$begingroup$
@MichaelSeifert Indeed. If $P$ is symmetric, then $C=DP$ is similar to $D^{-1/2}CD^{1/2}=D^{1/2}PD^{1/2}$, which is positive definite.
$endgroup$
– user1551
Nov 29 '18 at 11:41
$begingroup$
Why is $D^{1/2}PD{1/2}$ positive definite if P is symmetric? @user1551
$endgroup$
– Suhan Shetty
Nov 29 '18 at 12:26
$begingroup$
Why is $D^{1/2}PD{1/2}$ positive definite if P is symmetric? @user1551
$endgroup$
– Suhan Shetty
Nov 29 '18 at 12:26
$begingroup$
@SuhanShetty In your question, $P$ is assumed to be a real matrix with a positive spectrum. If it is also symmetric, it is positive definite. Hence $D^{1/2}PD^{1/2}$ (which is congruent to $P$) is positive definite too.
$endgroup$
– user1551
Nov 29 '18 at 12:42
$begingroup$
@SuhanShetty In your question, $P$ is assumed to be a real matrix with a positive spectrum. If it is also symmetric, it is positive definite. Hence $D^{1/2}PD^{1/2}$ (which is congruent to $P$) is positive definite too.
$endgroup$
– user1551
Nov 29 '18 at 12:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No. Random counterexample:
$$
P=pmatrix{3&2&5\ 4&5&1\ 3&3&4}, D=operatorname{diag}(8,1,7).
$$
The eigenvalues of $P$ are $10,1,1$ but the eigenvalues of $DP$ are $57.05$ and $-0.025pm3.13i$.
answered Nov 28 '18 at 13:58
user1551user1551
72k566126
$endgroup$
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%2f3017118%2feigen-values-of-a-product-of-two-matrices%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. Random counterexample:
$$
P=pmatrix{3&2&5\ 4&5&1\ 3&3&4}, D=operatorname{diag}(8,1,7).
$$
The eigenvalues of $P$ are $10,1,1$ but the eigenvalues of $DP$ are $57.05$ and $-0.025pm3.13i$.
answered Nov 28 '18 at 13:58
user1551user1551
72k566126
$endgroup$
add a comment |
$begingroup$
No. Random counterexample:
$$
P=pmatrix{3&2&5\ 4&5&1\ 3&3&4}, D=operatorname{diag}(8,1,7).
$$
The eigenvalues of $P$ are $10,1,1$ but the eigenvalues of $DP$ are $57.05$ and $-0.025pm3.13i$.
answered Nov 28 '18 at 13:58
user1551user1551
72k566126
$endgroup$
add a comment |
$begingroup$
No. Random counterexample:
$$
P=pmatrix{3&2&5\ 4&5&1\ 3&3&4}, D=operatorname{diag}(8,1,7).
$$
The eigenvalues of $P$ are $10,1,1$ but the eigenvalues of $DP$ are $57.05$ and $-0.025pm3.13i$.
answered Nov 28 '18 at 13:58
user1551user1551
72k566126
$endgroup$
No. Random counterexample:
$$
P=pmatrix{3&2&5\ 4&5&1\ 3&3&4}, D=operatorname{diag}(8,1,7).
$$
The eigenvalues of $P$ are $10,1,1$ but the eigenvalues of $DP$ are $57.05$ and $-0.025pm3.13i$.
answered Nov 28 '18 at 13:58
user1551user1551
72k566126
answered Nov 28 '18 at 13:58
user1551user1551
72k566126
answered Nov 28 '18 at 13:58
user1551user1551
72k566126
answered Nov 28 '18 at 13:58
user1551user1551
72k566126
72k566126
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%2f3017118%2feigen-values-of-a-product-of-two-matrices%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
$begingroup$
If the matrix $P$ is symmetric (in addition to the above conditions), I suspect that $C$ can be shown to have all positive eigenvalues. But I'm not 100% sure.
$endgroup$
– Michael Seifert
Nov 28 '18 at 14:01
$begingroup$
@MichaelSeifert Indeed. If $P$ is symmetric, then $C=DP$ is similar to $D^{-1/2}CD^{1/2}=D^{1/2}PD^{1/2}$, which is positive definite.
$endgroup$
– user1551
Nov 29 '18 at 11:41
$begingroup$
Why is $D^{1/2}PD{1/2}$ positive definite if P is symmetric? @user1551
$endgroup$
– Suhan Shetty
Nov 29 '18 at 12:26
$begingroup$
@SuhanShetty In your question, $P$ is assumed to be a real matrix with a positive spectrum. If it is also symmetric, it is positive definite. Hence $D^{1/2}PD^{1/2}$ (which is congruent to $P$) is positive definite too.
$endgroup$
– user1551
Nov 29 '18 at 12:42