Matrix proof : Show that if I − ED is nonsingular, then I − DE is nonsingular
up vote
0
down vote
favorite
I is the identity matrix.
Let D ∈ $R^{ntimes m}$ and E ∈ $R^{mtimes n}$
Show that if I − ED is nonsingular, then I − DE is nonsingular
and
$(I − DE)^{−1} = I + D(I − ED)^{−1}E$
i don't get how to this first of all i don't get how i could show the first proof. Because DE and ED are different dimensions i don't get how to you could ever compare them.
Second i dont get how to show that the equation is true because the I-ED is inbetween the E and D and with matrices you can't just multiple in any order i thought so i don't get how to get rid of it.
linear-algebra matrices
asked Nov 22 at 15:06
Danielvanheuven
457
add a comment |
up vote
0
down vote
favorite
I is the identity matrix.
Let D ∈ $R^{ntimes m}$ and E ∈ $R^{mtimes n}$
Show that if I − ED is nonsingular, then I − DE is nonsingular
and
$(I − DE)^{−1} = I + D(I − ED)^{−1}E$
i don't get how to this first of all i don't get how i could show the first proof. Because DE and ED are different dimensions i don't get how to you could ever compare them.
Second i dont get how to show that the equation is true because the I-ED is inbetween the E and D and with matrices you can't just multiple in any order i thought so i don't get how to get rid of it.
linear-algebra matrices
asked Nov 22 at 15:06
Danielvanheuven
457
This is a consequence of $det(I-DE)=det(I-ED)$ which is known as Sylvester's identity, math.stackexchange.com/q/17831
– Jean Marie
Nov 23 at 15:14
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I is the identity matrix.
Let D ∈ $R^{ntimes m}$ and E ∈ $R^{mtimes n}$
Show that if I − ED is nonsingular, then I − DE is nonsingular
and
$(I − DE)^{−1} = I + D(I − ED)^{−1}E$
i don't get how to this first of all i don't get how i could show the first proof. Because DE and ED are different dimensions i don't get how to you could ever compare them.
Second i dont get how to show that the equation is true because the I-ED is inbetween the E and D and with matrices you can't just multiple in any order i thought so i don't get how to get rid of it.
linear-algebra matrices
asked Nov 22 at 15:06
Danielvanheuven
457
I is the identity matrix.
Let D ∈ $R^{ntimes m}$ and E ∈ $R^{mtimes n}$
Show that if I − ED is nonsingular, then I − DE is nonsingular
and
$(I − DE)^{−1} = I + D(I − ED)^{−1}E$
i don't get how to this first of all i don't get how i could show the first proof. Because DE and ED are different dimensions i don't get how to you could ever compare them.
Second i dont get how to show that the equation is true because the I-ED is inbetween the E and D and with matrices you can't just multiple in any order i thought so i don't get how to get rid of it.
linear-algebra matrices
linear-algebra matrices
asked Nov 22 at 15:06
Danielvanheuven
457
asked Nov 22 at 15:06
Danielvanheuven
457
asked Nov 22 at 15:06
Danielvanheuven
457
asked Nov 22 at 15:06
Danielvanheuven
457
asked Nov 22 at 15:06
Danielvanheuven
457
457
This is a consequence of $det(I-DE)=det(I-ED)$ which is known as Sylvester's identity, math.stackexchange.com/q/17831
– Jean Marie
Nov 23 at 15:14
add a comment |
This is a consequence of $det(I-DE)=det(I-ED)$ which is known as Sylvester's identity, math.stackexchange.com/q/17831
– Jean Marie
Nov 23 at 15:14
This is a consequence of $det(I-DE)=det(I-ED)$ which is known as Sylvester's identity, math.stackexchange.com/q/17831
– Jean Marie
Nov 23 at 15:14
This is a consequence of $det(I-DE)=det(I-ED)$ which is known as Sylvester's identity, math.stackexchange.com/q/17831
– Jean Marie
Nov 23 at 15:14
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
Compute the product of the matrix $I-DE$ and its wannabe inverse $I+D(I-ED)^{-1}E$:
$$
(I+D(I-ED)^{-1}E)(I-DE)=I-DE + D(I-ED)^{-1}E(I-DE)\=I-DE + D(I-ED)^{-1}(I-ED)E=I-DE+DE=I.
$$
The product is the identity, which proves the claim.
answered Nov 22 at 15:12
daw
24k1544
add a comment |
up vote
1
down vote
We can show that $I-DE$ is nonsingular by contradiction. If it is singular, there is a vector $vinmathbb{R}^n$ such that
$$ (I-DE) v = 0 .$$
Consider $w = E vinmathbb{R}^m$.
$$(I-ED)w = E v - EDE v = E (I-DE) v = 0$$
so $I-ED$ is also singular (contradiction). Therefore, $I-DE$ is not singular.
The form of the inverse is verified with simple matrix algebra, see daw's answer.
answered Nov 22 at 15:24
fqq
29717
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',
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%2f3009254%2fmatrix-proof-show-that-if-i-%25e2%2588%2592-ed-is-nonsingular-then-i-%25e2%2588%2592-de-is-nonsingular%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
up vote
2
down vote
Compute the product of the matrix $I-DE$ and its wannabe inverse $I+D(I-ED)^{-1}E$:
$$
(I+D(I-ED)^{-1}E)(I-DE)=I-DE + D(I-ED)^{-1}E(I-DE)\=I-DE + D(I-ED)^{-1}(I-ED)E=I-DE+DE=I.
$$
The product is the identity, which proves the claim.
answered Nov 22 at 15:12
daw
24k1544
add a comment |
up vote
2
down vote
Compute the product of the matrix $I-DE$ and its wannabe inverse $I+D(I-ED)^{-1}E$:
$$
(I+D(I-ED)^{-1}E)(I-DE)=I-DE + D(I-ED)^{-1}E(I-DE)\=I-DE + D(I-ED)^{-1}(I-ED)E=I-DE+DE=I.
$$
The product is the identity, which proves the claim.
answered Nov 22 at 15:12
daw
24k1544
add a comment |
up vote
2
down vote
up vote
2
down vote
Compute the product of the matrix $I-DE$ and its wannabe inverse $I+D(I-ED)^{-1}E$:
$$
(I+D(I-ED)^{-1}E)(I-DE)=I-DE + D(I-ED)^{-1}E(I-DE)\=I-DE + D(I-ED)^{-1}(I-ED)E=I-DE+DE=I.
$$
The product is the identity, which proves the claim.
answered Nov 22 at 15:12
daw
24k1544
Compute the product of the matrix $I-DE$ and its wannabe inverse $I+D(I-ED)^{-1}E$:
$$
(I+D(I-ED)^{-1}E)(I-DE)=I-DE + D(I-ED)^{-1}E(I-DE)\=I-DE + D(I-ED)^{-1}(I-ED)E=I-DE+DE=I.
$$
The product is the identity, which proves the claim.
answered Nov 22 at 15:12
daw
24k1544
answered Nov 22 at 15:12
daw
24k1544
answered Nov 22 at 15:12
daw
24k1544
answered Nov 22 at 15:12
daw
24k1544
24k1544
add a comment |
add a comment |
up vote
1
down vote
We can show that $I-DE$ is nonsingular by contradiction. If it is singular, there is a vector $vinmathbb{R}^n$ such that
$$ (I-DE) v = 0 .$$
Consider $w = E vinmathbb{R}^m$.
$$(I-ED)w = E v - EDE v = E (I-DE) v = 0$$
so $I-ED$ is also singular (contradiction). Therefore, $I-DE$ is not singular.
The form of the inverse is verified with simple matrix algebra, see daw's answer.
answered Nov 22 at 15:24
fqq
29717
add a comment |
up vote
1
down vote
We can show that $I-DE$ is nonsingular by contradiction. If it is singular, there is a vector $vinmathbb{R}^n$ such that
$$ (I-DE) v = 0 .$$
Consider $w = E vinmathbb{R}^m$.
$$(I-ED)w = E v - EDE v = E (I-DE) v = 0$$
so $I-ED$ is also singular (contradiction). Therefore, $I-DE$ is not singular.
The form of the inverse is verified with simple matrix algebra, see daw's answer.
answered Nov 22 at 15:24
fqq
29717
add a comment |
up vote
1
down vote
up vote
1
down vote
We can show that $I-DE$ is nonsingular by contradiction. If it is singular, there is a vector $vinmathbb{R}^n$ such that
$$ (I-DE) v = 0 .$$
Consider $w = E vinmathbb{R}^m$.
$$(I-ED)w = E v - EDE v = E (I-DE) v = 0$$
so $I-ED$ is also singular (contradiction). Therefore, $I-DE$ is not singular.
The form of the inverse is verified with simple matrix algebra, see daw's answer.
answered Nov 22 at 15:24
fqq
29717
We can show that $I-DE$ is nonsingular by contradiction. If it is singular, there is a vector $vinmathbb{R}^n$ such that
$$ (I-DE) v = 0 .$$
Consider $w = E vinmathbb{R}^m$.
$$(I-ED)w = E v - EDE v = E (I-DE) v = 0$$
so $I-ED$ is also singular (contradiction). Therefore, $I-DE$ is not singular.
The form of the inverse is verified with simple matrix algebra, see daw's answer.
answered Nov 22 at 15:24
fqq
29717
answered Nov 22 at 15:24
fqq
29717
answered Nov 22 at 15:24
fqq
29717
answered Nov 22 at 15:24
fqq
29717
29717
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3009254%2fmatrix-proof-show-that-if-i-%25e2%2588%2592-ed-is-nonsingular-then-i-%25e2%2588%2592-de-is-nonsingular%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
This is a consequence of $det(I-DE)=det(I-ED)$ which is known as Sylvester's identity, math.stackexchange.com/q/17831
– Jean Marie
Nov 23 at 15:14