Mixing
invertible matrix and determinant
$begingroup$
Assume $(A + I_n)^m = 0$. Prove that $A$ is invertible and find $det(A)$.
I started by binomial expansion, and set it equal zero. is that correct? what would be the best approach?
characteristic-functions minimal-polynomials determinant-functions
edited Dec 3 '18 at 21:02
Bernard
119k740113
asked Oct 5 '18 at 2:14
user590734user590734
61
$endgroup$
add a comment |
$begingroup$
Assume $(A + I_n)^m = 0$. Prove that $A$ is invertible and find $det(A)$.
I started by binomial expansion, and set it equal zero. is that correct? what would be the best approach?
characteristic-functions minimal-polynomials determinant-functions
edited Dec 3 '18 at 21:02
Bernard
119k740113
asked Oct 5 '18 at 2:14
user590734user590734
61
$endgroup$
1
$begingroup$
Yes. The binomial expansion should give you an explicit formula for the inverse of A, thus proving invertability. For the determinant here is a hint: it is equal to the constant term of the characteristic polynomial of A.
$endgroup$
– Simon
Oct 5 '18 at 8:02
$begingroup$
Certainly not true when $n=0$.
$endgroup$
– Marc van Leeuwen
Oct 7 '18 at 8:43
$begingroup$
Um, I read $In$ as the scalar multiple more commonly written $nI$. You probably mean $I_n$ instead; try to learn the MathJax formatting.
$endgroup$
– Marc van Leeuwen
Oct 7 '18 at 10:40
$begingroup$
Write $A + I = N$, so that $N^m = 0$, hence $N$ is nilpotent. Then $A = N - I$ is the difference of a nilpotent and invertible matrix, and it's possible to write down very explicitly a formula for the inverse of $A$ (hint: $I - N^m = I$).
$endgroup$
– Qiaochu Yuan
Dec 3 '18 at 21:57
add a comment |
$begingroup$
Assume $(A + I_n)^m = 0$. Prove that $A$ is invertible and find $det(A)$.
I started by binomial expansion, and set it equal zero. is that correct? what would be the best approach?
characteristic-functions minimal-polynomials determinant-functions
edited Dec 3 '18 at 21:02
Bernard
119k740113
asked Oct 5 '18 at 2:14
user590734user590734
61
$endgroup$
Assume $(A + I_n)^m = 0$. Prove that $A$ is invertible and find $det(A)$.
I started by binomial expansion, and set it equal zero. is that correct? what would be the best approach?
characteristic-functions minimal-polynomials determinant-functions
characteristic-functions minimal-polynomials determinant-functions
edited Dec 3 '18 at 21:02
Bernard
119k740113
asked Oct 5 '18 at 2:14
user590734user590734
61
edited Dec 3 '18 at 21:02
Bernard
119k740113
asked Oct 5 '18 at 2:14
user590734user590734
61
edited Dec 3 '18 at 21:02
Bernard
119k740113
edited Dec 3 '18 at 21:02
Bernard
119k740113
edited Dec 3 '18 at 21:02
Bernard
119k740113
119k740113
asked Oct 5 '18 at 2:14
user590734user590734
61
asked Oct 5 '18 at 2:14
user590734user590734
61
asked Oct 5 '18 at 2:14
user590734user590734
61
61
1
$begingroup$
Yes. The binomial expansion should give you an explicit formula for the inverse of A, thus proving invertability. For the determinant here is a hint: it is equal to the constant term of the characteristic polynomial of A.
$endgroup$
– Simon
Oct 5 '18 at 8:02
$begingroup$
Certainly not true when $n=0$.
$endgroup$
– Marc van Leeuwen
Oct 7 '18 at 8:43
$begingroup$
Um, I read $In$ as the scalar multiple more commonly written $nI$. You probably mean $I_n$ instead; try to learn the MathJax formatting.
$endgroup$
– Marc van Leeuwen
Oct 7 '18 at 10:40
$begingroup$
Write $A + I = N$, so that $N^m = 0$, hence $N$ is nilpotent. Then $A = N - I$ is the difference of a nilpotent and invertible matrix, and it's possible to write down very explicitly a formula for the inverse of $A$ (hint: $I - N^m = I$).
$endgroup$
– Qiaochu Yuan
Dec 3 '18 at 21:57
add a comment |
1
$begingroup$
Yes. The binomial expansion should give you an explicit formula for the inverse of A, thus proving invertability. For the determinant here is a hint: it is equal to the constant term of the characteristic polynomial of A.
$endgroup$
– Simon
Oct 5 '18 at 8:02
$begingroup$
Certainly not true when $n=0$.
$endgroup$
– Marc van Leeuwen
Oct 7 '18 at 8:43
$begingroup$
Um, I read $In$ as the scalar multiple more commonly written $nI$. You probably mean $I_n$ instead; try to learn the MathJax formatting.
$endgroup$
– Marc van Leeuwen
Oct 7 '18 at 10:40
$begingroup$
Write $A + I = N$, so that $N^m = 0$, hence $N$ is nilpotent. Then $A = N - I$ is the difference of a nilpotent and invertible matrix, and it's possible to write down very explicitly a formula for the inverse of $A$ (hint: $I - N^m = I$).
$endgroup$
– Qiaochu Yuan
Dec 3 '18 at 21:57
1
1
$begingroup$
Yes. The binomial expansion should give you an explicit formula for the inverse of A, thus proving invertability. For the determinant here is a hint: it is equal to the constant term of the characteristic polynomial of A.
$endgroup$
– Simon
Oct 5 '18 at 8:02
$begingroup$
Yes. The binomial expansion should give you an explicit formula for the inverse of A, thus proving invertability. For the determinant here is a hint: it is equal to the constant term of the characteristic polynomial of A.
$endgroup$
– Simon
Oct 5 '18 at 8:02
$begingroup$
Certainly not true when $n=0$.
$endgroup$
– Marc van Leeuwen
Oct 7 '18 at 8:43
$begingroup$
Certainly not true when $n=0$.
$endgroup$
– Marc van Leeuwen
Oct 7 '18 at 8:43
$begingroup$
Um, I read $In$ as the scalar multiple more commonly written $nI$. You probably mean $I_n$ instead; try to learn the MathJax formatting.
$endgroup$
– Marc van Leeuwen
Oct 7 '18 at 10:40
$begingroup$
Um, I read $In$ as the scalar multiple more commonly written $nI$. You probably mean $I_n$ instead; try to learn the MathJax formatting.
$endgroup$
– Marc van Leeuwen
Oct 7 '18 at 10:40
$begingroup$
Write $A + I = N$, so that $N^m = 0$, hence $N$ is nilpotent. Then $A = N - I$ is the difference of a nilpotent and invertible matrix, and it's possible to write down very explicitly a formula for the inverse of $A$ (hint: $I - N^m = I$).
$endgroup$
– Qiaochu Yuan
Dec 3 '18 at 21:57
$begingroup$
Write $A + I = N$, so that $N^m = 0$, hence $N$ is nilpotent. Then $A = N - I$ is the difference of a nilpotent and invertible matrix, and it's possible to write down very explicitly a formula for the inverse of $A$ (hint: $I - N^m = I$).
$endgroup$
– Qiaochu Yuan
Dec 3 '18 at 21:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The equation says that $A+I_n$ is nilpotent, so it has characteristic polynomial $(X+1)^n$. Setting $X=0$ in it gives $det(-A)$, which apparently is $1$, so $det(A)=(-1)^n$ (alternatively, $det(A)$ is the product of all eigenvalues (counted with their algebraic multiplicity) and here they all are$~{-}1$).
answered Oct 7 '18 at 10:46
Marc van LeeuwenMarc van Leeuwen
86.6k5107222
$endgroup$
add a comment |
$begingroup$
Indeed, we can use the binomial expansion (since $A$ commutes with $I_n$) and we get
$$
0=sum_{k=0}^mbinom mk A^kI_n^{m-k}=I_n+sum_{k=1}^mbinom mk A^k
=I_n+Asum_{k=1}^mbinom mk A^{k-1}
$$
hence we got the equality
$$
left(-sum_{k=1}^mbinom mk A^{k-1}right)cdot A=I_n,
$$
which shows that $A$ is invertible, and its inverse is $-sum_{k=1}^mbinom mk A^{k-1}$.
answered Jan 4 at 15:00
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
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%2f2942884%2finvertible-matrix-and-determinant%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
$begingroup$
The equation says that $A+I_n$ is nilpotent, so it has characteristic polynomial $(X+1)^n$. Setting $X=0$ in it gives $det(-A)$, which apparently is $1$, so $det(A)=(-1)^n$ (alternatively, $det(A)$ is the product of all eigenvalues (counted with their algebraic multiplicity) and here they all are$~{-}1$).
answered Oct 7 '18 at 10:46
Marc van LeeuwenMarc van Leeuwen
86.6k5107222
$endgroup$
add a comment |
$begingroup$
The equation says that $A+I_n$ is nilpotent, so it has characteristic polynomial $(X+1)^n$. Setting $X=0$ in it gives $det(-A)$, which apparently is $1$, so $det(A)=(-1)^n$ (alternatively, $det(A)$ is the product of all eigenvalues (counted with their algebraic multiplicity) and here they all are$~{-}1$).
answered Oct 7 '18 at 10:46
Marc van LeeuwenMarc van Leeuwen
86.6k5107222
$endgroup$
add a comment |
$begingroup$
The equation says that $A+I_n$ is nilpotent, so it has characteristic polynomial $(X+1)^n$. Setting $X=0$ in it gives $det(-A)$, which apparently is $1$, so $det(A)=(-1)^n$ (alternatively, $det(A)$ is the product of all eigenvalues (counted with their algebraic multiplicity) and here they all are$~{-}1$).
answered Oct 7 '18 at 10:46
Marc van LeeuwenMarc van Leeuwen
86.6k5107222
$endgroup$
The equation says that $A+I_n$ is nilpotent, so it has characteristic polynomial $(X+1)^n$. Setting $X=0$ in it gives $det(-A)$, which apparently is $1$, so $det(A)=(-1)^n$ (alternatively, $det(A)$ is the product of all eigenvalues (counted with their algebraic multiplicity) and here they all are$~{-}1$).
answered Oct 7 '18 at 10:46
Marc van LeeuwenMarc van Leeuwen
86.6k5107222
edited Dec 3 '18 at 22:10
answered Oct 7 '18 at 10:46
Marc van LeeuwenMarc van Leeuwen
86.6k5107222
answered Oct 7 '18 at 10:46
Marc van LeeuwenMarc van Leeuwen
86.6k5107222
answered Oct 7 '18 at 10:46
Marc van LeeuwenMarc van Leeuwen
86.6k5107222
86.6k5107222
add a comment |
add a comment |
$begingroup$
Indeed, we can use the binomial expansion (since $A$ commutes with $I_n$) and we get
$$
0=sum_{k=0}^mbinom mk A^kI_n^{m-k}=I_n+sum_{k=1}^mbinom mk A^k
=I_n+Asum_{k=1}^mbinom mk A^{k-1}
$$
hence we got the equality
$$
left(-sum_{k=1}^mbinom mk A^{k-1}right)cdot A=I_n,
$$
which shows that $A$ is invertible, and its inverse is $-sum_{k=1}^mbinom mk A^{k-1}$.
answered Jan 4 at 15:00
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
add a comment |
$begingroup$
Indeed, we can use the binomial expansion (since $A$ commutes with $I_n$) and we get
$$
0=sum_{k=0}^mbinom mk A^kI_n^{m-k}=I_n+sum_{k=1}^mbinom mk A^k
=I_n+Asum_{k=1}^mbinom mk A^{k-1}
$$
hence we got the equality
$$
left(-sum_{k=1}^mbinom mk A^{k-1}right)cdot A=I_n,
$$
which shows that $A$ is invertible, and its inverse is $-sum_{k=1}^mbinom mk A^{k-1}$.
answered Jan 4 at 15:00
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
add a comment |
$begingroup$
Indeed, we can use the binomial expansion (since $A$ commutes with $I_n$) and we get
$$
0=sum_{k=0}^mbinom mk A^kI_n^{m-k}=I_n+sum_{k=1}^mbinom mk A^k
=I_n+Asum_{k=1}^mbinom mk A^{k-1}
$$
hence we got the equality
$$
left(-sum_{k=1}^mbinom mk A^{k-1}right)cdot A=I_n,
$$
which shows that $A$ is invertible, and its inverse is $-sum_{k=1}^mbinom mk A^{k-1}$.
answered Jan 4 at 15:00
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
Indeed, we can use the binomial expansion (since $A$ commutes with $I_n$) and we get
$$
0=sum_{k=0}^mbinom mk A^kI_n^{m-k}=I_n+sum_{k=1}^mbinom mk A^k
=I_n+Asum_{k=1}^mbinom mk A^{k-1}
$$
hence we got the equality
$$
left(-sum_{k=1}^mbinom mk A^{k-1}right)cdot A=I_n,
$$
which shows that $A$ is invertible, and its inverse is $-sum_{k=1}^mbinom mk A^{k-1}$.
answered Jan 4 at 15:00
Davide GiraudoDavide Giraudo
125k16150261
answered Jan 4 at 15:00
Davide GiraudoDavide Giraudo
125k16150261
answered Jan 4 at 15:00
Davide GiraudoDavide Giraudo
125k16150261
answered Jan 4 at 15:00
Davide GiraudoDavide Giraudo
125k16150261
125k16150261
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%2f2942884%2finvertible-matrix-and-determinant%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
1
$begingroup$
Yes. The binomial expansion should give you an explicit formula for the inverse of A, thus proving invertability. For the determinant here is a hint: it is equal to the constant term of the characteristic polynomial of A.
$endgroup$
– Simon
Oct 5 '18 at 8:02
$begingroup$
Certainly not true when $n=0$.
$endgroup$
– Marc van Leeuwen
Oct 7 '18 at 8:43
$begingroup$
Um, I read $In$ as the scalar multiple more commonly written $nI$. You probably mean $I_n$ instead; try to learn the MathJax formatting.
$endgroup$
– Marc van Leeuwen
Oct 7 '18 at 10:40
$begingroup$
Write $A + I = N$, so that $N^m = 0$, hence $N$ is nilpotent. Then $A = N - I$ is the difference of a nilpotent and invertible matrix, and it's possible to write down very explicitly a formula for the inverse of $A$ (hint: $I - N^m = I$).
$endgroup$
– Qiaochu Yuan
Dec 3 '18 at 21:57