Prove or disprove: if $A$ is nonzero $2 times 2$ matrix such that $A^2+A=0$, then A is invertible
$begingroup$
if $A$ is nonzero $2 times 2$ matrix such that $A^2+A=0$, then A is invertible
I really can't figure it out. I know it's true but don't know how to prove it
linear-algebra matrices
asked Nov 11 '13 at 4:10
12321232
111
$endgroup$
add a comment |
$begingroup$
if $A$ is nonzero $2 times 2$ matrix such that $A^2+A=0$, then A is invertible
I really can't figure it out. I know it's true but don't know how to prove it
linear-algebra matrices
asked Nov 11 '13 at 4:10
12321232
111
$endgroup$
$begingroup$
Prove is the verb, proof is the noun.
$endgroup$
– Pedro Tamaroff♦
Nov 11 '13 at 4:17
$begingroup$
I changed the quadratic equation, didn't check the grammar, but you are right though.
$endgroup$
– imranfat
Nov 11 '13 at 4:20
1
$begingroup$
"I know it's true..." how do you know this? You may want to be less certain of that.
$endgroup$
– alex.jordan
Nov 11 '13 at 8:20
add a comment |
$begingroup$
if $A$ is nonzero $2 times 2$ matrix such that $A^2+A=0$, then A is invertible
I really can't figure it out. I know it's true but don't know how to prove it
linear-algebra matrices
asked Nov 11 '13 at 4:10
12321232
111
$endgroup$
if $A$ is nonzero $2 times 2$ matrix such that $A^2+A=0$, then A is invertible
I really can't figure it out. I know it's true but don't know how to prove it
linear-algebra matrices
linear-algebra matrices
asked Nov 11 '13 at 4:10
12321232
111
asked Nov 11 '13 at 4:10
12321232
111
edited Nov 11 '13 at 4:22
user61527
asked Nov 11 '13 at 4:10
12321232
111
asked Nov 11 '13 at 4:10
12321232
111
asked Nov 11 '13 at 4:10
12321232
111
111
$begingroup$
Prove is the verb, proof is the noun.
$endgroup$
– Pedro Tamaroff♦
Nov 11 '13 at 4:17
$begingroup$
I changed the quadratic equation, didn't check the grammar, but you are right though.
$endgroup$
– imranfat
Nov 11 '13 at 4:20
1
$begingroup$
"I know it's true..." how do you know this? You may want to be less certain of that.
$endgroup$
– alex.jordan
Nov 11 '13 at 8:20
add a comment |
$begingroup$
Prove is the verb, proof is the noun.
$endgroup$
– Pedro Tamaroff♦
Nov 11 '13 at 4:17
$begingroup$
I changed the quadratic equation, didn't check the grammar, but you are right though.
$endgroup$
– imranfat
Nov 11 '13 at 4:20
1
$begingroup$
"I know it's true..." how do you know this? You may want to be less certain of that.
$endgroup$
– alex.jordan
Nov 11 '13 at 8:20
$begingroup$
Prove is the verb, proof is the noun.
$endgroup$
– Pedro Tamaroff♦
Nov 11 '13 at 4:17
$begingroup$
Prove is the verb, proof is the noun.
$endgroup$
– Pedro Tamaroff♦
Nov 11 '13 at 4:17
$begingroup$
I changed the quadratic equation, didn't check the grammar, but you are right though.
$endgroup$
– imranfat
Nov 11 '13 at 4:20
$begingroup$
I changed the quadratic equation, didn't check the grammar, but you are right though.
$endgroup$
– imranfat
Nov 11 '13 at 4:20
1
1
$begingroup$
"I know it's true..." how do you know this? You may want to be less certain of that.
$endgroup$
– alex.jordan
Nov 11 '13 at 8:20
$begingroup$
"I know it's true..." how do you know this? You may want to be less certain of that.
$endgroup$
– alex.jordan
Nov 11 '13 at 8:20
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
If $A$ had an inverse, then
$$I = A A^{-1} = -A^2 A^{-1} = -A^{-1}$$
So we conclude that $A^{-1} = -I$, implying that $A = -I$. Can you think of a matrix $A$ satisfying the given condition that is not $-I$?
$endgroup$
$begingroup$
yea just the inverse of the identity, can you clarify a little more plz
$endgroup$
– 1232
Nov 11 '13 at 4:19
$begingroup$
@1232 The problem is reduced to asking whether there is a matrix not equal to $-I$ such that $A^2 + A = 0$. Can you find such a matrix?
$endgroup$
– user61527
Nov 11 '13 at 4:21
$begingroup$
no i cannot find one
$endgroup$
– 1232
Nov 11 '13 at 4:26
4
$begingroup$
@1232 Look harder.
$endgroup$
– Will Nelson
Nov 11 '13 at 7:51
add a comment |
$begingroup$
Try a diagonal matrix $A$ that's not invertible and not zero, but for which $A^2+A=0$. It might be very easy to find one.
answered Nov 11 '13 at 7:05
Will NelsonWill Nelson
4,4541023
$endgroup$
add a comment |
$begingroup$
This is not true. For example
$$
A=begin{pmatrix}
-1&0\
0&0
end{pmatrix}
$$
answered Jan 11 '17 at 16:05
boazboaz
2,405614
$endgroup$
add a comment |
$begingroup$
The answer becomes evident, when you view it in terms of the minimal polynomial.
Now given the above equation A^2 + A = 0, this can also be written as A(A+I) = 0
Since A is not 0,
A either has A+I = 0, i.e, A = -I. (when A+I is the minimal polynomial of A)
or has A(A+I) = 0 as its minimal polynomial, in this case the eigenvalues of A are 0 and 1, which implies the determinant of A = 0.
Hence its not invertible.
So either A = -I or A is not invertible.
answered Dec 8 '18 at 19:37
manav gaddammanav gaddam
11
$endgroup$
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%2f562007%2fprove-or-disprove-if-a-is-nonzero-2-times-2-matrix-such-that-a2a-0-th%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $A$ had an inverse, then
$$I = A A^{-1} = -A^2 A^{-1} = -A^{-1}$$
So we conclude that $A^{-1} = -I$, implying that $A = -I$. Can you think of a matrix $A$ satisfying the given condition that is not $-I$?
$endgroup$
$begingroup$
yea just the inverse of the identity, can you clarify a little more plz
$endgroup$
– 1232
Nov 11 '13 at 4:19
$begingroup$
@1232 The problem is reduced to asking whether there is a matrix not equal to $-I$ such that $A^2 + A = 0$. Can you find such a matrix?
$endgroup$
– user61527
Nov 11 '13 at 4:21
$begingroup$
no i cannot find one
$endgroup$
– 1232
Nov 11 '13 at 4:26
4
$begingroup$
@1232 Look harder.
$endgroup$
– Will Nelson
Nov 11 '13 at 7:51
add a comment |
$begingroup$
If $A$ had an inverse, then
$$I = A A^{-1} = -A^2 A^{-1} = -A^{-1}$$
So we conclude that $A^{-1} = -I$, implying that $A = -I$. Can you think of a matrix $A$ satisfying the given condition that is not $-I$?
$endgroup$
$begingroup$
yea just the inverse of the identity, can you clarify a little more plz
$endgroup$
– 1232
Nov 11 '13 at 4:19
$begingroup$
@1232 The problem is reduced to asking whether there is a matrix not equal to $-I$ such that $A^2 + A = 0$. Can you find such a matrix?
$endgroup$
– user61527
Nov 11 '13 at 4:21
$begingroup$
no i cannot find one
$endgroup$
– 1232
Nov 11 '13 at 4:26
4
$begingroup$
@1232 Look harder.
$endgroup$
– Will Nelson
Nov 11 '13 at 7:51
add a comment |
$begingroup$
If $A$ had an inverse, then
$$I = A A^{-1} = -A^2 A^{-1} = -A^{-1}$$
So we conclude that $A^{-1} = -I$, implying that $A = -I$. Can you think of a matrix $A$ satisfying the given condition that is not $-I$?
$endgroup$
If $A$ had an inverse, then
$$I = A A^{-1} = -A^2 A^{-1} = -A^{-1}$$
So we conclude that $A^{-1} = -I$, implying that $A = -I$. Can you think of a matrix $A$ satisfying the given condition that is not $-I$?
answered Nov 11 '13 at 4:13
user61527
$begingroup$
yea just the inverse of the identity, can you clarify a little more plz
$endgroup$
– 1232
Nov 11 '13 at 4:19
$begingroup$
@1232 The problem is reduced to asking whether there is a matrix not equal to $-I$ such that $A^2 + A = 0$. Can you find such a matrix?
$endgroup$
– user61527
Nov 11 '13 at 4:21
$begingroup$
no i cannot find one
$endgroup$
– 1232
Nov 11 '13 at 4:26
4
$begingroup$
@1232 Look harder.
$endgroup$
– Will Nelson
Nov 11 '13 at 7:51
add a comment |
$begingroup$
yea just the inverse of the identity, can you clarify a little more plz
$endgroup$
– 1232
Nov 11 '13 at 4:19
$begingroup$
@1232 The problem is reduced to asking whether there is a matrix not equal to $-I$ such that $A^2 + A = 0$. Can you find such a matrix?
$endgroup$
– user61527
Nov 11 '13 at 4:21
$begingroup$
no i cannot find one
$endgroup$
– 1232
Nov 11 '13 at 4:26
4
$begingroup$
@1232 Look harder.
$endgroup$
– Will Nelson
Nov 11 '13 at 7:51
$begingroup$
yea just the inverse of the identity, can you clarify a little more plz
$endgroup$
– 1232
Nov 11 '13 at 4:19
$begingroup$
yea just the inverse of the identity, can you clarify a little more plz
$endgroup$
– 1232
Nov 11 '13 at 4:19
$begingroup$
@1232 The problem is reduced to asking whether there is a matrix not equal to $-I$ such that $A^2 + A = 0$. Can you find such a matrix?
$endgroup$
– user61527
Nov 11 '13 at 4:21
$begingroup$
@1232 The problem is reduced to asking whether there is a matrix not equal to $-I$ such that $A^2 + A = 0$. Can you find such a matrix?
$endgroup$
– user61527
Nov 11 '13 at 4:21
$begingroup$
no i cannot find one
$endgroup$
– 1232
Nov 11 '13 at 4:26
$begingroup$
no i cannot find one
$endgroup$
– 1232
Nov 11 '13 at 4:26
4
4
$begingroup$
@1232 Look harder.
$endgroup$
– Will Nelson
Nov 11 '13 at 7:51
$begingroup$
@1232 Look harder.
$endgroup$
– Will Nelson
Nov 11 '13 at 7:51
add a comment |
$begingroup$
Try a diagonal matrix $A$ that's not invertible and not zero, but for which $A^2+A=0$. It might be very easy to find one.
answered Nov 11 '13 at 7:05
Will NelsonWill Nelson
4,4541023
$endgroup$
add a comment |
$begingroup$
Try a diagonal matrix $A$ that's not invertible and not zero, but for which $A^2+A=0$. It might be very easy to find one.
answered Nov 11 '13 at 7:05
Will NelsonWill Nelson
4,4541023
$endgroup$
add a comment |
$begingroup$
Try a diagonal matrix $A$ that's not invertible and not zero, but for which $A^2+A=0$. It might be very easy to find one.
answered Nov 11 '13 at 7:05
Will NelsonWill Nelson
4,4541023
$endgroup$
Try a diagonal matrix $A$ that's not invertible and not zero, but for which $A^2+A=0$. It might be very easy to find one.
answered Nov 11 '13 at 7:05
Will NelsonWill Nelson
4,4541023
answered Nov 11 '13 at 7:05
Will NelsonWill Nelson
4,4541023
answered Nov 11 '13 at 7:05
Will NelsonWill Nelson
4,4541023
answered Nov 11 '13 at 7:05
Will NelsonWill Nelson
4,4541023
4,4541023
add a comment |
add a comment |
$begingroup$
This is not true. For example
$$
A=begin{pmatrix}
-1&0\
0&0
end{pmatrix}
$$
answered Jan 11 '17 at 16:05
boazboaz
2,405614
$endgroup$
add a comment |
$begingroup$
This is not true. For example
$$
A=begin{pmatrix}
-1&0\
0&0
end{pmatrix}
$$
answered Jan 11 '17 at 16:05
boazboaz
2,405614
$endgroup$
add a comment |
$begingroup$
This is not true. For example
$$
A=begin{pmatrix}
-1&0\
0&0
end{pmatrix}
$$
answered Jan 11 '17 at 16:05
boazboaz
2,405614
$endgroup$
This is not true. For example
$$
A=begin{pmatrix}
-1&0\
0&0
end{pmatrix}
$$
answered Jan 11 '17 at 16:05
boazboaz
2,405614
edited Jan 11 '17 at 16:18
answered Jan 11 '17 at 16:05
boazboaz
2,405614
answered Jan 11 '17 at 16:05
boazboaz
2,405614
answered Jan 11 '17 at 16:05
boazboaz
2,405614
2,405614
add a comment |
add a comment |
$begingroup$
The answer becomes evident, when you view it in terms of the minimal polynomial.
Now given the above equation A^2 + A = 0, this can also be written as A(A+I) = 0
Since A is not 0,
A either has A+I = 0, i.e, A = -I. (when A+I is the minimal polynomial of A)
or has A(A+I) = 0 as its minimal polynomial, in this case the eigenvalues of A are 0 and 1, which implies the determinant of A = 0.
Hence its not invertible.
So either A = -I or A is not invertible.
answered Dec 8 '18 at 19:37
manav gaddammanav gaddam
11
$endgroup$
add a comment |
$begingroup$
The answer becomes evident, when you view it in terms of the minimal polynomial.
Now given the above equation A^2 + A = 0, this can also be written as A(A+I) = 0
Since A is not 0,
A either has A+I = 0, i.e, A = -I. (when A+I is the minimal polynomial of A)
or has A(A+I) = 0 as its minimal polynomial, in this case the eigenvalues of A are 0 and 1, which implies the determinant of A = 0.
Hence its not invertible.
So either A = -I or A is not invertible.
answered Dec 8 '18 at 19:37
manav gaddammanav gaddam
11
$endgroup$
add a comment |
$begingroup$
The answer becomes evident, when you view it in terms of the minimal polynomial.
Now given the above equation A^2 + A = 0, this can also be written as A(A+I) = 0
Since A is not 0,
A either has A+I = 0, i.e, A = -I. (when A+I is the minimal polynomial of A)
or has A(A+I) = 0 as its minimal polynomial, in this case the eigenvalues of A are 0 and 1, which implies the determinant of A = 0.
Hence its not invertible.
So either A = -I or A is not invertible.
answered Dec 8 '18 at 19:37
manav gaddammanav gaddam
11
$endgroup$
The answer becomes evident, when you view it in terms of the minimal polynomial.
Now given the above equation A^2 + A = 0, this can also be written as A(A+I) = 0
Since A is not 0,
A either has A+I = 0, i.e, A = -I. (when A+I is the minimal polynomial of A)
or has A(A+I) = 0 as its minimal polynomial, in this case the eigenvalues of A are 0 and 1, which implies the determinant of A = 0.
Hence its not invertible.
So either A = -I or A is not invertible.
answered Dec 8 '18 at 19:37
manav gaddammanav gaddam
11
answered Dec 8 '18 at 19:37
manav gaddammanav gaddam
11
answered Dec 8 '18 at 19:37
manav gaddammanav gaddam
11
answered Dec 8 '18 at 19:37
manav gaddammanav gaddam
11
11
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%2f562007%2fprove-or-disprove-if-a-is-nonzero-2-times-2-matrix-such-that-a2a-0-th%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
$begingroup$
Prove is the verb, proof is the noun.
$endgroup$
– Pedro Tamaroff♦
Nov 11 '13 at 4:17
$begingroup$
I changed the quadratic equation, didn't check the grammar, but you are right though.
$endgroup$
– imranfat
Nov 11 '13 at 4:20
1
$begingroup$
"I know it's true..." how do you know this? You may want to be less certain of that.
$endgroup$
– alex.jordan
Nov 11 '13 at 8:20