Mixing
Can $A^3=0$ imply $|I+A|=0$?
$begingroup$
Suppose $A$ is a non-zero matrix such that $A^3=0$. Prove the following assertions or provide counter examples:-
$(1) A^2$ is a zero matrix
$(2) A+A^2$ can have zero trace
$(3) A-A^2$ can have zero trace
$(4) I+A$ is singular.
My Attempt:- I know if $A^3=0$ then $A^2=0$ can be true (though not always). I have no idea whether $tr(A+A^2)=0$ or $tr(A-A^2)=0$ is possible or not if $A^3=0$. But when I looked closely at $|I+A|$ then I found that
$$|I+A|=0$$
For $2times2$ matrix, we have
$$Rightarrow |A|+tr(A)+1=0 $$
$$Rightarrow lambda_1lambda_2+lambda_1+lambda_2+1=0 $$
where $lambda_1$ and $lambda_2$ are the two eigenvalues of $A$
$$Rightarrow lambda_1(lambda_2+1)+1.(lambda_2+1)=0 $$
$$Rightarrow (lambda_2+1)(lambda_1+1)=0 $$
$$Rightarrow lambda_2=-1, lambda_1=-1 tag1$$
But we have $$A^3=0$$
$$Rightarrow |A^3|=0$$
$$Rightarrow |A|^3=0$$
$$Rightarrow |A|=0$$
So,
either $lambda_1=0$ or $lambda_2=0$ (or both may be zero) which contradicts with equation $(1)$. So, $I+A$ is non singular.
Am I Correct ?
linear-algebra matrices determinant
$endgroup$
add a comment |
$begingroup$
Suppose $A$ is a non-zero matrix such that $A^3=0$. Prove the following assertions or provide counter examples:-
$(1) A^2$ is a zero matrix
$(2) A+A^2$ can have zero trace
$(3) A-A^2$ can have zero trace
$(4) I+A$ is singular.
My Attempt:- I know if $A^3=0$ then $A^2=0$ can be true (though not always). I have no idea whether $tr(A+A^2)=0$ or $tr(A-A^2)=0$ is possible or not if $A^3=0$. But when I looked closely at $|I+A|$ then I found that
$$|I+A|=0$$
For $2times2$ matrix, we have
$$Rightarrow |A|+tr(A)+1=0 $$
$$Rightarrow lambda_1lambda_2+lambda_1+lambda_2+1=0 $$
where $lambda_1$ and $lambda_2$ are the two eigenvalues of $A$
$$Rightarrow lambda_1(lambda_2+1)+1.(lambda_2+1)=0 $$
$$Rightarrow (lambda_2+1)(lambda_1+1)=0 $$
$$Rightarrow lambda_2=-1, lambda_1=-1 tag1$$
But we have $$A^3=0$$
$$Rightarrow |A^3|=0$$
$$Rightarrow |A|^3=0$$
$$Rightarrow |A|=0$$
So,
either $lambda_1=0$ or $lambda_2=0$ (or both may be zero) which contradicts with equation $(1)$. So, $I+A$ is non singular.
Am I Correct ?
linear-algebra matrices determinant
$endgroup$
$begingroup$
Hint to much easier approach for (4): Consider $(I+A)(I-A+A^2)$
$endgroup$
– JMoravitz
Mar 3 '18 at 1:12
1
$begingroup$
For (2), (3), any such $A$ you try will show these options can be ruled out. So every example is a counterexample! That is because when $A^3=0$, all eigenvalues of $A$ exist and are equal to zero. Then the same goes for all eigenvalues of the powers $A^k$ of $A$. And the trace is the sum of the eigenvalues, and the trace is linear, $mathrm{tr}(X+Y)=mathrm{tr}(X)+mathrm{tr}(Y)$.
$endgroup$
– Jeppe Stig Nielsen
Mar 3 '18 at 2:23
add a comment |
$begingroup$
Suppose $A$ is a non-zero matrix such that $A^3=0$. Prove the following assertions or provide counter examples:-
$(1) A^2$ is a zero matrix
$(2) A+A^2$ can have zero trace
$(3) A-A^2$ can have zero trace
$(4) I+A$ is singular.
My Attempt:- I know if $A^3=0$ then $A^2=0$ can be true (though not always). I have no idea whether $tr(A+A^2)=0$ or $tr(A-A^2)=0$ is possible or not if $A^3=0$. But when I looked closely at $|I+A|$ then I found that
$$|I+A|=0$$
For $2times2$ matrix, we have
$$Rightarrow |A|+tr(A)+1=0 $$
$$Rightarrow lambda_1lambda_2+lambda_1+lambda_2+1=0 $$
where $lambda_1$ and $lambda_2$ are the two eigenvalues of $A$
$$Rightarrow lambda_1(lambda_2+1)+1.(lambda_2+1)=0 $$
$$Rightarrow (lambda_2+1)(lambda_1+1)=0 $$
$$Rightarrow lambda_2=-1, lambda_1=-1 tag1$$
But we have $$A^3=0$$
$$Rightarrow |A^3|=0$$
$$Rightarrow |A|^3=0$$
$$Rightarrow |A|=0$$
So,
either $lambda_1=0$ or $lambda_2=0$ (or both may be zero) which contradicts with equation $(1)$. So, $I+A$ is non singular.
Am I Correct ?
linear-algebra matrices determinant
$endgroup$
Suppose $A$ is a non-zero matrix such that $A^3=0$. Prove the following assertions or provide counter examples:-
$(1) A^2$ is a zero matrix
$(2) A+A^2$ can have zero trace
$(3) A-A^2$ can have zero trace
$(4) I+A$ is singular.
My Attempt:- I know if $A^3=0$ then $A^2=0$ can be true (though not always). I have no idea whether $tr(A+A^2)=0$ or $tr(A-A^2)=0$ is possible or not if $A^3=0$. But when I looked closely at $|I+A|$ then I found that
$$|I+A|=0$$
For $2times2$ matrix, we have
$$Rightarrow |A|+tr(A)+1=0 $$
$$Rightarrow lambda_1lambda_2+lambda_1+lambda_2+1=0 $$
where $lambda_1$ and $lambda_2$ are the two eigenvalues of $A$
$$Rightarrow lambda_1(lambda_2+1)+1.(lambda_2+1)=0 $$
$$Rightarrow (lambda_2+1)(lambda_1+1)=0 $$
$$Rightarrow lambda_2=-1, lambda_1=-1 tag1$$
But we have $$A^3=0$$
$$Rightarrow |A^3|=0$$
$$Rightarrow |A|^3=0$$
$$Rightarrow |A|=0$$
So,
either $lambda_1=0$ or $lambda_2=0$ (or both may be zero) which contradicts with equation $(1)$. So, $I+A$ is non singular.
Am I Correct ?
linear-algebra matrices determinant
linear-algebra matrices determinant
edited Jan 31 at 4:01
asked Mar 3 '18 at 1:01
user440191
$begingroup$
Hint to much easier approach for (4): Consider $(I+A)(I-A+A^2)$
$endgroup$
– JMoravitz
Mar 3 '18 at 1:12
1
$begingroup$
For (2), (3), any such $A$ you try will show these options can be ruled out. So every example is a counterexample! That is because when $A^3=0$, all eigenvalues of $A$ exist and are equal to zero. Then the same goes for all eigenvalues of the powers $A^k$ of $A$. And the trace is the sum of the eigenvalues, and the trace is linear, $mathrm{tr}(X+Y)=mathrm{tr}(X)+mathrm{tr}(Y)$.
$endgroup$
– Jeppe Stig Nielsen
Mar 3 '18 at 2:23
add a comment |
$begingroup$
Hint to much easier approach for (4): Consider $(I+A)(I-A+A^2)$
$endgroup$
– JMoravitz
Mar 3 '18 at 1:12
1
$begingroup$
For (2), (3), any such $A$ you try will show these options can be ruled out. So every example is a counterexample! That is because when $A^3=0$, all eigenvalues of $A$ exist and are equal to zero. Then the same goes for all eigenvalues of the powers $A^k$ of $A$. And the trace is the sum of the eigenvalues, and the trace is linear, $mathrm{tr}(X+Y)=mathrm{tr}(X)+mathrm{tr}(Y)$.
$endgroup$
– Jeppe Stig Nielsen
Mar 3 '18 at 2:23
$begingroup$
Hint to much easier approach for (4): Consider $(I+A)(I-A+A^2)$
$endgroup$
– JMoravitz
Mar 3 '18 at 1:12
$begingroup$
Hint to much easier approach for (4): Consider $(I+A)(I-A+A^2)$
$endgroup$
– JMoravitz
Mar 3 '18 at 1:12
1
1
$begingroup$
For (2), (3), any such $A$ you try will show these options can be ruled out. So every example is a counterexample! That is because when $A^3=0$, all eigenvalues of $A$ exist and are equal to zero. Then the same goes for all eigenvalues of the powers $A^k$ of $A$. And the trace is the sum of the eigenvalues, and the trace is linear, $mathrm{tr}(X+Y)=mathrm{tr}(X)+mathrm{tr}(Y)$.
$endgroup$
– Jeppe Stig Nielsen
Mar 3 '18 at 2:23
$begingroup$
For (2), (3), any such $A$ you try will show these options can be ruled out. So every example is a counterexample! That is because when $A^3=0$, all eigenvalues of $A$ exist and are equal to zero. Then the same goes for all eigenvalues of the powers $A^k$ of $A$. And the trace is the sum of the eigenvalues, and the trace is linear, $mathrm{tr}(X+Y)=mathrm{tr}(X)+mathrm{tr}(Y)$.
$endgroup$
– Jeppe Stig Nielsen
Mar 3 '18 at 2:23
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You're on the right track.
Your first sentence about option (1) is correct, it can be ruled out.
Your thoughts about $2times2$ matrices and option (4) are also correct.
However, here we might deal with bigger matrices. Especially because for a $2times2$ matrix $A$, we do have $A^3=0 implies A^2=0$.
It's generally a good practice to think about examples. A typical example for $A^3=0$ is
$$pmatrix{0&1&0\0&0&1\0&0&0}$$
This example rules out options (2) and (3), so we're indeed only left with (4).
For this, can you find an inverse for $I+A$, knowing $A^3=0$?
answered Mar 3 '18 at 1:12
BerciBerci
61.9k23776
$endgroup$
add a comment |
$begingroup$
Notice that $$(I+A)(I-A+A^2)=I,$$ hence $I+A$ is nonsingular.
For $(2), (3)$, try the matrix $begin{bmatrix} 0 & 1 \ 0 & 0end{bmatrix}$
answered Mar 3 '18 at 1:12
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
add a comment |
$begingroup$
$(a)$ can be true, set $A = pmatrix{ 0 & 1 \ 0 & 0}$.
$A^3 = 0$ implies that $sigma(A) = {0}$, because the minimal polynomial of $A$ divides $x^3$.
This in turn means that $sigma(A^k) = {0}$ so $operatorname{Tr} A^k = 0$ for all $k in mathbb{N}$. Therefore $(b)$ and $(c)$ are always true:
$$operatorname{Tr}(A pm A^2) = operatorname{Tr} A pm operatorname{Tr} A^2 = 0$$
Also, $sigma(I + A) = {1}$ so $det(I + A) = 1 ne 0$, so $(d)$ is certainly false.
answered Mar 3 '18 at 1:13
mechanodroidmechanodroid
28.9k62648
$endgroup$
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%2f2674385%2fcan-a3-0-imply-ia-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You're on the right track.
Your first sentence about option (1) is correct, it can be ruled out.
Your thoughts about $2times2$ matrices and option (4) are also correct.
However, here we might deal with bigger matrices. Especially because for a $2times2$ matrix $A$, we do have $A^3=0 implies A^2=0$.
It's generally a good practice to think about examples. A typical example for $A^3=0$ is
$$pmatrix{0&1&0\0&0&1\0&0&0}$$
This example rules out options (2) and (3), so we're indeed only left with (4).
For this, can you find an inverse for $I+A$, knowing $A^3=0$?
answered Mar 3 '18 at 1:12
BerciBerci
61.9k23776
$endgroup$
add a comment |
$begingroup$
You're on the right track.
Your first sentence about option (1) is correct, it can be ruled out.
Your thoughts about $2times2$ matrices and option (4) are also correct.
However, here we might deal with bigger matrices. Especially because for a $2times2$ matrix $A$, we do have $A^3=0 implies A^2=0$.
It's generally a good practice to think about examples. A typical example for $A^3=0$ is
$$pmatrix{0&1&0\0&0&1\0&0&0}$$
This example rules out options (2) and (3), so we're indeed only left with (4).
For this, can you find an inverse for $I+A$, knowing $A^3=0$?
answered Mar 3 '18 at 1:12
BerciBerci
61.9k23776
$endgroup$
add a comment |
$begingroup$
You're on the right track.
Your first sentence about option (1) is correct, it can be ruled out.
Your thoughts about $2times2$ matrices and option (4) are also correct.
However, here we might deal with bigger matrices. Especially because for a $2times2$ matrix $A$, we do have $A^3=0 implies A^2=0$.
It's generally a good practice to think about examples. A typical example for $A^3=0$ is
$$pmatrix{0&1&0\0&0&1\0&0&0}$$
This example rules out options (2) and (3), so we're indeed only left with (4).
For this, can you find an inverse for $I+A$, knowing $A^3=0$?
answered Mar 3 '18 at 1:12
BerciBerci
61.9k23776
$endgroup$
You're on the right track.
Your first sentence about option (1) is correct, it can be ruled out.
Your thoughts about $2times2$ matrices and option (4) are also correct.
However, here we might deal with bigger matrices. Especially because for a $2times2$ matrix $A$, we do have $A^3=0 implies A^2=0$.
It's generally a good practice to think about examples. A typical example for $A^3=0$ is
$$pmatrix{0&1&0\0&0&1\0&0&0}$$
This example rules out options (2) and (3), so we're indeed only left with (4).
For this, can you find an inverse for $I+A$, knowing $A^3=0$?
answered Mar 3 '18 at 1:12
BerciBerci
61.9k23776
answered Mar 3 '18 at 1:12
BerciBerci
61.9k23776
answered Mar 3 '18 at 1:12
BerciBerci
61.9k23776
answered Mar 3 '18 at 1:12
BerciBerci
61.9k23776
61.9k23776
add a comment |
add a comment |
$begingroup$
Notice that $$(I+A)(I-A+A^2)=I,$$ hence $I+A$ is nonsingular.
For $(2), (3)$, try the matrix $begin{bmatrix} 0 & 1 \ 0 & 0end{bmatrix}$
answered Mar 3 '18 at 1:12
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
add a comment |
$begingroup$
Notice that $$(I+A)(I-A+A^2)=I,$$ hence $I+A$ is nonsingular.
For $(2), (3)$, try the matrix $begin{bmatrix} 0 & 1 \ 0 & 0end{bmatrix}$
answered Mar 3 '18 at 1:12
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
add a comment |
$begingroup$
Notice that $$(I+A)(I-A+A^2)=I,$$ hence $I+A$ is nonsingular.
For $(2), (3)$, try the matrix $begin{bmatrix} 0 & 1 \ 0 & 0end{bmatrix}$
answered Mar 3 '18 at 1:12
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
Notice that $$(I+A)(I-A+A^2)=I,$$ hence $I+A$ is nonsingular.
For $(2), (3)$, try the matrix $begin{bmatrix} 0 & 1 \ 0 & 0end{bmatrix}$
answered Mar 3 '18 at 1:12
Siong Thye GohSiong Thye Goh
104k1468120
answered Mar 3 '18 at 1:12
Siong Thye GohSiong Thye Goh
104k1468120
answered Mar 3 '18 at 1:12
Siong Thye GohSiong Thye Goh
104k1468120
answered Mar 3 '18 at 1:12
Siong Thye GohSiong Thye Goh
104k1468120
104k1468120
add a comment |
add a comment |
$begingroup$
$(a)$ can be true, set $A = pmatrix{ 0 & 1 \ 0 & 0}$.
$A^3 = 0$ implies that $sigma(A) = {0}$, because the minimal polynomial of $A$ divides $x^3$.
This in turn means that $sigma(A^k) = {0}$ so $operatorname{Tr} A^k = 0$ for all $k in mathbb{N}$. Therefore $(b)$ and $(c)$ are always true:
$$operatorname{Tr}(A pm A^2) = operatorname{Tr} A pm operatorname{Tr} A^2 = 0$$
Also, $sigma(I + A) = {1}$ so $det(I + A) = 1 ne 0$, so $(d)$ is certainly false.
answered Mar 3 '18 at 1:13
mechanodroidmechanodroid
28.9k62648
$endgroup$
add a comment |
$begingroup$
$(a)$ can be true, set $A = pmatrix{ 0 & 1 \ 0 & 0}$.
$A^3 = 0$ implies that $sigma(A) = {0}$, because the minimal polynomial of $A$ divides $x^3$.
This in turn means that $sigma(A^k) = {0}$ so $operatorname{Tr} A^k = 0$ for all $k in mathbb{N}$. Therefore $(b)$ and $(c)$ are always true:
$$operatorname{Tr}(A pm A^2) = operatorname{Tr} A pm operatorname{Tr} A^2 = 0$$
Also, $sigma(I + A) = {1}$ so $det(I + A) = 1 ne 0$, so $(d)$ is certainly false.
answered Mar 3 '18 at 1:13
mechanodroidmechanodroid
28.9k62648
$endgroup$
add a comment |
$begingroup$
$(a)$ can be true, set $A = pmatrix{ 0 & 1 \ 0 & 0}$.
$A^3 = 0$ implies that $sigma(A) = {0}$, because the minimal polynomial of $A$ divides $x^3$.
This in turn means that $sigma(A^k) = {0}$ so $operatorname{Tr} A^k = 0$ for all $k in mathbb{N}$. Therefore $(b)$ and $(c)$ are always true:
$$operatorname{Tr}(A pm A^2) = operatorname{Tr} A pm operatorname{Tr} A^2 = 0$$
Also, $sigma(I + A) = {1}$ so $det(I + A) = 1 ne 0$, so $(d)$ is certainly false.
answered Mar 3 '18 at 1:13
mechanodroidmechanodroid
28.9k62648
$endgroup$
$(a)$ can be true, set $A = pmatrix{ 0 & 1 \ 0 & 0}$.
$A^3 = 0$ implies that $sigma(A) = {0}$, because the minimal polynomial of $A$ divides $x^3$.
This in turn means that $sigma(A^k) = {0}$ so $operatorname{Tr} A^k = 0$ for all $k in mathbb{N}$. Therefore $(b)$ and $(c)$ are always true:
$$operatorname{Tr}(A pm A^2) = operatorname{Tr} A pm operatorname{Tr} A^2 = 0$$
Also, $sigma(I + A) = {1}$ so $det(I + A) = 1 ne 0$, so $(d)$ is certainly false.
answered Mar 3 '18 at 1:13
mechanodroidmechanodroid
28.9k62648
answered Mar 3 '18 at 1:13
mechanodroidmechanodroid
28.9k62648
answered Mar 3 '18 at 1:13
mechanodroidmechanodroid
28.9k62648
answered Mar 3 '18 at 1:13
mechanodroidmechanodroid
28.9k62648
28.9k62648
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%2f2674385%2fcan-a3-0-imply-ia-0%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
$begingroup$
Hint to much easier approach for (4): Consider $(I+A)(I-A+A^2)$
$endgroup$
– JMoravitz
Mar 3 '18 at 1:12
1
$begingroup$
For (2), (3), any such $A$ you try will show these options can be ruled out. So every example is a counterexample! That is because when $A^3=0$, all eigenvalues of $A$ exist and are equal to zero. Then the same goes for all eigenvalues of the powers $A^k$ of $A$. And the trace is the sum of the eigenvalues, and the trace is linear, $mathrm{tr}(X+Y)=mathrm{tr}(X)+mathrm{tr}(Y)$.
$endgroup$
– Jeppe Stig Nielsen
Mar 3 '18 at 2:23