Mixing
On...
$begingroup$
Consider the $4times4$ matrices $A=bigg[begin{matrix}
E&O_2\O_2&F
end{matrix}bigg]$, where $E,F$ are any nilpotent $2times2$ matrices, and $B_b=left[begin{matrix}2b&-1&0&-1\0&b&0&0\0&-1&0&-1\0&1&0&b
end{matrix}right]$ with $binBbb R$. For what values of $b$ are $A$ and $B_b$ similar?
I first checked a necessary condition: the characteristic polynomials must be equal.
By the nilpotence of $E$ and $F$,
$$det(A-lambda I)=det(E-lambda I) cdot det(F-lambda I)=lambda^2cdotlambda^2=lambda^4$$and $$det(B_b-lambda I)=lambda(lambda-2b)(lambda-b)^2 $$so they are equal if and only if $b=0$.
Now, minimal polynomials should also be the same, and I noticed $A^2=bigg[begin{matrix}
E^2&O_2\O_2&F^2
end{matrix}bigg]=0,$ so the minimal polynomial of $A$ is $x^2$. But then $(B_0)^2$ is not the null matrix, the minimal polynomial of $B_0$ appears to be $x^3$, which would then mean $A$ and $B_b$ are not similar for any $binBbb R$.
But it seems strange, did I make a mistake?
linear-algebra matrices proof-verification minimal-polynomials nilpotence
asked Jan 19 at 23:55
LearnerLearner
17510
$endgroup$
add a comment |
$begingroup$
Consider the $4times4$ matrices $A=bigg[begin{matrix}
E&O_2\O_2&F
end{matrix}bigg]$, where $E,F$ are any nilpotent $2times2$ matrices, and $B_b=left[begin{matrix}2b&-1&0&-1\0&b&0&0\0&-1&0&-1\0&1&0&b
end{matrix}right]$ with $binBbb R$. For what values of $b$ are $A$ and $B_b$ similar?
I first checked a necessary condition: the characteristic polynomials must be equal.
By the nilpotence of $E$ and $F$,
$$det(A-lambda I)=det(E-lambda I) cdot det(F-lambda I)=lambda^2cdotlambda^2=lambda^4$$and $$det(B_b-lambda I)=lambda(lambda-2b)(lambda-b)^2 $$so they are equal if and only if $b=0$.
Now, minimal polynomials should also be the same, and I noticed $A^2=bigg[begin{matrix}
E^2&O_2\O_2&F^2
end{matrix}bigg]=0,$ so the minimal polynomial of $A$ is $x^2$. But then $(B_0)^2$ is not the null matrix, the minimal polynomial of $B_0$ appears to be $x^3$, which would then mean $A$ and $B_b$ are not similar for any $binBbb R$.
But it seems strange, did I make a mistake?
linear-algebra matrices proof-verification minimal-polynomials nilpotence
asked Jan 19 at 23:55
LearnerLearner
17510
$endgroup$
$begingroup$
Why does it seem strange?
$endgroup$
– Jacky Chong
Jan 20 at 1:36
$begingroup$
@JackyChong Because it's from an exam, and I hadn't previously seen from this professor an exercise where the answer is "no value of the parameter works". Guess there's always a first time!
$endgroup$
– Learner
Jan 20 at 9:40
add a comment |
$begingroup$
Consider the $4times4$ matrices $A=bigg[begin{matrix}
E&O_2\O_2&F
end{matrix}bigg]$, where $E,F$ are any nilpotent $2times2$ matrices, and $B_b=left[begin{matrix}2b&-1&0&-1\0&b&0&0\0&-1&0&-1\0&1&0&b
end{matrix}right]$ with $binBbb R$. For what values of $b$ are $A$ and $B_b$ similar?
I first checked a necessary condition: the characteristic polynomials must be equal.
By the nilpotence of $E$ and $F$,
$$det(A-lambda I)=det(E-lambda I) cdot det(F-lambda I)=lambda^2cdotlambda^2=lambda^4$$and $$det(B_b-lambda I)=lambda(lambda-2b)(lambda-b)^2 $$so they are equal if and only if $b=0$.
Now, minimal polynomials should also be the same, and I noticed $A^2=bigg[begin{matrix}
E^2&O_2\O_2&F^2
end{matrix}bigg]=0,$ so the minimal polynomial of $A$ is $x^2$. But then $(B_0)^2$ is not the null matrix, the minimal polynomial of $B_0$ appears to be $x^3$, which would then mean $A$ and $B_b$ are not similar for any $binBbb R$.
But it seems strange, did I make a mistake?
linear-algebra matrices proof-verification minimal-polynomials nilpotence
asked Jan 19 at 23:55
LearnerLearner
17510
$endgroup$
Consider the $4times4$ matrices $A=bigg[begin{matrix}
E&O_2\O_2&F
end{matrix}bigg]$, where $E,F$ are any nilpotent $2times2$ matrices, and $B_b=left[begin{matrix}2b&-1&0&-1\0&b&0&0\0&-1&0&-1\0&1&0&b
end{matrix}right]$ with $binBbb R$. For what values of $b$ are $A$ and $B_b$ similar?
I first checked a necessary condition: the characteristic polynomials must be equal.
By the nilpotence of $E$ and $F$,
$$det(A-lambda I)=det(E-lambda I) cdot det(F-lambda I)=lambda^2cdotlambda^2=lambda^4$$and $$det(B_b-lambda I)=lambda(lambda-2b)(lambda-b)^2 $$so they are equal if and only if $b=0$.
Now, minimal polynomials should also be the same, and I noticed $A^2=bigg[begin{matrix}
E^2&O_2\O_2&F^2
end{matrix}bigg]=0,$ so the minimal polynomial of $A$ is $x^2$. But then $(B_0)^2$ is not the null matrix, the minimal polynomial of $B_0$ appears to be $x^3$, which would then mean $A$ and $B_b$ are not similar for any $binBbb R$.
But it seems strange, did I make a mistake?
linear-algebra matrices proof-verification minimal-polynomials nilpotence
linear-algebra matrices proof-verification minimal-polynomials nilpotence
asked Jan 19 at 23:55
LearnerLearner
17510
asked Jan 19 at 23:55
LearnerLearner
17510
asked Jan 19 at 23:55
LearnerLearner
17510
asked Jan 19 at 23:55
LearnerLearner
17510
asked Jan 19 at 23:55
LearnerLearner
17510
17510
$begingroup$
Why does it seem strange?
$endgroup$
– Jacky Chong
Jan 20 at 1:36
$begingroup$
@JackyChong Because it's from an exam, and I hadn't previously seen from this professor an exercise where the answer is "no value of the parameter works". Guess there's always a first time!
$endgroup$
– Learner
Jan 20 at 9:40
add a comment |
$begingroup$
Why does it seem strange?
$endgroup$
– Jacky Chong
Jan 20 at 1:36
$begingroup$
@JackyChong Because it's from an exam, and I hadn't previously seen from this professor an exercise where the answer is "no value of the parameter works". Guess there's always a first time!
$endgroup$
– Learner
Jan 20 at 9:40
$begingroup$
Why does it seem strange?
$endgroup$
– Jacky Chong
Jan 20 at 1:36
$begingroup$
Why does it seem strange?
$endgroup$
– Jacky Chong
Jan 20 at 1:36
$begingroup$
@JackyChong Because it's from an exam, and I hadn't previously seen from this professor an exercise where the answer is "no value of the parameter works". Guess there's always a first time!
$endgroup$
– Learner
Jan 20 at 9:40
$begingroup$
@JackyChong Because it's from an exam, and I hadn't previously seen from this professor an exercise where the answer is "no value of the parameter works". Guess there's always a first time!
$endgroup$
– Learner
Jan 20 at 9:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You could replace $B_b$ by a similar triangular matrix to see somewhat better what is going on. Here you could just permute the standard basis vectors into the order $[e_3,e_1,e_4,e_2]$ to give a matrix
$$ B'_b=pmatrix{0&0&-1&-1\0&2b&-1&-1\0&0&b&1\0&0&0&b} $$
similar to $B_b$, and which is nilpotent only for $b=0$. Since the final standard basis vector now survives two multiplications by $B'_0$ before becoming zero: $$(0,0,0,1)mapsto(-1,-1,1,0)mapsto(-1,-1,0,0)mapsto(0,0,0,0),$$ it is clear that $B'_0$, unlike $A$, is not annihilated by the polynomial $X^2$, so $A$ and $B'_0$ are not similar. Your conclusion that $A$ is not similar to any $B_b$ is correct.
answered Jan 23 at 9:20
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
$endgroup$
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%2f3079986%2fon-bigg-beginmatrixeo-of-endmatrix-bigg-b-b-left-beginmatrix2b-1%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
$begingroup$
You could replace $B_b$ by a similar triangular matrix to see somewhat better what is going on. Here you could just permute the standard basis vectors into the order $[e_3,e_1,e_4,e_2]$ to give a matrix
$$ B'_b=pmatrix{0&0&-1&-1\0&2b&-1&-1\0&0&b&1\0&0&0&b} $$
similar to $B_b$, and which is nilpotent only for $b=0$. Since the final standard basis vector now survives two multiplications by $B'_0$ before becoming zero: $$(0,0,0,1)mapsto(-1,-1,1,0)mapsto(-1,-1,0,0)mapsto(0,0,0,0),$$ it is clear that $B'_0$, unlike $A$, is not annihilated by the polynomial $X^2$, so $A$ and $B'_0$ are not similar. Your conclusion that $A$ is not similar to any $B_b$ is correct.
answered Jan 23 at 9:20
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
$endgroup$
add a comment |
$begingroup$
You could replace $B_b$ by a similar triangular matrix to see somewhat better what is going on. Here you could just permute the standard basis vectors into the order $[e_3,e_1,e_4,e_2]$ to give a matrix
$$ B'_b=pmatrix{0&0&-1&-1\0&2b&-1&-1\0&0&b&1\0&0&0&b} $$
similar to $B_b$, and which is nilpotent only for $b=0$. Since the final standard basis vector now survives two multiplications by $B'_0$ before becoming zero: $$(0,0,0,1)mapsto(-1,-1,1,0)mapsto(-1,-1,0,0)mapsto(0,0,0,0),$$ it is clear that $B'_0$, unlike $A$, is not annihilated by the polynomial $X^2$, so $A$ and $B'_0$ are not similar. Your conclusion that $A$ is not similar to any $B_b$ is correct.
answered Jan 23 at 9:20
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
$endgroup$
add a comment |
$begingroup$
You could replace $B_b$ by a similar triangular matrix to see somewhat better what is going on. Here you could just permute the standard basis vectors into the order $[e_3,e_1,e_4,e_2]$ to give a matrix
$$ B'_b=pmatrix{0&0&-1&-1\0&2b&-1&-1\0&0&b&1\0&0&0&b} $$
similar to $B_b$, and which is nilpotent only for $b=0$. Since the final standard basis vector now survives two multiplications by $B'_0$ before becoming zero: $$(0,0,0,1)mapsto(-1,-1,1,0)mapsto(-1,-1,0,0)mapsto(0,0,0,0),$$ it is clear that $B'_0$, unlike $A$, is not annihilated by the polynomial $X^2$, so $A$ and $B'_0$ are not similar. Your conclusion that $A$ is not similar to any $B_b$ is correct.
answered Jan 23 at 9:20
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
$endgroup$
You could replace $B_b$ by a similar triangular matrix to see somewhat better what is going on. Here you could just permute the standard basis vectors into the order $[e_3,e_1,e_4,e_2]$ to give a matrix
$$ B'_b=pmatrix{0&0&-1&-1\0&2b&-1&-1\0&0&b&1\0&0&0&b} $$
similar to $B_b$, and which is nilpotent only for $b=0$. Since the final standard basis vector now survives two multiplications by $B'_0$ before becoming zero: $$(0,0,0,1)mapsto(-1,-1,1,0)mapsto(-1,-1,0,0)mapsto(0,0,0,0),$$ it is clear that $B'_0$, unlike $A$, is not annihilated by the polynomial $X^2$, so $A$ and $B'_0$ are not similar. Your conclusion that $A$ is not similar to any $B_b$ is correct.
answered Jan 23 at 9:20
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
answered Jan 23 at 9:20
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
answered Jan 23 at 9:20
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
answered Jan 23 at 9:20
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
87.6k5110225
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%2f3079986%2fon-bigg-beginmatrixeo-of-endmatrix-bigg-b-b-left-beginmatrix2b-1%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
$begingroup$
Why does it seem strange?
$endgroup$
– Jacky Chong
Jan 20 at 1:36
$begingroup$
@JackyChong Because it's from an exam, and I hadn't previously seen from this professor an exercise where the answer is "no value of the parameter works". Guess there's always a first time!
$endgroup$
– Learner
Jan 20 at 9:40