prove that $A$ is diagonalizable if $A^{3}-3A^{2}-A+3I_{n} = 0$Hermitian Matrices are DiagonalizableA...
How can saying a song's name be a copyright violation?
Brothers & sisters
I would say: "You are another teacher", but she is a woman and I am a man
Intersection of two sorted vectors in C++
Why does Arabsat 6A need a Falcon Heavy to launch
Alternative to sending password over mail?
How much of data wrangling is a data scientist's job?
1960's book about a plague that kills all white people
How to model explosives?
Facing a paradox: Earnshaw's theorem in one dimension
What about the virus in 12 Monkeys?
Why do I get two different answers for this counting problem?
What is the word for reserving something for yourself before others do?
Why is it a bad idea to hire a hitman to eliminate most corrupt politicians?
What mechanic is there to disable a threat instead of killing it?
How to prevent "they're falling in love" trope
Doing something right before you need it - expression for this?
Why "Having chlorophyll without photosynthesis is actually very dangerous" and "like living with a bomb"?
Anagram holiday
SSH "lag" in LAN on some machines, mixed distros
Why can't we play rap on piano?
Is it inappropriate for a student to attend their mentor's dissertation defense?
How to draw the figure with four pentagons?
Today is the Center
prove that $A$ is diagonalizable if $A^{3}-3A^{2}-A+3I_{n} = 0$
Hermitian Matrices are DiagonalizableA question about diagonalizable.Show that matrix $A$ is NOT diagonalizable.Prove that A … is not diagonalizableProve a matrix is not diagonalizableHow to use inner products in C(n) to prove normal matrix is unitarily diagonalizable after knowing that normal matrix is diagonalizable?Prove that $A$ is diagonalizable.Prove that a general matrix is diagonalizableDetermine $a$ to make matrix $A$ diagonalizableDiagonalizable block-diagonal matrix
$begingroup$
We have :
$A^{3}-3A^{2}-A+3I_{n} = 0 $
how can i prove that A is diagonalizable .
I don't know how to do when A is written this way
linear-algebra matrices eigenvalues-eigenvectors diagonalization
edited 1 hour ago
user21820
40k544161
asked 6 hours ago
JoshuaKJoshuaK
264
$endgroup$
add a comment |
$begingroup$
We have :
$A^{3}-3A^{2}-A+3I_{n} = 0 $
how can i prove that A is diagonalizable .
I don't know how to do when A is written this way
linear-algebra matrices eigenvalues-eigenvectors diagonalization
edited 1 hour ago
user21820
40k544161
asked 6 hours ago
JoshuaKJoshuaK
264
$endgroup$
2
$begingroup$
Note that if $A=operatorname{Id}_n$, then $A^3-3A^2-A+3operatorname{Id}_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatorname{Id}_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
6 hours ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
6 hours ago
$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
2 hours ago
add a comment |
$begingroup$
We have :
$A^{3}-3A^{2}-A+3I_{n} = 0 $
how can i prove that A is diagonalizable .
I don't know how to do when A is written this way
linear-algebra matrices eigenvalues-eigenvectors diagonalization
edited 1 hour ago
user21820
40k544161
asked 6 hours ago
JoshuaKJoshuaK
264
$endgroup$
We have :
$A^{3}-3A^{2}-A+3I_{n} = 0 $
how can i prove that A is diagonalizable .
I don't know how to do when A is written this way
linear-algebra matrices eigenvalues-eigenvectors diagonalization
linear-algebra matrices eigenvalues-eigenvectors diagonalization
edited 1 hour ago
user21820
40k544161
asked 6 hours ago
JoshuaKJoshuaK
264
edited 1 hour ago
user21820
40k544161
asked 6 hours ago
JoshuaKJoshuaK
264
edited 1 hour ago
user21820
40k544161
edited 1 hour ago
user21820
40k544161
edited 1 hour ago
user21820
40k544161
40k544161
asked 6 hours ago
JoshuaKJoshuaK
264
asked 6 hours ago
JoshuaKJoshuaK
264
asked 6 hours ago
JoshuaKJoshuaK
264
264
2
$begingroup$
Note that if $A=operatorname{Id}_n$, then $A^3-3A^2-A+3operatorname{Id}_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatorname{Id}_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
6 hours ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
6 hours ago
$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
2 hours ago
add a comment |
2
$begingroup$
Note that if $A=operatorname{Id}_n$, then $A^3-3A^2-A+3operatorname{Id}_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatorname{Id}_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
6 hours ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
6 hours ago
$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
2 hours ago
2
2
$begingroup$
Note that if $A=operatorname{Id}_n$, then $A^3-3A^2-A+3operatorname{Id}_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatorname{Id}_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
6 hours ago
$begingroup$
Note that if $A=operatorname{Id}_n$, then $A^3-3A^2-A+3operatorname{Id}_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatorname{Id}_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
6 hours ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
6 hours ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
6 hours ago
$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
2 hours ago
$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
2 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbb{R}$.
answered 6 hours ago
TheSilverDoeTheSilverDoe
5,324215
$endgroup$
3
$begingroup$
I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
$endgroup$
– Acccumulation
4 hours ago
add a comment |
$begingroup$
Solving a simpler example, $A-cI_n=0$, it's clear that $a_{(i,i)}=c$ because $a_{(i,j)} - cI_{(i,j)} = 0$ for all $i,j in {1,dots,n}$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.
answered 6 hours ago
EricEric
513
$endgroup$
add a comment |
$begingroup$
We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.
answered 6 hours ago
GSoferGSofer
8631313
$endgroup$
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
6 hours ago
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%2f3175144%2fprove-that-a-is-diagonalizable-if-a3-3a2-a3i-n-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
$begingroup$
The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbb{R}$.
answered 6 hours ago
TheSilverDoeTheSilverDoe
5,324215
$endgroup$
3
$begingroup$
I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
$endgroup$
– Acccumulation
4 hours ago
add a comment |
$begingroup$
The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbb{R}$.
answered 6 hours ago
TheSilverDoeTheSilverDoe
5,324215
$endgroup$
3
$begingroup$
I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
$endgroup$
– Acccumulation
4 hours ago
add a comment |
$begingroup$
The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbb{R}$.
answered 6 hours ago
TheSilverDoeTheSilverDoe
5,324215
$endgroup$
The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbb{R}$.
answered 6 hours ago
TheSilverDoeTheSilverDoe
5,324215
answered 6 hours ago
TheSilverDoeTheSilverDoe
5,324215
answered 6 hours ago
TheSilverDoeTheSilverDoe
5,324215
answered 6 hours ago
TheSilverDoeTheSilverDoe
5,324215
5,324215
3
$begingroup$
I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
$endgroup$
– Acccumulation
4 hours ago
add a comment |
3
$begingroup$
I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
$endgroup$
– Acccumulation
4 hours ago
3
3
$begingroup$
I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
$endgroup$
– Acccumulation
4 hours ago
$begingroup$
I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
$endgroup$
– Acccumulation
4 hours ago
add a comment |
$begingroup$
Solving a simpler example, $A-cI_n=0$, it's clear that $a_{(i,i)}=c$ because $a_{(i,j)} - cI_{(i,j)} = 0$ for all $i,j in {1,dots,n}$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.
answered 6 hours ago
EricEric
513
$endgroup$
add a comment |
$begingroup$
Solving a simpler example, $A-cI_n=0$, it's clear that $a_{(i,i)}=c$ because $a_{(i,j)} - cI_{(i,j)} = 0$ for all $i,j in {1,dots,n}$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.
answered 6 hours ago
EricEric
513
$endgroup$
add a comment |
$begingroup$
Solving a simpler example, $A-cI_n=0$, it's clear that $a_{(i,i)}=c$ because $a_{(i,j)} - cI_{(i,j)} = 0$ for all $i,j in {1,dots,n}$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.
answered 6 hours ago
EricEric
513
$endgroup$
Solving a simpler example, $A-cI_n=0$, it's clear that $a_{(i,i)}=c$ because $a_{(i,j)} - cI_{(i,j)} = 0$ for all $i,j in {1,dots,n}$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.
answered 6 hours ago
EricEric
513
answered 6 hours ago
EricEric
513
answered 6 hours ago
EricEric
513
answered 6 hours ago
EricEric
513
513
add a comment |
add a comment |
$begingroup$
We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.
answered 6 hours ago
GSoferGSofer
8631313
$endgroup$
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
6 hours ago
add a comment |
$begingroup$
We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.
answered 6 hours ago
GSoferGSofer
8631313
$endgroup$
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
6 hours ago
add a comment |
$begingroup$
We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.
answered 6 hours ago
GSoferGSofer
8631313
$endgroup$
We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.
answered 6 hours ago
GSoferGSofer
8631313
answered 6 hours ago
GSoferGSofer
8631313
answered 6 hours ago
GSoferGSofer
8631313
answered 6 hours ago
GSoferGSofer
8631313
8631313
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
6 hours ago
add a comment |
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
6 hours ago
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
6 hours ago
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
6 hours ago
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%2f3175144%2fprove-that-a-is-diagonalizable-if-a3-3a2-a3i-n-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
2
$begingroup$
Note that if $A=operatorname{Id}_n$, then $A^3-3A^2-A+3operatorname{Id}_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatorname{Id}_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
6 hours ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
6 hours ago
$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
2 hours ago