More efficient methods to show that something is the characteristic polynomial of a matrix
$begingroup$
Consider the $5times 5$ matrix $A$, $$A = begin{pmatrix}-2 & -1 & -1 & 3 & 2 \ -4 & 1 & -1 & 3 & 2 \ 1 & 1 & 0 & -3 & -2 \ -4 & -2 & -1 & 5 & 1 \ 1 & 1 & -2 & -3 & 0 end{pmatrix}$$ Show that $A$ has characteristic polynomial $(t+1)^2(t-2)^3$.
I already know how to determine the characteristic polynomial of $A$ from scratch, but I am interested in whether there is a more efficient method to check whether a given polynomial is the characteristic polynomial of a matrix. The best idea I could come up with is to check that the dimension of $lambda=-1$ is $2$ and the dimension of $lambda=2$ is $3$, but this method involves a lot of row reductions. Is there are a more efficient method?
linear-algebra
asked Dec 4 '18 at 22:06
Ryan GreylingRyan Greyling
1827
$endgroup$
add a comment |
$begingroup$
Consider the $5times 5$ matrix $A$, $$A = begin{pmatrix}-2 & -1 & -1 & 3 & 2 \ -4 & 1 & -1 & 3 & 2 \ 1 & 1 & 0 & -3 & -2 \ -4 & -2 & -1 & 5 & 1 \ 1 & 1 & -2 & -3 & 0 end{pmatrix}$$ Show that $A$ has characteristic polynomial $(t+1)^2(t-2)^3$.
I already know how to determine the characteristic polynomial of $A$ from scratch, but I am interested in whether there is a more efficient method to check whether a given polynomial is the characteristic polynomial of a matrix. The best idea I could come up with is to check that the dimension of $lambda=-1$ is $2$ and the dimension of $lambda=2$ is $3$, but this method involves a lot of row reductions. Is there are a more efficient method?
linear-algebra
asked Dec 4 '18 at 22:06
Ryan GreylingRyan Greyling
1827
$endgroup$
$begingroup$
When you say "dimension", do you mean the dimension of the eigenspaces? If yes, then you probably get a wrong conclusion, since the dimension of the eigenspace, i.e. the geometric multiplicities is always less than or equal to the index of $t-c$ in the characteristic polynomial, i.e. the algebraic multiplicity, and the $<$ could happen.
$endgroup$
– xbh
Dec 5 '18 at 2:06
$begingroup$
Yes, I do mean the dimension of the eigenspaces. That is a good point, so then the method I came up with wouldn't work.
$endgroup$
– Ryan Greyling
Dec 5 '18 at 3:58
add a comment |
$begingroup$
Consider the $5times 5$ matrix $A$, $$A = begin{pmatrix}-2 & -1 & -1 & 3 & 2 \ -4 & 1 & -1 & 3 & 2 \ 1 & 1 & 0 & -3 & -2 \ -4 & -2 & -1 & 5 & 1 \ 1 & 1 & -2 & -3 & 0 end{pmatrix}$$ Show that $A$ has characteristic polynomial $(t+1)^2(t-2)^3$.
I already know how to determine the characteristic polynomial of $A$ from scratch, but I am interested in whether there is a more efficient method to check whether a given polynomial is the characteristic polynomial of a matrix. The best idea I could come up with is to check that the dimension of $lambda=-1$ is $2$ and the dimension of $lambda=2$ is $3$, but this method involves a lot of row reductions. Is there are a more efficient method?
linear-algebra
asked Dec 4 '18 at 22:06
Ryan GreylingRyan Greyling
1827
$endgroup$
Consider the $5times 5$ matrix $A$, $$A = begin{pmatrix}-2 & -1 & -1 & 3 & 2 \ -4 & 1 & -1 & 3 & 2 \ 1 & 1 & 0 & -3 & -2 \ -4 & -2 & -1 & 5 & 1 \ 1 & 1 & -2 & -3 & 0 end{pmatrix}$$ Show that $A$ has characteristic polynomial $(t+1)^2(t-2)^3$.
I already know how to determine the characteristic polynomial of $A$ from scratch, but I am interested in whether there is a more efficient method to check whether a given polynomial is the characteristic polynomial of a matrix. The best idea I could come up with is to check that the dimension of $lambda=-1$ is $2$ and the dimension of $lambda=2$ is $3$, but this method involves a lot of row reductions. Is there are a more efficient method?
linear-algebra
linear-algebra
asked Dec 4 '18 at 22:06
Ryan GreylingRyan Greyling
1827
asked Dec 4 '18 at 22:06
Ryan GreylingRyan Greyling
1827
asked Dec 4 '18 at 22:06
Ryan GreylingRyan Greyling
1827
asked Dec 4 '18 at 22:06
Ryan GreylingRyan Greyling
1827
asked Dec 4 '18 at 22:06
Ryan GreylingRyan Greyling
1827
1827
$begingroup$
When you say "dimension", do you mean the dimension of the eigenspaces? If yes, then you probably get a wrong conclusion, since the dimension of the eigenspace, i.e. the geometric multiplicities is always less than or equal to the index of $t-c$ in the characteristic polynomial, i.e. the algebraic multiplicity, and the $<$ could happen.
$endgroup$
– xbh
Dec 5 '18 at 2:06
$begingroup$
Yes, I do mean the dimension of the eigenspaces. That is a good point, so then the method I came up with wouldn't work.
$endgroup$
– Ryan Greyling
Dec 5 '18 at 3:58
add a comment |
$begingroup$
When you say "dimension", do you mean the dimension of the eigenspaces? If yes, then you probably get a wrong conclusion, since the dimension of the eigenspace, i.e. the geometric multiplicities is always less than or equal to the index of $t-c$ in the characteristic polynomial, i.e. the algebraic multiplicity, and the $<$ could happen.
$endgroup$
– xbh
Dec 5 '18 at 2:06
$begingroup$
Yes, I do mean the dimension of the eigenspaces. That is a good point, so then the method I came up with wouldn't work.
$endgroup$
– Ryan Greyling
Dec 5 '18 at 3:58
$begingroup$
When you say "dimension", do you mean the dimension of the eigenspaces? If yes, then you probably get a wrong conclusion, since the dimension of the eigenspace, i.e. the geometric multiplicities is always less than or equal to the index of $t-c$ in the characteristic polynomial, i.e. the algebraic multiplicity, and the $<$ could happen.
$endgroup$
– xbh
Dec 5 '18 at 2:06
$begingroup$
When you say "dimension", do you mean the dimension of the eigenspaces? If yes, then you probably get a wrong conclusion, since the dimension of the eigenspace, i.e. the geometric multiplicities is always less than or equal to the index of $t-c$ in the characteristic polynomial, i.e. the algebraic multiplicity, and the $<$ could happen.
$endgroup$
– xbh
Dec 5 '18 at 2:06
$begingroup$
Yes, I do mean the dimension of the eigenspaces. That is a good point, so then the method I came up with wouldn't work.
$endgroup$
– Ryan Greyling
Dec 5 '18 at 3:58
$begingroup$
Yes, I do mean the dimension of the eigenspaces. That is a good point, so then the method I came up with wouldn't work.
$endgroup$
– Ryan Greyling
Dec 5 '18 at 3:58
add a comment |
0
active
oldest
votes
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%2f3026246%2fmore-efficient-methods-to-show-that-something-is-the-characteristic-polynomial-o%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3026246%2fmore-efficient-methods-to-show-that-something-is-the-characteristic-polynomial-o%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
$begingroup$
When you say "dimension", do you mean the dimension of the eigenspaces? If yes, then you probably get a wrong conclusion, since the dimension of the eigenspace, i.e. the geometric multiplicities is always less than or equal to the index of $t-c$ in the characteristic polynomial, i.e. the algebraic multiplicity, and the $<$ could happen.
$endgroup$
– xbh
Dec 5 '18 at 2:06
$begingroup$
Yes, I do mean the dimension of the eigenspaces. That is a good point, so then the method I came up with wouldn't work.
$endgroup$
– Ryan Greyling
Dec 5 '18 at 3:58