Mixing
Construct a 2x2 matrix with real eigenvalues that is not diagonalizable
$begingroup$
I've been banging my head against the table with this one for a while. I know that for it not to be diagonalizable that the columns can't be linearly independent but can't quite seem to come up with one.
linear-algebra eigenvalues-eigenvectors diagonalization
asked Apr 3 '17 at 3:01
uRockNinjauRockNinja
112
$endgroup$
add a comment |
$begingroup$
I've been banging my head against the table with this one for a while. I know that for it not to be diagonalizable that the columns can't be linearly independent but can't quite seem to come up with one.
linear-algebra eigenvalues-eigenvectors diagonalization
asked Apr 3 '17 at 3:01
uRockNinjauRockNinja
112
$endgroup$
4
$begingroup$
You might want to consider a $2times 2$ Jordan block.
$endgroup$
– thanasissdr
Apr 3 '17 at 3:05
2
$begingroup$
What you think you know might not be correct.
$endgroup$
– Brian Borchers
Apr 3 '17 at 3:08
$begingroup$
For it not to be diagonalizable all you need is to have an eigenvalue with multiplicity greater than the dimension of the eigenspace. (i.e not enough eigenvectors). This is not related to independence
$endgroup$
– B.A
Apr 3 '17 at 3:23
$begingroup$
Also complex matrices can always be put on a form with either 1s or 0s off the main diagonal and eigenvalues on the diagonal. For real matrices you can be sure to not have to have more than at most blocks of size 2x2 along the diagonal.
$endgroup$
– mathreadler
Apr 3 '17 at 9:11
add a comment |
$begingroup$
I've been banging my head against the table with this one for a while. I know that for it not to be diagonalizable that the columns can't be linearly independent but can't quite seem to come up with one.
linear-algebra eigenvalues-eigenvectors diagonalization
asked Apr 3 '17 at 3:01
uRockNinjauRockNinja
112
$endgroup$
I've been banging my head against the table with this one for a while. I know that for it not to be diagonalizable that the columns can't be linearly independent but can't quite seem to come up with one.
linear-algebra eigenvalues-eigenvectors diagonalization
linear-algebra eigenvalues-eigenvectors diagonalization
asked Apr 3 '17 at 3:01
uRockNinjauRockNinja
112
asked Apr 3 '17 at 3:01
uRockNinjauRockNinja
112
asked Apr 3 '17 at 3:01
uRockNinjauRockNinja
112
asked Apr 3 '17 at 3:01
uRockNinjauRockNinja
112
asked Apr 3 '17 at 3:01
uRockNinjauRockNinja
112
112
4
$begingroup$
You might want to consider a $2times 2$ Jordan block.
$endgroup$
– thanasissdr
Apr 3 '17 at 3:05
2
$begingroup$
What you think you know might not be correct.
$endgroup$
– Brian Borchers
Apr 3 '17 at 3:08
$begingroup$
For it not to be diagonalizable all you need is to have an eigenvalue with multiplicity greater than the dimension of the eigenspace. (i.e not enough eigenvectors). This is not related to independence
$endgroup$
– B.A
Apr 3 '17 at 3:23
$begingroup$
Also complex matrices can always be put on a form with either 1s or 0s off the main diagonal and eigenvalues on the diagonal. For real matrices you can be sure to not have to have more than at most blocks of size 2x2 along the diagonal.
$endgroup$
– mathreadler
Apr 3 '17 at 9:11
add a comment |
4
$begingroup$
You might want to consider a $2times 2$ Jordan block.
$endgroup$
– thanasissdr
Apr 3 '17 at 3:05
2
$begingroup$
What you think you know might not be correct.
$endgroup$
– Brian Borchers
Apr 3 '17 at 3:08
$begingroup$
For it not to be diagonalizable all you need is to have an eigenvalue with multiplicity greater than the dimension of the eigenspace. (i.e not enough eigenvectors). This is not related to independence
$endgroup$
– B.A
Apr 3 '17 at 3:23
$begingroup$
Also complex matrices can always be put on a form with either 1s or 0s off the main diagonal and eigenvalues on the diagonal. For real matrices you can be sure to not have to have more than at most blocks of size 2x2 along the diagonal.
$endgroup$
– mathreadler
Apr 3 '17 at 9:11
4
4
$begingroup$
You might want to consider a $2times 2$ Jordan block.
$endgroup$
– thanasissdr
Apr 3 '17 at 3:05
$begingroup$
You might want to consider a $2times 2$ Jordan block.
$endgroup$
– thanasissdr
Apr 3 '17 at 3:05
2
2
$begingroup$
What you think you know might not be correct.
$endgroup$
– Brian Borchers
Apr 3 '17 at 3:08
$begingroup$
What you think you know might not be correct.
$endgroup$
– Brian Borchers
Apr 3 '17 at 3:08
$begingroup$
For it not to be diagonalizable all you need is to have an eigenvalue with multiplicity greater than the dimension of the eigenspace. (i.e not enough eigenvectors). This is not related to independence
$endgroup$
– B.A
Apr 3 '17 at 3:23
$begingroup$
For it not to be diagonalizable all you need is to have an eigenvalue with multiplicity greater than the dimension of the eigenspace. (i.e not enough eigenvectors). This is not related to independence
$endgroup$
– B.A
Apr 3 '17 at 3:23
$begingroup$
Also complex matrices can always be put on a form with either 1s or 0s off the main diagonal and eigenvalues on the diagonal. For real matrices you can be sure to not have to have more than at most blocks of size 2x2 along the diagonal.
$endgroup$
– mathreadler
Apr 3 '17 at 9:11
$begingroup$
Also complex matrices can always be put on a form with either 1s or 0s off the main diagonal and eigenvalues on the diagonal. For real matrices you can be sure to not have to have more than at most blocks of size 2x2 along the diagonal.
$endgroup$
– mathreadler
Apr 3 '17 at 9:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint A matrix $A$ with geometric multiplicity equal to its algebraic multiplicity is diagonalizable, so any nondiagonalizable $2 times 2$ matrix must have a single eigenvalue, say, $lambda$ of algebraic multiplicity $2$ but geometric multiplicity $1$.
Additional hint If some invertible matrix $P$ diagonalized $A - lambda I$, it would diagonalize $A$, a contradiction. Thus, if $A$ is an example, so is $A - lambda I$, which has eigenvalue $0$ of algebraic multiplicity $2$, and thus we may as well restrict our search to the case $lambda = 0$.
Since $A$ satisfies its own characteristic polynomial, $A^2 = 0$, but we cannot have $A = 0$, because $0$ is diagonalizable. Thus, there is a vector ${bf v} in Bbb R^2$ such that $A {bf v} neq 0$, so $B = (A {bf v}, {bf v})$ is a basis of $Bbb R^2$. With respect to $B$, the transformation matrix is $$J = pmatrix{0&1\0&0} .$$ (This is the $2 times 2$ Jordan block of eigenvalue $0$.) On the other hand, the sole eigenspace of $J$ is spanned by $pmatrix{1\0}$, so $J$ and thus $A$ has geometric multiplicity $1$. It follows from our construction that up to similarity and addition of multiples of $I$ this example is unique.
answered Dec 20 '18 at 7:45
TravisTravis
63.5k769150
$endgroup$
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%2f2215444%2fconstruct-a-2x2-matrix-with-real-eigenvalues-that-is-not-diagonalizable%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
$begingroup$
Hint A matrix $A$ with geometric multiplicity equal to its algebraic multiplicity is diagonalizable, so any nondiagonalizable $2 times 2$ matrix must have a single eigenvalue, say, $lambda$ of algebraic multiplicity $2$ but geometric multiplicity $1$.
Additional hint If some invertible matrix $P$ diagonalized $A - lambda I$, it would diagonalize $A$, a contradiction. Thus, if $A$ is an example, so is $A - lambda I$, which has eigenvalue $0$ of algebraic multiplicity $2$, and thus we may as well restrict our search to the case $lambda = 0$.
Since $A$ satisfies its own characteristic polynomial, $A^2 = 0$, but we cannot have $A = 0$, because $0$ is diagonalizable. Thus, there is a vector ${bf v} in Bbb R^2$ such that $A {bf v} neq 0$, so $B = (A {bf v}, {bf v})$ is a basis of $Bbb R^2$. With respect to $B$, the transformation matrix is $$J = pmatrix{0&1\0&0} .$$ (This is the $2 times 2$ Jordan block of eigenvalue $0$.) On the other hand, the sole eigenspace of $J$ is spanned by $pmatrix{1\0}$, so $J$ and thus $A$ has geometric multiplicity $1$. It follows from our construction that up to similarity and addition of multiples of $I$ this example is unique.
answered Dec 20 '18 at 7:45
TravisTravis
63.5k769150
$endgroup$
add a comment |
$begingroup$
Hint A matrix $A$ with geometric multiplicity equal to its algebraic multiplicity is diagonalizable, so any nondiagonalizable $2 times 2$ matrix must have a single eigenvalue, say, $lambda$ of algebraic multiplicity $2$ but geometric multiplicity $1$.
Additional hint If some invertible matrix $P$ diagonalized $A - lambda I$, it would diagonalize $A$, a contradiction. Thus, if $A$ is an example, so is $A - lambda I$, which has eigenvalue $0$ of algebraic multiplicity $2$, and thus we may as well restrict our search to the case $lambda = 0$.
Since $A$ satisfies its own characteristic polynomial, $A^2 = 0$, but we cannot have $A = 0$, because $0$ is diagonalizable. Thus, there is a vector ${bf v} in Bbb R^2$ such that $A {bf v} neq 0$, so $B = (A {bf v}, {bf v})$ is a basis of $Bbb R^2$. With respect to $B$, the transformation matrix is $$J = pmatrix{0&1\0&0} .$$ (This is the $2 times 2$ Jordan block of eigenvalue $0$.) On the other hand, the sole eigenspace of $J$ is spanned by $pmatrix{1\0}$, so $J$ and thus $A$ has geometric multiplicity $1$. It follows from our construction that up to similarity and addition of multiples of $I$ this example is unique.
answered Dec 20 '18 at 7:45
TravisTravis
63.5k769150
$endgroup$
add a comment |
$begingroup$
Hint A matrix $A$ with geometric multiplicity equal to its algebraic multiplicity is diagonalizable, so any nondiagonalizable $2 times 2$ matrix must have a single eigenvalue, say, $lambda$ of algebraic multiplicity $2$ but geometric multiplicity $1$.
Additional hint If some invertible matrix $P$ diagonalized $A - lambda I$, it would diagonalize $A$, a contradiction. Thus, if $A$ is an example, so is $A - lambda I$, which has eigenvalue $0$ of algebraic multiplicity $2$, and thus we may as well restrict our search to the case $lambda = 0$.
Since $A$ satisfies its own characteristic polynomial, $A^2 = 0$, but we cannot have $A = 0$, because $0$ is diagonalizable. Thus, there is a vector ${bf v} in Bbb R^2$ such that $A {bf v} neq 0$, so $B = (A {bf v}, {bf v})$ is a basis of $Bbb R^2$. With respect to $B$, the transformation matrix is $$J = pmatrix{0&1\0&0} .$$ (This is the $2 times 2$ Jordan block of eigenvalue $0$.) On the other hand, the sole eigenspace of $J$ is spanned by $pmatrix{1\0}$, so $J$ and thus $A$ has geometric multiplicity $1$. It follows from our construction that up to similarity and addition of multiples of $I$ this example is unique.
answered Dec 20 '18 at 7:45
TravisTravis
63.5k769150
$endgroup$
Hint A matrix $A$ with geometric multiplicity equal to its algebraic multiplicity is diagonalizable, so any nondiagonalizable $2 times 2$ matrix must have a single eigenvalue, say, $lambda$ of algebraic multiplicity $2$ but geometric multiplicity $1$.
Additional hint If some invertible matrix $P$ diagonalized $A - lambda I$, it would diagonalize $A$, a contradiction. Thus, if $A$ is an example, so is $A - lambda I$, which has eigenvalue $0$ of algebraic multiplicity $2$, and thus we may as well restrict our search to the case $lambda = 0$.
Since $A$ satisfies its own characteristic polynomial, $A^2 = 0$, but we cannot have $A = 0$, because $0$ is diagonalizable. Thus, there is a vector ${bf v} in Bbb R^2$ such that $A {bf v} neq 0$, so $B = (A {bf v}, {bf v})$ is a basis of $Bbb R^2$. With respect to $B$, the transformation matrix is $$J = pmatrix{0&1\0&0} .$$ (This is the $2 times 2$ Jordan block of eigenvalue $0$.) On the other hand, the sole eigenspace of $J$ is spanned by $pmatrix{1\0}$, so $J$ and thus $A$ has geometric multiplicity $1$. It follows from our construction that up to similarity and addition of multiples of $I$ this example is unique.
answered Dec 20 '18 at 7:45
TravisTravis
63.5k769150
answered Dec 20 '18 at 7:45
TravisTravis
63.5k769150
answered Dec 20 '18 at 7:45
TravisTravis
63.5k769150
answered Dec 20 '18 at 7:45
TravisTravis
63.5k769150
63.5k769150
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%2f2215444%2fconstruct-a-2x2-matrix-with-real-eigenvalues-that-is-not-diagonalizable%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
4
$begingroup$
You might want to consider a $2times 2$ Jordan block.
$endgroup$
– thanasissdr
Apr 3 '17 at 3:05
2
$begingroup$
What you think you know might not be correct.
$endgroup$
– Brian Borchers
Apr 3 '17 at 3:08
$begingroup$
For it not to be diagonalizable all you need is to have an eigenvalue with multiplicity greater than the dimension of the eigenspace. (i.e not enough eigenvectors). This is not related to independence
$endgroup$
– B.A
Apr 3 '17 at 3:23
$begingroup$
Also complex matrices can always be put on a form with either 1s or 0s off the main diagonal and eigenvalues on the diagonal. For real matrices you can be sure to not have to have more than at most blocks of size 2x2 along the diagonal.
$endgroup$
– mathreadler
Apr 3 '17 at 9:11