Diagonalizability of $A in M_{n}(Bbb{R})$ (Upper traingular matrix) with all diagonal entries 1 and $A neq...
If $A in M_{n}(Bbb{R})$ is an Upper triangular matrix with diagonal entries $1$ such that $A neq I$, then what can we say about the diagonalizability of $A$ ?
I know that if the matrix has distinct eigenvalues or the set of eigenvectors are linearly independent then the matrix is diagonalizable.
And that the eigenvalues of the triangular matrices are given by the diagonal elements like here and here, but they work nocely if we had distinct elements on the main diagonal.
But in my case I have same value 1 on the main diagonal, how can I approach about the diagonalizability of the matrix?
linear-algebra matrices diagonalization
asked Dec 20 '17 at 4:37
BAYMAX
2,83221123
add a comment |
If $A in M_{n}(Bbb{R})$ is an Upper triangular matrix with diagonal entries $1$ such that $A neq I$, then what can we say about the diagonalizability of $A$ ?
I know that if the matrix has distinct eigenvalues or the set of eigenvectors are linearly independent then the matrix is diagonalizable.
And that the eigenvalues of the triangular matrices are given by the diagonal elements like here and here, but they work nocely if we had distinct elements on the main diagonal.
But in my case I have same value 1 on the main diagonal, how can I approach about the diagonalizability of the matrix?
linear-algebra matrices diagonalization
asked Dec 20 '17 at 4:37
BAYMAX
2,83221123
It depends. Refer to: en.wikipedia.org/wiki/Jordan_normal_form
– Brian Cheung
Dec 20 '17 at 4:39
add a comment |
If $A in M_{n}(Bbb{R})$ is an Upper triangular matrix with diagonal entries $1$ such that $A neq I$, then what can we say about the diagonalizability of $A$ ?
I know that if the matrix has distinct eigenvalues or the set of eigenvectors are linearly independent then the matrix is diagonalizable.
And that the eigenvalues of the triangular matrices are given by the diagonal elements like here and here, but they work nocely if we had distinct elements on the main diagonal.
But in my case I have same value 1 on the main diagonal, how can I approach about the diagonalizability of the matrix?
linear-algebra matrices diagonalization
asked Dec 20 '17 at 4:37
BAYMAX
2,83221123
If $A in M_{n}(Bbb{R})$ is an Upper triangular matrix with diagonal entries $1$ such that $A neq I$, then what can we say about the diagonalizability of $A$ ?
I know that if the matrix has distinct eigenvalues or the set of eigenvectors are linearly independent then the matrix is diagonalizable.
And that the eigenvalues of the triangular matrices are given by the diagonal elements like here and here, but they work nocely if we had distinct elements on the main diagonal.
But in my case I have same value 1 on the main diagonal, how can I approach about the diagonalizability of the matrix?
linear-algebra matrices diagonalization
linear-algebra matrices diagonalization
asked Dec 20 '17 at 4:37
BAYMAX
2,83221123
asked Dec 20 '17 at 4:37
BAYMAX
2,83221123
asked Dec 20 '17 at 4:37
BAYMAX
2,83221123
asked Dec 20 '17 at 4:37
BAYMAX
2,83221123
asked Dec 20 '17 at 4:37
BAYMAX
2,83221123
2,83221123
It depends. Refer to: en.wikipedia.org/wiki/Jordan_normal_form
– Brian Cheung
Dec 20 '17 at 4:39
add a comment |
It depends. Refer to: en.wikipedia.org/wiki/Jordan_normal_form
– Brian Cheung
Dec 20 '17 at 4:39
It depends. Refer to: en.wikipedia.org/wiki/Jordan_normal_form
– Brian Cheung
Dec 20 '17 at 4:39
It depends. Refer to: en.wikipedia.org/wiki/Jordan_normal_form
– Brian Cheung
Dec 20 '17 at 4:39
add a comment |
2 Answers
2
active
oldest
votes
It isn't diagonalizable. Your matrix is of the form $I+N$ where $I$ is the identity matrix and $N$ is strictly upper triangular. You can check that being strictly upper triangular means that $N^n=0$, since if the $e_i$ are the basis vectors, and $E_m$ is the subspace spanned by $e_1,ldots,e_m$, with $E_0=0$ then you can check that $NE_m subseteq E_{m-1}$, so $N^nE_n = E_0=0$. However since $Ane I$, $Nne 0$. Now suppose $M$ diagonalizes $A$, so $MAM^{-1} = D$ for $D$ a diagonal matrix. Then $M(I+N)M^{-1}=MIM^{-1}+MNM^{-1} = D$, but $MIM^{-1}=I$, so we in fact have $MNM^{-1}=D-I$, so in fact $N$ must be diagonalizable. However we still have $(MNM^{-1})^n=0$, so if $D-I$ has elements $lambda_1,ldots,lambda_n$ on the diagonal, then $lambda_i^n=0$, which implies $lambda_i=0$, hence $D-I=0$. Thus we would have that $D=I$, and $N=0$, contradicting the assumption that $Ane I$, so $Nne 0$.
answered Dec 20 '17 at 4:46
jgon
12.4k21940
$N^{n} = 0$ for some $n$ means that it is Idempotent , then does this mean that we can give Upper triangular matrices as examples of Idempotent matrices?
– BAYMAX
Dec 20 '17 at 5:43
Err, $N$ is nilpotent (nilpotence means that some power of something is 0). Idempotence is the property that $P^2=P$, where say $P$ is some matrix. Examples of idempotent operators are projections (say $(a,b)mapsto (a,0)$). But strictly upper triangular matrices are nilpotent, if that's the question you were asking, then yes.
– jgon
Dec 20 '17 at 5:45
Here $n$ is the order of the matrix ? then is it possible that $N^{k} =0$ for $k < n$?
– BAYMAX
Dec 20 '17 at 5:47
Here $n$ is the dimension of the vector space. It is certainly possible that $N^k=0$ for $k<n$, but we can guarantee that it will become 0 by the time we reach $N^n$. An example with $N^k = 0$ for $k<n$ is $begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ end{pmatrix}$, which squares to 0.
– jgon
Dec 20 '17 at 5:48
add a comment |
Since the matrix is upper triangular with all diagonal entries $1$, its characteristic equation is $(1-lambda)^n=0$. This give an eigenvalue of $1$ with algebraic multiplicity $n$. Also, finding the corresponding eigenspace through the null space of $A-lambda I$ tells us that it is generated by a single vector $[1,0,...,0]^T$. Thus its geometric multiplicity is $1$, different from $n$ and hence it is not diagonalizable.
answered Nov 30 at 15:47
Rhaldryn
192213
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%2f2574046%2fdiagonalizability-of-a-in-m-n-bbbr-upper-traingular-matrix-with-all-d%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
It isn't diagonalizable. Your matrix is of the form $I+N$ where $I$ is the identity matrix and $N$ is strictly upper triangular. You can check that being strictly upper triangular means that $N^n=0$, since if the $e_i$ are the basis vectors, and $E_m$ is the subspace spanned by $e_1,ldots,e_m$, with $E_0=0$ then you can check that $NE_m subseteq E_{m-1}$, so $N^nE_n = E_0=0$. However since $Ane I$, $Nne 0$. Now suppose $M$ diagonalizes $A$, so $MAM^{-1} = D$ for $D$ a diagonal matrix. Then $M(I+N)M^{-1}=MIM^{-1}+MNM^{-1} = D$, but $MIM^{-1}=I$, so we in fact have $MNM^{-1}=D-I$, so in fact $N$ must be diagonalizable. However we still have $(MNM^{-1})^n=0$, so if $D-I$ has elements $lambda_1,ldots,lambda_n$ on the diagonal, then $lambda_i^n=0$, which implies $lambda_i=0$, hence $D-I=0$. Thus we would have that $D=I$, and $N=0$, contradicting the assumption that $Ane I$, so $Nne 0$.
answered Dec 20 '17 at 4:46
jgon
12.4k21940
$N^{n} = 0$ for some $n$ means that it is Idempotent , then does this mean that we can give Upper triangular matrices as examples of Idempotent matrices?
– BAYMAX
Dec 20 '17 at 5:43
Err, $N$ is nilpotent (nilpotence means that some power of something is 0). Idempotence is the property that $P^2=P$, where say $P$ is some matrix. Examples of idempotent operators are projections (say $(a,b)mapsto (a,0)$). But strictly upper triangular matrices are nilpotent, if that's the question you were asking, then yes.
– jgon
Dec 20 '17 at 5:45
Here $n$ is the order of the matrix ? then is it possible that $N^{k} =0$ for $k < n$?
– BAYMAX
Dec 20 '17 at 5:47
Here $n$ is the dimension of the vector space. It is certainly possible that $N^k=0$ for $k<n$, but we can guarantee that it will become 0 by the time we reach $N^n$. An example with $N^k = 0$ for $k<n$ is $begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ end{pmatrix}$, which squares to 0.
– jgon
Dec 20 '17 at 5:48
add a comment |
It isn't diagonalizable. Your matrix is of the form $I+N$ where $I$ is the identity matrix and $N$ is strictly upper triangular. You can check that being strictly upper triangular means that $N^n=0$, since if the $e_i$ are the basis vectors, and $E_m$ is the subspace spanned by $e_1,ldots,e_m$, with $E_0=0$ then you can check that $NE_m subseteq E_{m-1}$, so $N^nE_n = E_0=0$. However since $Ane I$, $Nne 0$. Now suppose $M$ diagonalizes $A$, so $MAM^{-1} = D$ for $D$ a diagonal matrix. Then $M(I+N)M^{-1}=MIM^{-1}+MNM^{-1} = D$, but $MIM^{-1}=I$, so we in fact have $MNM^{-1}=D-I$, so in fact $N$ must be diagonalizable. However we still have $(MNM^{-1})^n=0$, so if $D-I$ has elements $lambda_1,ldots,lambda_n$ on the diagonal, then $lambda_i^n=0$, which implies $lambda_i=0$, hence $D-I=0$. Thus we would have that $D=I$, and $N=0$, contradicting the assumption that $Ane I$, so $Nne 0$.
answered Dec 20 '17 at 4:46
jgon
12.4k21940
$N^{n} = 0$ for some $n$ means that it is Idempotent , then does this mean that we can give Upper triangular matrices as examples of Idempotent matrices?
– BAYMAX
Dec 20 '17 at 5:43
Err, $N$ is nilpotent (nilpotence means that some power of something is 0). Idempotence is the property that $P^2=P$, where say $P$ is some matrix. Examples of idempotent operators are projections (say $(a,b)mapsto (a,0)$). But strictly upper triangular matrices are nilpotent, if that's the question you were asking, then yes.
– jgon
Dec 20 '17 at 5:45
Here $n$ is the order of the matrix ? then is it possible that $N^{k} =0$ for $k < n$?
– BAYMAX
Dec 20 '17 at 5:47
Here $n$ is the dimension of the vector space. It is certainly possible that $N^k=0$ for $k<n$, but we can guarantee that it will become 0 by the time we reach $N^n$. An example with $N^k = 0$ for $k<n$ is $begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ end{pmatrix}$, which squares to 0.
– jgon
Dec 20 '17 at 5:48
add a comment |
It isn't diagonalizable. Your matrix is of the form $I+N$ where $I$ is the identity matrix and $N$ is strictly upper triangular. You can check that being strictly upper triangular means that $N^n=0$, since if the $e_i$ are the basis vectors, and $E_m$ is the subspace spanned by $e_1,ldots,e_m$, with $E_0=0$ then you can check that $NE_m subseteq E_{m-1}$, so $N^nE_n = E_0=0$. However since $Ane I$, $Nne 0$. Now suppose $M$ diagonalizes $A$, so $MAM^{-1} = D$ for $D$ a diagonal matrix. Then $M(I+N)M^{-1}=MIM^{-1}+MNM^{-1} = D$, but $MIM^{-1}=I$, so we in fact have $MNM^{-1}=D-I$, so in fact $N$ must be diagonalizable. However we still have $(MNM^{-1})^n=0$, so if $D-I$ has elements $lambda_1,ldots,lambda_n$ on the diagonal, then $lambda_i^n=0$, which implies $lambda_i=0$, hence $D-I=0$. Thus we would have that $D=I$, and $N=0$, contradicting the assumption that $Ane I$, so $Nne 0$.
answered Dec 20 '17 at 4:46
jgon
12.4k21940
It isn't diagonalizable. Your matrix is of the form $I+N$ where $I$ is the identity matrix and $N$ is strictly upper triangular. You can check that being strictly upper triangular means that $N^n=0$, since if the $e_i$ are the basis vectors, and $E_m$ is the subspace spanned by $e_1,ldots,e_m$, with $E_0=0$ then you can check that $NE_m subseteq E_{m-1}$, so $N^nE_n = E_0=0$. However since $Ane I$, $Nne 0$. Now suppose $M$ diagonalizes $A$, so $MAM^{-1} = D$ for $D$ a diagonal matrix. Then $M(I+N)M^{-1}=MIM^{-1}+MNM^{-1} = D$, but $MIM^{-1}=I$, so we in fact have $MNM^{-1}=D-I$, so in fact $N$ must be diagonalizable. However we still have $(MNM^{-1})^n=0$, so if $D-I$ has elements $lambda_1,ldots,lambda_n$ on the diagonal, then $lambda_i^n=0$, which implies $lambda_i=0$, hence $D-I=0$. Thus we would have that $D=I$, and $N=0$, contradicting the assumption that $Ane I$, so $Nne 0$.
answered Dec 20 '17 at 4:46
jgon
12.4k21940
answered Dec 20 '17 at 4:46
jgon
12.4k21940
answered Dec 20 '17 at 4:46
jgon
12.4k21940
answered Dec 20 '17 at 4:46
jgon
12.4k21940
12.4k21940
$N^{n} = 0$ for some $n$ means that it is Idempotent , then does this mean that we can give Upper triangular matrices as examples of Idempotent matrices?
– BAYMAX
Dec 20 '17 at 5:43
Err, $N$ is nilpotent (nilpotence means that some power of something is 0). Idempotence is the property that $P^2=P$, where say $P$ is some matrix. Examples of idempotent operators are projections (say $(a,b)mapsto (a,0)$). But strictly upper triangular matrices are nilpotent, if that's the question you were asking, then yes.
– jgon
Dec 20 '17 at 5:45
Here $n$ is the order of the matrix ? then is it possible that $N^{k} =0$ for $k < n$?
– BAYMAX
Dec 20 '17 at 5:47
Here $n$ is the dimension of the vector space. It is certainly possible that $N^k=0$ for $k<n$, but we can guarantee that it will become 0 by the time we reach $N^n$. An example with $N^k = 0$ for $k<n$ is $begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ end{pmatrix}$, which squares to 0.
– jgon
Dec 20 '17 at 5:48
add a comment |
$N^{n} = 0$ for some $n$ means that it is Idempotent , then does this mean that we can give Upper triangular matrices as examples of Idempotent matrices?
– BAYMAX
Dec 20 '17 at 5:43
Err, $N$ is nilpotent (nilpotence means that some power of something is 0). Idempotence is the property that $P^2=P$, where say $P$ is some matrix. Examples of idempotent operators are projections (say $(a,b)mapsto (a,0)$). But strictly upper triangular matrices are nilpotent, if that's the question you were asking, then yes.
– jgon
Dec 20 '17 at 5:45
Here $n$ is the order of the matrix ? then is it possible that $N^{k} =0$ for $k < n$?
– BAYMAX
Dec 20 '17 at 5:47
Here $n$ is the dimension of the vector space. It is certainly possible that $N^k=0$ for $k<n$, but we can guarantee that it will become 0 by the time we reach $N^n$. An example with $N^k = 0$ for $k<n$ is $begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ end{pmatrix}$, which squares to 0.
– jgon
Dec 20 '17 at 5:48
$N^{n} = 0$ for some $n$ means that it is Idempotent , then does this mean that we can give Upper triangular matrices as examples of Idempotent matrices?
– BAYMAX
Dec 20 '17 at 5:43
$N^{n} = 0$ for some $n$ means that it is Idempotent , then does this mean that we can give Upper triangular matrices as examples of Idempotent matrices?
– BAYMAX
Dec 20 '17 at 5:43
Err, $N$ is nilpotent (nilpotence means that some power of something is 0). Idempotence is the property that $P^2=P$, where say $P$ is some matrix. Examples of idempotent operators are projections (say $(a,b)mapsto (a,0)$). But strictly upper triangular matrices are nilpotent, if that's the question you were asking, then yes.
– jgon
Dec 20 '17 at 5:45
Err, $N$ is nilpotent (nilpotence means that some power of something is 0). Idempotence is the property that $P^2=P$, where say $P$ is some matrix. Examples of idempotent operators are projections (say $(a,b)mapsto (a,0)$). But strictly upper triangular matrices are nilpotent, if that's the question you were asking, then yes.
– jgon
Dec 20 '17 at 5:45
Here $n$ is the order of the matrix ? then is it possible that $N^{k} =0$ for $k < n$?
– BAYMAX
Dec 20 '17 at 5:47
Here $n$ is the order of the matrix ? then is it possible that $N^{k} =0$ for $k < n$?
– BAYMAX
Dec 20 '17 at 5:47
Here $n$ is the dimension of the vector space. It is certainly possible that $N^k=0$ for $k<n$, but we can guarantee that it will become 0 by the time we reach $N^n$. An example with $N^k = 0$ for $k<n$ is $begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ end{pmatrix}$, which squares to 0.
– jgon
Dec 20 '17 at 5:48
Here $n$ is the dimension of the vector space. It is certainly possible that $N^k=0$ for $k<n$, but we can guarantee that it will become 0 by the time we reach $N^n$. An example with $N^k = 0$ for $k<n$ is $begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ end{pmatrix}$, which squares to 0.
– jgon
Dec 20 '17 at 5:48
add a comment |
Since the matrix is upper triangular with all diagonal entries $1$, its characteristic equation is $(1-lambda)^n=0$. This give an eigenvalue of $1$ with algebraic multiplicity $n$. Also, finding the corresponding eigenspace through the null space of $A-lambda I$ tells us that it is generated by a single vector $[1,0,...,0]^T$. Thus its geometric multiplicity is $1$, different from $n$ and hence it is not diagonalizable.
answered Nov 30 at 15:47
Rhaldryn
192213
add a comment |
Since the matrix is upper triangular with all diagonal entries $1$, its characteristic equation is $(1-lambda)^n=0$. This give an eigenvalue of $1$ with algebraic multiplicity $n$. Also, finding the corresponding eigenspace through the null space of $A-lambda I$ tells us that it is generated by a single vector $[1,0,...,0]^T$. Thus its geometric multiplicity is $1$, different from $n$ and hence it is not diagonalizable.
answered Nov 30 at 15:47
Rhaldryn
192213
add a comment |
Since the matrix is upper triangular with all diagonal entries $1$, its characteristic equation is $(1-lambda)^n=0$. This give an eigenvalue of $1$ with algebraic multiplicity $n$. Also, finding the corresponding eigenspace through the null space of $A-lambda I$ tells us that it is generated by a single vector $[1,0,...,0]^T$. Thus its geometric multiplicity is $1$, different from $n$ and hence it is not diagonalizable.
answered Nov 30 at 15:47
Rhaldryn
192213
Since the matrix is upper triangular with all diagonal entries $1$, its characteristic equation is $(1-lambda)^n=0$. This give an eigenvalue of $1$ with algebraic multiplicity $n$. Also, finding the corresponding eigenspace through the null space of $A-lambda I$ tells us that it is generated by a single vector $[1,0,...,0]^T$. Thus its geometric multiplicity is $1$, different from $n$ and hence it is not diagonalizable.
answered Nov 30 at 15:47
Rhaldryn
192213
answered Nov 30 at 15:47
Rhaldryn
192213
answered Nov 30 at 15:47
Rhaldryn
192213
answered Nov 30 at 15:47
Rhaldryn
192213
192213
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f2574046%2fdiagonalizability-of-a-in-m-n-bbbr-upper-traingular-matrix-with-all-d%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
It depends. Refer to: en.wikipedia.org/wiki/Jordan_normal_form
– Brian Cheung
Dec 20 '17 at 4:39