Mixing
How many realizations are there then in a structured set of matrices yielding characteristic polynomial...
$begingroup$
Let us consider a subset $S$ of $M_4(mathbb R)$ which has following form
begin{align*}
begin{pmatrix}
0 & * & 0 & * \
1 & * & 0 & * \
0 & * & 0 & * \
0 & * & 1 & *
end{pmatrix},
end{align*}
where $*$ can assume any real number.
Is there a systemic way to determine how many realizations there are of $p(t) = (t+1)^4$ in $S$ up to Jordan forms? That is, I would like to know whether
all Jordan blocks with diagonals to be $-1$'s are realizable in $S$? Obviously,
begin{align*}
J_1 = begin{pmatrix}
-1 & 0 & 0 & 0 \
1 & -1 & 0 & 0 \
0 & 0 & -1 & 0 \
0 & 0 & 1 & -1
end{pmatrix}, quad J_2 = begin{pmatrix}
-1 & 0 & 0 & 0 \
1 & -1 & 0 & 0 \
0 & 1 & -1 & 0 \
0 & 0 & 1 & -1
end{pmatrix}
end{align*}
are realizable by choosing
begin{align*}
A_1 = begin{pmatrix}
0 & -1 & 0 & 0 \
1 & -2 & 0 & 0 \
0 & 0 & 0 & -1\
0 & 0 & 1 & -2
end{pmatrix}, quad
A_2 = begin{pmatrix}
0 & 0 & 0 & -1 \
1 & 0 & 0 & -4 \
0 & 1 & 0 & -6\
0 & 0 & 1 & -4
end{pmatrix},
end{align*}
For other cases, I am essentially experimenting and try. But with no luck to get any result.
linear-algebra matrices jordan-normal-form
asked Aug 6 '18 at 18:20
MyCindy2012MyCindy2012
9210
$endgroup$
|
show 1 more comment
$begingroup$
Let us consider a subset $S$ of $M_4(mathbb R)$ which has following form
begin{align*}
begin{pmatrix}
0 & * & 0 & * \
1 & * & 0 & * \
0 & * & 0 & * \
0 & * & 1 & *
end{pmatrix},
end{align*}
where $*$ can assume any real number.
Is there a systemic way to determine how many realizations there are of $p(t) = (t+1)^4$ in $S$ up to Jordan forms? That is, I would like to know whether
all Jordan blocks with diagonals to be $-1$'s are realizable in $S$? Obviously,
begin{align*}
J_1 = begin{pmatrix}
-1 & 0 & 0 & 0 \
1 & -1 & 0 & 0 \
0 & 0 & -1 & 0 \
0 & 0 & 1 & -1
end{pmatrix}, quad J_2 = begin{pmatrix}
-1 & 0 & 0 & 0 \
1 & -1 & 0 & 0 \
0 & 1 & -1 & 0 \
0 & 0 & 1 & -1
end{pmatrix}
end{align*}
are realizable by choosing
begin{align*}
A_1 = begin{pmatrix}
0 & -1 & 0 & 0 \
1 & -2 & 0 & 0 \
0 & 0 & 0 & -1\
0 & 0 & 1 & -2
end{pmatrix}, quad
A_2 = begin{pmatrix}
0 & 0 & 0 & -1 \
1 & 0 & 0 & -4 \
0 & 1 & 0 & -6\
0 & 0 & 1 & -4
end{pmatrix},
end{align*}
For other cases, I am essentially experimenting and try. But with no luck to get any result.
linear-algebra matrices jordan-normal-form
asked Aug 6 '18 at 18:20
MyCindy2012MyCindy2012
9210
$endgroup$
$begingroup$
You mean Jordan blocks with diagonals to be all $-1$, right?
$endgroup$
– amarney
Aug 6 '18 at 18:54
$begingroup$
Yes. Thanks for spotting the typo.
$endgroup$
– MyCindy2012
Aug 6 '18 at 18:59
1
$begingroup$
My guess is to look at geometric multiplicities of your eigenvalues, since they are what determines the size and number of your Jordan blocks. Maybe look at the equation begin{align} (A - lambda I)x = begin{bmatrix} -lambda & * & 0 & * \ 1 & *-lambda & 0 & * \ 0 & * & - lambda & * \ 0 & * & 1 & *-lambda end{bmatrix}x =0. end{align}
$endgroup$
– amarney
Aug 6 '18 at 19:06
$begingroup$
I edited my partial answer and it is now a full one. Using that the diagonal part of the Jordan form commutes with everything is a very handy feature here, and it simplifies the calculations so that they can be carried out explicitly :)
$endgroup$
– Lukas Miristwhisky
Aug 6 '18 at 20:45
$begingroup$
Update: I noticed my "Case 3" was wrong, because I assumed that a nilpotent matrix of rank 2 has $N^2 = 0$. This is wrong, as can be seen in the case of a Jordan block of size 3. Instead, I found an example for this case. It should now be complete and correct. Sorry!
$endgroup$
– Lukas Miristwhisky
Aug 7 '18 at 12:32
|
show 1 more comment
$begingroup$
Let us consider a subset $S$ of $M_4(mathbb R)$ which has following form
begin{align*}
begin{pmatrix}
0 & * & 0 & * \
1 & * & 0 & * \
0 & * & 0 & * \
0 & * & 1 & *
end{pmatrix},
end{align*}
where $*$ can assume any real number.
Is there a systemic way to determine how many realizations there are of $p(t) = (t+1)^4$ in $S$ up to Jordan forms? That is, I would like to know whether
all Jordan blocks with diagonals to be $-1$'s are realizable in $S$? Obviously,
begin{align*}
J_1 = begin{pmatrix}
-1 & 0 & 0 & 0 \
1 & -1 & 0 & 0 \
0 & 0 & -1 & 0 \
0 & 0 & 1 & -1
end{pmatrix}, quad J_2 = begin{pmatrix}
-1 & 0 & 0 & 0 \
1 & -1 & 0 & 0 \
0 & 1 & -1 & 0 \
0 & 0 & 1 & -1
end{pmatrix}
end{align*}
are realizable by choosing
begin{align*}
A_1 = begin{pmatrix}
0 & -1 & 0 & 0 \
1 & -2 & 0 & 0 \
0 & 0 & 0 & -1\
0 & 0 & 1 & -2
end{pmatrix}, quad
A_2 = begin{pmatrix}
0 & 0 & 0 & -1 \
1 & 0 & 0 & -4 \
0 & 1 & 0 & -6\
0 & 0 & 1 & -4
end{pmatrix},
end{align*}
For other cases, I am essentially experimenting and try. But with no luck to get any result.
linear-algebra matrices jordan-normal-form
asked Aug 6 '18 at 18:20
MyCindy2012MyCindy2012
9210
$endgroup$
Let us consider a subset $S$ of $M_4(mathbb R)$ which has following form
begin{align*}
begin{pmatrix}
0 & * & 0 & * \
1 & * & 0 & * \
0 & * & 0 & * \
0 & * & 1 & *
end{pmatrix},
end{align*}
where $*$ can assume any real number.
Is there a systemic way to determine how many realizations there are of $p(t) = (t+1)^4$ in $S$ up to Jordan forms? That is, I would like to know whether
all Jordan blocks with diagonals to be $-1$'s are realizable in $S$? Obviously,
begin{align*}
J_1 = begin{pmatrix}
-1 & 0 & 0 & 0 \
1 & -1 & 0 & 0 \
0 & 0 & -1 & 0 \
0 & 0 & 1 & -1
end{pmatrix}, quad J_2 = begin{pmatrix}
-1 & 0 & 0 & 0 \
1 & -1 & 0 & 0 \
0 & 1 & -1 & 0 \
0 & 0 & 1 & -1
end{pmatrix}
end{align*}
are realizable by choosing
begin{align*}
A_1 = begin{pmatrix}
0 & -1 & 0 & 0 \
1 & -2 & 0 & 0 \
0 & 0 & 0 & -1\
0 & 0 & 1 & -2
end{pmatrix}, quad
A_2 = begin{pmatrix}
0 & 0 & 0 & -1 \
1 & 0 & 0 & -4 \
0 & 1 & 0 & -6\
0 & 0 & 1 & -4
end{pmatrix},
end{align*}
For other cases, I am essentially experimenting and try. But with no luck to get any result.
linear-algebra matrices jordan-normal-form
linear-algebra matrices jordan-normal-form
asked Aug 6 '18 at 18:20
MyCindy2012MyCindy2012
9210
asked Aug 6 '18 at 18:20
MyCindy2012MyCindy2012
9210
edited Aug 6 '18 at 18:59
MyCindy2012
asked Aug 6 '18 at 18:20
MyCindy2012MyCindy2012
9210
asked Aug 6 '18 at 18:20
MyCindy2012MyCindy2012
9210
asked Aug 6 '18 at 18:20
MyCindy2012MyCindy2012
9210
9210
$begingroup$
You mean Jordan blocks with diagonals to be all $-1$, right?
$endgroup$
– amarney
Aug 6 '18 at 18:54
$begingroup$
Yes. Thanks for spotting the typo.
$endgroup$
– MyCindy2012
Aug 6 '18 at 18:59
1
$begingroup$
My guess is to look at geometric multiplicities of your eigenvalues, since they are what determines the size and number of your Jordan blocks. Maybe look at the equation begin{align} (A - lambda I)x = begin{bmatrix} -lambda & * & 0 & * \ 1 & *-lambda & 0 & * \ 0 & * & - lambda & * \ 0 & * & 1 & *-lambda end{bmatrix}x =0. end{align}
$endgroup$
– amarney
Aug 6 '18 at 19:06
$begingroup$
I edited my partial answer and it is now a full one. Using that the diagonal part of the Jordan form commutes with everything is a very handy feature here, and it simplifies the calculations so that they can be carried out explicitly :)
$endgroup$
– Lukas Miristwhisky
Aug 6 '18 at 20:45
$begingroup$
Update: I noticed my "Case 3" was wrong, because I assumed that a nilpotent matrix of rank 2 has $N^2 = 0$. This is wrong, as can be seen in the case of a Jordan block of size 3. Instead, I found an example for this case. It should now be complete and correct. Sorry!
$endgroup$
– Lukas Miristwhisky
Aug 7 '18 at 12:32
|
show 1 more comment
$begingroup$
You mean Jordan blocks with diagonals to be all $-1$, right?
$endgroup$
– amarney
Aug 6 '18 at 18:54
$begingroup$
Yes. Thanks for spotting the typo.
$endgroup$
– MyCindy2012
Aug 6 '18 at 18:59
1
$begingroup$
My guess is to look at geometric multiplicities of your eigenvalues, since they are what determines the size and number of your Jordan blocks. Maybe look at the equation begin{align} (A - lambda I)x = begin{bmatrix} -lambda & * & 0 & * \ 1 & *-lambda & 0 & * \ 0 & * & - lambda & * \ 0 & * & 1 & *-lambda end{bmatrix}x =0. end{align}
$endgroup$
– amarney
Aug 6 '18 at 19:06
$begingroup$
I edited my partial answer and it is now a full one. Using that the diagonal part of the Jordan form commutes with everything is a very handy feature here, and it simplifies the calculations so that they can be carried out explicitly :)
$endgroup$
– Lukas Miristwhisky
Aug 6 '18 at 20:45
$begingroup$
Update: I noticed my "Case 3" was wrong, because I assumed that a nilpotent matrix of rank 2 has $N^2 = 0$. This is wrong, as can be seen in the case of a Jordan block of size 3. Instead, I found an example for this case. It should now be complete and correct. Sorry!
$endgroup$
– Lukas Miristwhisky
Aug 7 '18 at 12:32
$begingroup$
You mean Jordan blocks with diagonals to be all $-1$, right?
$endgroup$
– amarney
Aug 6 '18 at 18:54
$begingroup$
You mean Jordan blocks with diagonals to be all $-1$, right?
$endgroup$
– amarney
Aug 6 '18 at 18:54
$begingroup$
Yes. Thanks for spotting the typo.
$endgroup$
– MyCindy2012
Aug 6 '18 at 18:59
$begingroup$
Yes. Thanks for spotting the typo.
$endgroup$
– MyCindy2012
Aug 6 '18 at 18:59
1
1
$begingroup$
My guess is to look at geometric multiplicities of your eigenvalues, since they are what determines the size and number of your Jordan blocks. Maybe look at the equation begin{align} (A - lambda I)x = begin{bmatrix} -lambda & * & 0 & * \ 1 & *-lambda & 0 & * \ 0 & * & - lambda & * \ 0 & * & 1 & *-lambda end{bmatrix}x =0. end{align}
$endgroup$
– amarney
Aug 6 '18 at 19:06
$begingroup$
My guess is to look at geometric multiplicities of your eigenvalues, since they are what determines the size and number of your Jordan blocks. Maybe look at the equation begin{align} (A - lambda I)x = begin{bmatrix} -lambda & * & 0 & * \ 1 & *-lambda & 0 & * \ 0 & * & - lambda & * \ 0 & * & 1 & *-lambda end{bmatrix}x =0. end{align}
$endgroup$
– amarney
Aug 6 '18 at 19:06
$begingroup$
I edited my partial answer and it is now a full one. Using that the diagonal part of the Jordan form commutes with everything is a very handy feature here, and it simplifies the calculations so that they can be carried out explicitly :)
$endgroup$
– Lukas Miristwhisky
Aug 6 '18 at 20:45
$begingroup$
I edited my partial answer and it is now a full one. Using that the diagonal part of the Jordan form commutes with everything is a very handy feature here, and it simplifies the calculations so that they can be carried out explicitly :)
$endgroup$
– Lukas Miristwhisky
Aug 6 '18 at 20:45
$begingroup$
Update: I noticed my "Case 3" was wrong, because I assumed that a nilpotent matrix of rank 2 has $N^2 = 0$. This is wrong, as can be seen in the case of a Jordan block of size 3. Instead, I found an example for this case. It should now be complete and correct. Sorry!
$endgroup$
– Lukas Miristwhisky
Aug 7 '18 at 12:32
$begingroup$
Update: I noticed my "Case 3" was wrong, because I assumed that a nilpotent matrix of rank 2 has $N^2 = 0$. This is wrong, as can be seen in the case of a Jordan block of size 3. Instead, I found an example for this case. It should now be complete and correct. Sorry!
$endgroup$
– Lukas Miristwhisky
Aug 7 '18 at 12:32
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
The Jordan form of a matrix of your desired shape with characteristic polynomial $(1 + t)^4$ has nilpotent part of rank 2 or higher, and all such Jordan forms can be realized by such a matrix.
Split up the Jordan form $J$ into a diagonal part $J_d$ and a lower triangular nilpotent part $N$:
$$J = J_d + N,$$
where $J_d = text{diag}(-1,-1,-1,-1)$. This is super handy, because we have the special case where $J_d$ commutes with every matrix. I am going to divide this up into a few cases depending on the rank of the matrix.
Case 1: $N = 0$. Then, it is not possible for $J$ to be similar to a matrix of your desired form; the only matrix similar to $J = J_d$ is $J_d$ itself, which cannot be represented in the form you seek.
Case 2: $N$ is a rank 1 matrix. A such representation is in this case also not possible: Let $M in mathbb{R}^{4 times 4}$ be a matrix of your desired shape and $M = A cdot J cdot A^{-1}$ with some matrix $A in GL_4(mathbb{R})$. This similarity would imply:
$$ M = J_d + A N A^{-1}.$$
Since $N$ is of rank 1, so is $A N A^{-1}$, so it can be represented in the form
$$ANA^{-1} = v cdot w^T$$
with vectors $v,w in mathbb{R}^4.$ Writing out the equality $M = J_d + v cdot w^T$ yields an incompatible set of equations in the first and third column: Taking the equation for the $(1,1)$-,$(3,1)$- and $(3,3)$-entries for example yields
$$1 = v_1 w_1, quad
0 = v_1 w_3, quad
1 = v_3 w_3.$$
From these follow $w_1 neq 0, w_3 neq 0$, and thus simultaneously $v_1 = 0$ and $v_1 = frac{1}{w_1}$.
Case 3: $N$ is of rank 2 with one Jordan block of size 3 and one of size 1. For this case, consider for example the matrix
$$begin{pmatrix}
0 &1 &0 &-2 \
1 &0 &0 &-2 \
0 &2 &0 &-3 \
0 &2 &1 &-4 \
end{pmatrix},$$
which is in your desired form and similar to a matrix with one Jordan block of size 3 and one of size 1. I give an explanation for how I found this example below. All other cases were already handled in the OP. $$tag*{$square$}$$
How I derived the last example: The case where $N$ possesses Jordan blocks of size 1 and 3 is equivalent to demanding $(M-J_d)^2 neq 0$ but $(M-J_d)^3 = 0$.
Remember:
$$M - J_d = begin{pmatrix}
1 &* &0 &* \
1 &* &0 &* \
0 &* &1 &* \
0 &* &1 &* \
end{pmatrix}.$$
Note that the matrix $M - J_d = ANA^{-1}$ is of rank 2, as $N$ is of rank 2, and $M - J_d$ already possesses two linearly independent vectors in column 1 and 3, so column 2 and 4 need to be linear combinations of those. As such, the most general form for $M - J_d$ is
$$M - J_d = begin{pmatrix}
1 &a &0 &c \
1 &a &0 &c \
0 &b &1 &d \
0 &b &1 &d \
end{pmatrix}$$
with parameters $a,b,c,d in mathbb{R}$. By writing the equations down explicitly, one easily receives that $(M - J_d)^2 = 0$ is equivalent to the conditions $a = d = -1, b = c = 0$.
The equations for $(M - J_d)^3 = 0$ are a fair bit more convoluted (click here), but you can play around with these entries a bit, and for example demand that $d = -a - 2$, from which you then simply get the easy condition
$$(a + 1)^2 + bc = 0.$$
If this is fulfilled, $(M - J_d)^3$ vanishes, and now you just need to pick the parameters in a way that $(M - J_d)^2$ does not vanish. In the matrix example I gave above, I, for example, picked the parameters $a = 1,, b = 2,, c = -2,, d = -3$, fulfilling the equation.
answered Aug 6 '18 at 19:50
Lukas MiristwhiskyLukas Miristwhisky
367112
$endgroup$
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%2f2874181%2fhow-many-realizations-are-there-then-in-a-structured-set-of-matrices-yielding-ch%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
$begingroup$
The Jordan form of a matrix of your desired shape with characteristic polynomial $(1 + t)^4$ has nilpotent part of rank 2 or higher, and all such Jordan forms can be realized by such a matrix.
Split up the Jordan form $J$ into a diagonal part $J_d$ and a lower triangular nilpotent part $N$:
$$J = J_d + N,$$
where $J_d = text{diag}(-1,-1,-1,-1)$. This is super handy, because we have the special case where $J_d$ commutes with every matrix. I am going to divide this up into a few cases depending on the rank of the matrix.
Case 1: $N = 0$. Then, it is not possible for $J$ to be similar to a matrix of your desired form; the only matrix similar to $J = J_d$ is $J_d$ itself, which cannot be represented in the form you seek.
Case 2: $N$ is a rank 1 matrix. A such representation is in this case also not possible: Let $M in mathbb{R}^{4 times 4}$ be a matrix of your desired shape and $M = A cdot J cdot A^{-1}$ with some matrix $A in GL_4(mathbb{R})$. This similarity would imply:
$$ M = J_d + A N A^{-1}.$$
Since $N$ is of rank 1, so is $A N A^{-1}$, so it can be represented in the form
$$ANA^{-1} = v cdot w^T$$
with vectors $v,w in mathbb{R}^4.$ Writing out the equality $M = J_d + v cdot w^T$ yields an incompatible set of equations in the first and third column: Taking the equation for the $(1,1)$-,$(3,1)$- and $(3,3)$-entries for example yields
$$1 = v_1 w_1, quad
0 = v_1 w_3, quad
1 = v_3 w_3.$$
From these follow $w_1 neq 0, w_3 neq 0$, and thus simultaneously $v_1 = 0$ and $v_1 = frac{1}{w_1}$.
Case 3: $N$ is of rank 2 with one Jordan block of size 3 and one of size 1. For this case, consider for example the matrix
$$begin{pmatrix}
0 &1 &0 &-2 \
1 &0 &0 &-2 \
0 &2 &0 &-3 \
0 &2 &1 &-4 \
end{pmatrix},$$
which is in your desired form and similar to a matrix with one Jordan block of size 3 and one of size 1. I give an explanation for how I found this example below. All other cases were already handled in the OP. $$tag*{$square$}$$
How I derived the last example: The case where $N$ possesses Jordan blocks of size 1 and 3 is equivalent to demanding $(M-J_d)^2 neq 0$ but $(M-J_d)^3 = 0$.
Remember:
$$M - J_d = begin{pmatrix}
1 &* &0 &* \
1 &* &0 &* \
0 &* &1 &* \
0 &* &1 &* \
end{pmatrix}.$$
Note that the matrix $M - J_d = ANA^{-1}$ is of rank 2, as $N$ is of rank 2, and $M - J_d$ already possesses two linearly independent vectors in column 1 and 3, so column 2 and 4 need to be linear combinations of those. As such, the most general form for $M - J_d$ is
$$M - J_d = begin{pmatrix}
1 &a &0 &c \
1 &a &0 &c \
0 &b &1 &d \
0 &b &1 &d \
end{pmatrix}$$
with parameters $a,b,c,d in mathbb{R}$. By writing the equations down explicitly, one easily receives that $(M - J_d)^2 = 0$ is equivalent to the conditions $a = d = -1, b = c = 0$.
The equations for $(M - J_d)^3 = 0$ are a fair bit more convoluted (click here), but you can play around with these entries a bit, and for example demand that $d = -a - 2$, from which you then simply get the easy condition
$$(a + 1)^2 + bc = 0.$$
If this is fulfilled, $(M - J_d)^3$ vanishes, and now you just need to pick the parameters in a way that $(M - J_d)^2$ does not vanish. In the matrix example I gave above, I, for example, picked the parameters $a = 1,, b = 2,, c = -2,, d = -3$, fulfilling the equation.
answered Aug 6 '18 at 19:50
Lukas MiristwhiskyLukas Miristwhisky
367112
$endgroup$
add a comment |
$begingroup$
The Jordan form of a matrix of your desired shape with characteristic polynomial $(1 + t)^4$ has nilpotent part of rank 2 or higher, and all such Jordan forms can be realized by such a matrix.
Split up the Jordan form $J$ into a diagonal part $J_d$ and a lower triangular nilpotent part $N$:
$$J = J_d + N,$$
where $J_d = text{diag}(-1,-1,-1,-1)$. This is super handy, because we have the special case where $J_d$ commutes with every matrix. I am going to divide this up into a few cases depending on the rank of the matrix.
Case 1: $N = 0$. Then, it is not possible for $J$ to be similar to a matrix of your desired form; the only matrix similar to $J = J_d$ is $J_d$ itself, which cannot be represented in the form you seek.
Case 2: $N$ is a rank 1 matrix. A such representation is in this case also not possible: Let $M in mathbb{R}^{4 times 4}$ be a matrix of your desired shape and $M = A cdot J cdot A^{-1}$ with some matrix $A in GL_4(mathbb{R})$. This similarity would imply:
$$ M = J_d + A N A^{-1}.$$
Since $N$ is of rank 1, so is $A N A^{-1}$, so it can be represented in the form
$$ANA^{-1} = v cdot w^T$$
with vectors $v,w in mathbb{R}^4.$ Writing out the equality $M = J_d + v cdot w^T$ yields an incompatible set of equations in the first and third column: Taking the equation for the $(1,1)$-,$(3,1)$- and $(3,3)$-entries for example yields
$$1 = v_1 w_1, quad
0 = v_1 w_3, quad
1 = v_3 w_3.$$
From these follow $w_1 neq 0, w_3 neq 0$, and thus simultaneously $v_1 = 0$ and $v_1 = frac{1}{w_1}$.
Case 3: $N$ is of rank 2 with one Jordan block of size 3 and one of size 1. For this case, consider for example the matrix
$$begin{pmatrix}
0 &1 &0 &-2 \
1 &0 &0 &-2 \
0 &2 &0 &-3 \
0 &2 &1 &-4 \
end{pmatrix},$$
which is in your desired form and similar to a matrix with one Jordan block of size 3 and one of size 1. I give an explanation for how I found this example below. All other cases were already handled in the OP. $$tag*{$square$}$$
How I derived the last example: The case where $N$ possesses Jordan blocks of size 1 and 3 is equivalent to demanding $(M-J_d)^2 neq 0$ but $(M-J_d)^3 = 0$.
Remember:
$$M - J_d = begin{pmatrix}
1 &* &0 &* \
1 &* &0 &* \
0 &* &1 &* \
0 &* &1 &* \
end{pmatrix}.$$
Note that the matrix $M - J_d = ANA^{-1}$ is of rank 2, as $N$ is of rank 2, and $M - J_d$ already possesses two linearly independent vectors in column 1 and 3, so column 2 and 4 need to be linear combinations of those. As such, the most general form for $M - J_d$ is
$$M - J_d = begin{pmatrix}
1 &a &0 &c \
1 &a &0 &c \
0 &b &1 &d \
0 &b &1 &d \
end{pmatrix}$$
with parameters $a,b,c,d in mathbb{R}$. By writing the equations down explicitly, one easily receives that $(M - J_d)^2 = 0$ is equivalent to the conditions $a = d = -1, b = c = 0$.
The equations for $(M - J_d)^3 = 0$ are a fair bit more convoluted (click here), but you can play around with these entries a bit, and for example demand that $d = -a - 2$, from which you then simply get the easy condition
$$(a + 1)^2 + bc = 0.$$
If this is fulfilled, $(M - J_d)^3$ vanishes, and now you just need to pick the parameters in a way that $(M - J_d)^2$ does not vanish. In the matrix example I gave above, I, for example, picked the parameters $a = 1,, b = 2,, c = -2,, d = -3$, fulfilling the equation.
answered Aug 6 '18 at 19:50
Lukas MiristwhiskyLukas Miristwhisky
367112
$endgroup$
add a comment |
$begingroup$
The Jordan form of a matrix of your desired shape with characteristic polynomial $(1 + t)^4$ has nilpotent part of rank 2 or higher, and all such Jordan forms can be realized by such a matrix.
Split up the Jordan form $J$ into a diagonal part $J_d$ and a lower triangular nilpotent part $N$:
$$J = J_d + N,$$
where $J_d = text{diag}(-1,-1,-1,-1)$. This is super handy, because we have the special case where $J_d$ commutes with every matrix. I am going to divide this up into a few cases depending on the rank of the matrix.
Case 1: $N = 0$. Then, it is not possible for $J$ to be similar to a matrix of your desired form; the only matrix similar to $J = J_d$ is $J_d$ itself, which cannot be represented in the form you seek.
Case 2: $N$ is a rank 1 matrix. A such representation is in this case also not possible: Let $M in mathbb{R}^{4 times 4}$ be a matrix of your desired shape and $M = A cdot J cdot A^{-1}$ with some matrix $A in GL_4(mathbb{R})$. This similarity would imply:
$$ M = J_d + A N A^{-1}.$$
Since $N$ is of rank 1, so is $A N A^{-1}$, so it can be represented in the form
$$ANA^{-1} = v cdot w^T$$
with vectors $v,w in mathbb{R}^4.$ Writing out the equality $M = J_d + v cdot w^T$ yields an incompatible set of equations in the first and third column: Taking the equation for the $(1,1)$-,$(3,1)$- and $(3,3)$-entries for example yields
$$1 = v_1 w_1, quad
0 = v_1 w_3, quad
1 = v_3 w_3.$$
From these follow $w_1 neq 0, w_3 neq 0$, and thus simultaneously $v_1 = 0$ and $v_1 = frac{1}{w_1}$.
Case 3: $N$ is of rank 2 with one Jordan block of size 3 and one of size 1. For this case, consider for example the matrix
$$begin{pmatrix}
0 &1 &0 &-2 \
1 &0 &0 &-2 \
0 &2 &0 &-3 \
0 &2 &1 &-4 \
end{pmatrix},$$
which is in your desired form and similar to a matrix with one Jordan block of size 3 and one of size 1. I give an explanation for how I found this example below. All other cases were already handled in the OP. $$tag*{$square$}$$
How I derived the last example: The case where $N$ possesses Jordan blocks of size 1 and 3 is equivalent to demanding $(M-J_d)^2 neq 0$ but $(M-J_d)^3 = 0$.
Remember:
$$M - J_d = begin{pmatrix}
1 &* &0 &* \
1 &* &0 &* \
0 &* &1 &* \
0 &* &1 &* \
end{pmatrix}.$$
Note that the matrix $M - J_d = ANA^{-1}$ is of rank 2, as $N$ is of rank 2, and $M - J_d$ already possesses two linearly independent vectors in column 1 and 3, so column 2 and 4 need to be linear combinations of those. As such, the most general form for $M - J_d$ is
$$M - J_d = begin{pmatrix}
1 &a &0 &c \
1 &a &0 &c \
0 &b &1 &d \
0 &b &1 &d \
end{pmatrix}$$
with parameters $a,b,c,d in mathbb{R}$. By writing the equations down explicitly, one easily receives that $(M - J_d)^2 = 0$ is equivalent to the conditions $a = d = -1, b = c = 0$.
The equations for $(M - J_d)^3 = 0$ are a fair bit more convoluted (click here), but you can play around with these entries a bit, and for example demand that $d = -a - 2$, from which you then simply get the easy condition
$$(a + 1)^2 + bc = 0.$$
If this is fulfilled, $(M - J_d)^3$ vanishes, and now you just need to pick the parameters in a way that $(M - J_d)^2$ does not vanish. In the matrix example I gave above, I, for example, picked the parameters $a = 1,, b = 2,, c = -2,, d = -3$, fulfilling the equation.
answered Aug 6 '18 at 19:50
Lukas MiristwhiskyLukas Miristwhisky
367112
$endgroup$
The Jordan form of a matrix of your desired shape with characteristic polynomial $(1 + t)^4$ has nilpotent part of rank 2 or higher, and all such Jordan forms can be realized by such a matrix.
Split up the Jordan form $J$ into a diagonal part $J_d$ and a lower triangular nilpotent part $N$:
$$J = J_d + N,$$
where $J_d = text{diag}(-1,-1,-1,-1)$. This is super handy, because we have the special case where $J_d$ commutes with every matrix. I am going to divide this up into a few cases depending on the rank of the matrix.
Case 1: $N = 0$. Then, it is not possible for $J$ to be similar to a matrix of your desired form; the only matrix similar to $J = J_d$ is $J_d$ itself, which cannot be represented in the form you seek.
Case 2: $N$ is a rank 1 matrix. A such representation is in this case also not possible: Let $M in mathbb{R}^{4 times 4}$ be a matrix of your desired shape and $M = A cdot J cdot A^{-1}$ with some matrix $A in GL_4(mathbb{R})$. This similarity would imply:
$$ M = J_d + A N A^{-1}.$$
Since $N$ is of rank 1, so is $A N A^{-1}$, so it can be represented in the form
$$ANA^{-1} = v cdot w^T$$
with vectors $v,w in mathbb{R}^4.$ Writing out the equality $M = J_d + v cdot w^T$ yields an incompatible set of equations in the first and third column: Taking the equation for the $(1,1)$-,$(3,1)$- and $(3,3)$-entries for example yields
$$1 = v_1 w_1, quad
0 = v_1 w_3, quad
1 = v_3 w_3.$$
From these follow $w_1 neq 0, w_3 neq 0$, and thus simultaneously $v_1 = 0$ and $v_1 = frac{1}{w_1}$.
Case 3: $N$ is of rank 2 with one Jordan block of size 3 and one of size 1. For this case, consider for example the matrix
$$begin{pmatrix}
0 &1 &0 &-2 \
1 &0 &0 &-2 \
0 &2 &0 &-3 \
0 &2 &1 &-4 \
end{pmatrix},$$
which is in your desired form and similar to a matrix with one Jordan block of size 3 and one of size 1. I give an explanation for how I found this example below. All other cases were already handled in the OP. $$tag*{$square$}$$
How I derived the last example: The case where $N$ possesses Jordan blocks of size 1 and 3 is equivalent to demanding $(M-J_d)^2 neq 0$ but $(M-J_d)^3 = 0$.
Remember:
$$M - J_d = begin{pmatrix}
1 &* &0 &* \
1 &* &0 &* \
0 &* &1 &* \
0 &* &1 &* \
end{pmatrix}.$$
Note that the matrix $M - J_d = ANA^{-1}$ is of rank 2, as $N$ is of rank 2, and $M - J_d$ already possesses two linearly independent vectors in column 1 and 3, so column 2 and 4 need to be linear combinations of those. As such, the most general form for $M - J_d$ is
$$M - J_d = begin{pmatrix}
1 &a &0 &c \
1 &a &0 &c \
0 &b &1 &d \
0 &b &1 &d \
end{pmatrix}$$
with parameters $a,b,c,d in mathbb{R}$. By writing the equations down explicitly, one easily receives that $(M - J_d)^2 = 0$ is equivalent to the conditions $a = d = -1, b = c = 0$.
The equations for $(M - J_d)^3 = 0$ are a fair bit more convoluted (click here), but you can play around with these entries a bit, and for example demand that $d = -a - 2$, from which you then simply get the easy condition
$$(a + 1)^2 + bc = 0.$$
If this is fulfilled, $(M - J_d)^3$ vanishes, and now you just need to pick the parameters in a way that $(M - J_d)^2$ does not vanish. In the matrix example I gave above, I, for example, picked the parameters $a = 1,, b = 2,, c = -2,, d = -3$, fulfilling the equation.
answered Aug 6 '18 at 19:50
Lukas MiristwhiskyLukas Miristwhisky
367112
edited Jan 25 at 9:46
answered Aug 6 '18 at 19:50
Lukas MiristwhiskyLukas Miristwhisky
367112
answered Aug 6 '18 at 19:50
Lukas MiristwhiskyLukas Miristwhisky
367112
answered Aug 6 '18 at 19:50
Lukas MiristwhiskyLukas Miristwhisky
367112
367112
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%2f2874181%2fhow-many-realizations-are-there-then-in-a-structured-set-of-matrices-yielding-ch%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
$begingroup$
You mean Jordan blocks with diagonals to be all $-1$, right?
$endgroup$
– amarney
Aug 6 '18 at 18:54
$begingroup$
Yes. Thanks for spotting the typo.
$endgroup$
– MyCindy2012
Aug 6 '18 at 18:59
1
$begingroup$
My guess is to look at geometric multiplicities of your eigenvalues, since they are what determines the size and number of your Jordan blocks. Maybe look at the equation begin{align} (A - lambda I)x = begin{bmatrix} -lambda & * & 0 & * \ 1 & *-lambda & 0 & * \ 0 & * & - lambda & * \ 0 & * & 1 & *-lambda end{bmatrix}x =0. end{align}
$endgroup$
– amarney
Aug 6 '18 at 19:06
$begingroup$
I edited my partial answer and it is now a full one. Using that the diagonal part of the Jordan form commutes with everything is a very handy feature here, and it simplifies the calculations so that they can be carried out explicitly :)
$endgroup$
– Lukas Miristwhisky
Aug 6 '18 at 20:45
$begingroup$
Update: I noticed my "Case 3" was wrong, because I assumed that a nilpotent matrix of rank 2 has $N^2 = 0$. This is wrong, as can be seen in the case of a Jordan block of size 3. Instead, I found an example for this case. It should now be complete and correct. Sorry!
$endgroup$
– Lukas Miristwhisky
Aug 7 '18 at 12:32