Linear Algebra : Jordan Canonical form (Jordan blocks and the Super-Diagonal)
$begingroup$
In terms of Jordan Canonical Form, and more specifically
about Jordan Blocks.
When there is a definition about Jordan Blocks they say the eigenvalues go on the principle diagonal and the diagonal above it(usually called the super-diagonal) contains the number 1!
I want to understand why there are 1's in the superdiagonal.
I have looked all over and no body tries to explain why there is this mysterious 1's in the super-diagonal.
I hope that someone here on this forum can explain this.
Here is a link to PDF:
http://ckottke.ncf.edu/docs/jordan.pdf
linear-algebra jordan-normal-form
asked Dec 22 '18 at 5:53
PaluPalu
3422822
$endgroup$
add a comment |
$begingroup$
In terms of Jordan Canonical Form, and more specifically
about Jordan Blocks.
When there is a definition about Jordan Blocks they say the eigenvalues go on the principle diagonal and the diagonal above it(usually called the super-diagonal) contains the number 1!
I want to understand why there are 1's in the superdiagonal.
I have looked all over and no body tries to explain why there is this mysterious 1's in the super-diagonal.
I hope that someone here on this forum can explain this.
Here is a link to PDF:
http://ckottke.ncf.edu/docs/jordan.pdf
linear-algebra jordan-normal-form
asked Dec 22 '18 at 5:53
PaluPalu
3422822
$endgroup$
$begingroup$
"I have looked all over": Have you looked at a proof that every (complex) matrix has a Jordan form?
$endgroup$
– David C. Ullrich
Dec 22 '18 at 16:03
$begingroup$
Hi David, i have seen things about complex matrix has Jordan form, but when they discuss it they assume the Ones on the Super Diagonal as though this is a fundamental axiom without need of explanation.
$endgroup$
– Palu
Dec 22 '18 at 19:57
$begingroup$
I just added a link to a pdf article I want to illustrate, how they explain things, where they just assume it to be a fact. See above.
$endgroup$
– Palu
Dec 22 '18 at 20:00
add a comment |
$begingroup$
In terms of Jordan Canonical Form, and more specifically
about Jordan Blocks.
When there is a definition about Jordan Blocks they say the eigenvalues go on the principle diagonal and the diagonal above it(usually called the super-diagonal) contains the number 1!
I want to understand why there are 1's in the superdiagonal.
I have looked all over and no body tries to explain why there is this mysterious 1's in the super-diagonal.
I hope that someone here on this forum can explain this.
Here is a link to PDF:
http://ckottke.ncf.edu/docs/jordan.pdf
linear-algebra jordan-normal-form
asked Dec 22 '18 at 5:53
PaluPalu
3422822
$endgroup$
In terms of Jordan Canonical Form, and more specifically
about Jordan Blocks.
When there is a definition about Jordan Blocks they say the eigenvalues go on the principle diagonal and the diagonal above it(usually called the super-diagonal) contains the number 1!
I want to understand why there are 1's in the superdiagonal.
I have looked all over and no body tries to explain why there is this mysterious 1's in the super-diagonal.
I hope that someone here on this forum can explain this.
Here is a link to PDF:
http://ckottke.ncf.edu/docs/jordan.pdf
linear-algebra jordan-normal-form
linear-algebra jordan-normal-form
asked Dec 22 '18 at 5:53
PaluPalu
3422822
asked Dec 22 '18 at 5:53
PaluPalu
3422822
edited Dec 22 '18 at 19:59
Palu
asked Dec 22 '18 at 5:53
PaluPalu
3422822
asked Dec 22 '18 at 5:53
PaluPalu
3422822
asked Dec 22 '18 at 5:53
PaluPalu
3422822
3422822
$begingroup$
"I have looked all over": Have you looked at a proof that every (complex) matrix has a Jordan form?
$endgroup$
– David C. Ullrich
Dec 22 '18 at 16:03
$begingroup$
Hi David, i have seen things about complex matrix has Jordan form, but when they discuss it they assume the Ones on the Super Diagonal as though this is a fundamental axiom without need of explanation.
$endgroup$
– Palu
Dec 22 '18 at 19:57
$begingroup$
I just added a link to a pdf article I want to illustrate, how they explain things, where they just assume it to be a fact. See above.
$endgroup$
– Palu
Dec 22 '18 at 20:00
add a comment |
$begingroup$
"I have looked all over": Have you looked at a proof that every (complex) matrix has a Jordan form?
$endgroup$
– David C. Ullrich
Dec 22 '18 at 16:03
$begingroup$
Hi David, i have seen things about complex matrix has Jordan form, but when they discuss it they assume the Ones on the Super Diagonal as though this is a fundamental axiom without need of explanation.
$endgroup$
– Palu
Dec 22 '18 at 19:57
$begingroup$
I just added a link to a pdf article I want to illustrate, how they explain things, where they just assume it to be a fact. See above.
$endgroup$
– Palu
Dec 22 '18 at 20:00
$begingroup$
"I have looked all over": Have you looked at a proof that every (complex) matrix has a Jordan form?
$endgroup$
– David C. Ullrich
Dec 22 '18 at 16:03
$begingroup$
"I have looked all over": Have you looked at a proof that every (complex) matrix has a Jordan form?
$endgroup$
– David C. Ullrich
Dec 22 '18 at 16:03
$begingroup$
Hi David, i have seen things about complex matrix has Jordan form, but when they discuss it they assume the Ones on the Super Diagonal as though this is a fundamental axiom without need of explanation.
$endgroup$
– Palu
Dec 22 '18 at 19:57
$begingroup$
Hi David, i have seen things about complex matrix has Jordan form, but when they discuss it they assume the Ones on the Super Diagonal as though this is a fundamental axiom without need of explanation.
$endgroup$
– Palu
Dec 22 '18 at 19:57
$begingroup$
I just added a link to a pdf article I want to illustrate, how they explain things, where they just assume it to be a fact. See above.
$endgroup$
– Palu
Dec 22 '18 at 20:00
$begingroup$
I just added a link to a pdf article I want to illustrate, how they explain things, where they just assume it to be a fact. See above.
$endgroup$
– Palu
Dec 22 '18 at 20:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you have a non-zero vector $x$ such that $(A-lambda I)^{n-1}x ne 0$ and $(A-lambda I)^{n}x=0$, then
$$
{x,(A-lambda I)x,cdots,(A-lambda I)^{n-1}x}
$$
is a linearly independent set of vectors. Let
begin{align}
v_1 & = (A-lambda I)^{n-1}x,\v_2 & =(A-lambda I)^{n-2}x,\&cdots\v_{n-1} & =(A-lambda I)^{1}xend{align}
Then $A-lambda I$ has the following matrix representation with respect to this basis:
$$
left[begin{array}{cccc}0 & 1 & 0 & cdots & 0 \
0 & 0 & 1 & cdots & 0\
vdots & vdots & vdots & ddots & vdots \
0 & 0 & 0 & cdots & 1
\ 0 & 0 & 0 & cdots & 0
end{array}right]
$$
And $A-lambda I$ has the representation $lambda I$ plus the above, which is a Jordan block.
$$
left[begin{array}{cccc}lambda & 1 & 0 & cdots & 0 \
0 & lambda & 1 & cdots & 0\
vdots & vdots & vdots & ddots & vdots \
0 & 0 & lambda & cdots & 1
\ 0 & 0 & 0 & cdots & lambda
end{array}right]
$$
So the answer to your question is that you can always choose the basis so that you get $1$'s on the diagonal above the main diagonal for a Jordan block.
answered Dec 23 '18 at 2:20
DisintegratingByPartsDisintegratingByParts
59.2k42580
$endgroup$
$begingroup$
Hi DisintegratingByParts, i like what you have done here. When you talk about "Then A−λI has the following matrix representation with respect to this basis: " and you get that matrix that has the ones in the SuperDiagonal, i need to try to look at this more carefully, it does not jump out immediately to me. I will try to look at this carefully. Thanks for what you have provided, it looks really interesting.
$endgroup$
– Palu
Dec 23 '18 at 22:03
$begingroup$
Hi DisintegratingByParts, I am having difficulty with the algebra manipulation of the e basis vectors. Is e1, e2, e3 etc going to be the column vectors of the Matrix, then would i factor out the A−λI with power out. Hope you can help with this.
$endgroup$
– Palu
Dec 24 '18 at 0:43
$begingroup$
By a change of basis, the k-th basis element becomes $e_k$. Jordan form is achieved by a change of basis
$endgroup$
– DisintegratingByParts
Dec 24 '18 at 2:49
$begingroup$
So when you use $e_k$ i guess you mean they are the standard basis of $R^n$. For example (1,0,0) , (0,1,0) and (0,0,1) for $R^3$.
$endgroup$
– Palu
Dec 24 '18 at 16:07
add a comment |
Your Answer
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1 Answer
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$begingroup$
If you have a non-zero vector $x$ such that $(A-lambda I)^{n-1}x ne 0$ and $(A-lambda I)^{n}x=0$, then
$$
{x,(A-lambda I)x,cdots,(A-lambda I)^{n-1}x}
$$
is a linearly independent set of vectors. Let
begin{align}
v_1 & = (A-lambda I)^{n-1}x,\v_2 & =(A-lambda I)^{n-2}x,\&cdots\v_{n-1} & =(A-lambda I)^{1}xend{align}
Then $A-lambda I$ has the following matrix representation with respect to this basis:
$$
left[begin{array}{cccc}0 & 1 & 0 & cdots & 0 \
0 & 0 & 1 & cdots & 0\
vdots & vdots & vdots & ddots & vdots \
0 & 0 & 0 & cdots & 1
\ 0 & 0 & 0 & cdots & 0
end{array}right]
$$
And $A-lambda I$ has the representation $lambda I$ plus the above, which is a Jordan block.
$$
left[begin{array}{cccc}lambda & 1 & 0 & cdots & 0 \
0 & lambda & 1 & cdots & 0\
vdots & vdots & vdots & ddots & vdots \
0 & 0 & lambda & cdots & 1
\ 0 & 0 & 0 & cdots & lambda
end{array}right]
$$
So the answer to your question is that you can always choose the basis so that you get $1$'s on the diagonal above the main diagonal for a Jordan block.
answered Dec 23 '18 at 2:20
DisintegratingByPartsDisintegratingByParts
59.2k42580
$endgroup$
$begingroup$
Hi DisintegratingByParts, i like what you have done here. When you talk about "Then A−λI has the following matrix representation with respect to this basis: " and you get that matrix that has the ones in the SuperDiagonal, i need to try to look at this more carefully, it does not jump out immediately to me. I will try to look at this carefully. Thanks for what you have provided, it looks really interesting.
$endgroup$
– Palu
Dec 23 '18 at 22:03
$begingroup$
Hi DisintegratingByParts, I am having difficulty with the algebra manipulation of the e basis vectors. Is e1, e2, e3 etc going to be the column vectors of the Matrix, then would i factor out the A−λI with power out. Hope you can help with this.
$endgroup$
– Palu
Dec 24 '18 at 0:43
$begingroup$
By a change of basis, the k-th basis element becomes $e_k$. Jordan form is achieved by a change of basis
$endgroup$
– DisintegratingByParts
Dec 24 '18 at 2:49
$begingroup$
So when you use $e_k$ i guess you mean they are the standard basis of $R^n$. For example (1,0,0) , (0,1,0) and (0,0,1) for $R^3$.
$endgroup$
– Palu
Dec 24 '18 at 16:07
add a comment |
$begingroup$
If you have a non-zero vector $x$ such that $(A-lambda I)^{n-1}x ne 0$ and $(A-lambda I)^{n}x=0$, then
$$
{x,(A-lambda I)x,cdots,(A-lambda I)^{n-1}x}
$$
is a linearly independent set of vectors. Let
begin{align}
v_1 & = (A-lambda I)^{n-1}x,\v_2 & =(A-lambda I)^{n-2}x,\&cdots\v_{n-1} & =(A-lambda I)^{1}xend{align}
Then $A-lambda I$ has the following matrix representation with respect to this basis:
$$
left[begin{array}{cccc}0 & 1 & 0 & cdots & 0 \
0 & 0 & 1 & cdots & 0\
vdots & vdots & vdots & ddots & vdots \
0 & 0 & 0 & cdots & 1
\ 0 & 0 & 0 & cdots & 0
end{array}right]
$$
And $A-lambda I$ has the representation $lambda I$ plus the above, which is a Jordan block.
$$
left[begin{array}{cccc}lambda & 1 & 0 & cdots & 0 \
0 & lambda & 1 & cdots & 0\
vdots & vdots & vdots & ddots & vdots \
0 & 0 & lambda & cdots & 1
\ 0 & 0 & 0 & cdots & lambda
end{array}right]
$$
So the answer to your question is that you can always choose the basis so that you get $1$'s on the diagonal above the main diagonal for a Jordan block.
answered Dec 23 '18 at 2:20
DisintegratingByPartsDisintegratingByParts
59.2k42580
$endgroup$
$begingroup$
Hi DisintegratingByParts, i like what you have done here. When you talk about "Then A−λI has the following matrix representation with respect to this basis: " and you get that matrix that has the ones in the SuperDiagonal, i need to try to look at this more carefully, it does not jump out immediately to me. I will try to look at this carefully. Thanks for what you have provided, it looks really interesting.
$endgroup$
– Palu
Dec 23 '18 at 22:03
$begingroup$
Hi DisintegratingByParts, I am having difficulty with the algebra manipulation of the e basis vectors. Is e1, e2, e3 etc going to be the column vectors of the Matrix, then would i factor out the A−λI with power out. Hope you can help with this.
$endgroup$
– Palu
Dec 24 '18 at 0:43
$begingroup$
By a change of basis, the k-th basis element becomes $e_k$. Jordan form is achieved by a change of basis
$endgroup$
– DisintegratingByParts
Dec 24 '18 at 2:49
$begingroup$
So when you use $e_k$ i guess you mean they are the standard basis of $R^n$. For example (1,0,0) , (0,1,0) and (0,0,1) for $R^3$.
$endgroup$
– Palu
Dec 24 '18 at 16:07
add a comment |
$begingroup$
If you have a non-zero vector $x$ such that $(A-lambda I)^{n-1}x ne 0$ and $(A-lambda I)^{n}x=0$, then
$$
{x,(A-lambda I)x,cdots,(A-lambda I)^{n-1}x}
$$
is a linearly independent set of vectors. Let
begin{align}
v_1 & = (A-lambda I)^{n-1}x,\v_2 & =(A-lambda I)^{n-2}x,\&cdots\v_{n-1} & =(A-lambda I)^{1}xend{align}
Then $A-lambda I$ has the following matrix representation with respect to this basis:
$$
left[begin{array}{cccc}0 & 1 & 0 & cdots & 0 \
0 & 0 & 1 & cdots & 0\
vdots & vdots & vdots & ddots & vdots \
0 & 0 & 0 & cdots & 1
\ 0 & 0 & 0 & cdots & 0
end{array}right]
$$
And $A-lambda I$ has the representation $lambda I$ plus the above, which is a Jordan block.
$$
left[begin{array}{cccc}lambda & 1 & 0 & cdots & 0 \
0 & lambda & 1 & cdots & 0\
vdots & vdots & vdots & ddots & vdots \
0 & 0 & lambda & cdots & 1
\ 0 & 0 & 0 & cdots & lambda
end{array}right]
$$
So the answer to your question is that you can always choose the basis so that you get $1$'s on the diagonal above the main diagonal for a Jordan block.
answered Dec 23 '18 at 2:20
DisintegratingByPartsDisintegratingByParts
59.2k42580
$endgroup$
If you have a non-zero vector $x$ such that $(A-lambda I)^{n-1}x ne 0$ and $(A-lambda I)^{n}x=0$, then
$$
{x,(A-lambda I)x,cdots,(A-lambda I)^{n-1}x}
$$
is a linearly independent set of vectors. Let
begin{align}
v_1 & = (A-lambda I)^{n-1}x,\v_2 & =(A-lambda I)^{n-2}x,\&cdots\v_{n-1} & =(A-lambda I)^{1}xend{align}
Then $A-lambda I$ has the following matrix representation with respect to this basis:
$$
left[begin{array}{cccc}0 & 1 & 0 & cdots & 0 \
0 & 0 & 1 & cdots & 0\
vdots & vdots & vdots & ddots & vdots \
0 & 0 & 0 & cdots & 1
\ 0 & 0 & 0 & cdots & 0
end{array}right]
$$
And $A-lambda I$ has the representation $lambda I$ plus the above, which is a Jordan block.
$$
left[begin{array}{cccc}lambda & 1 & 0 & cdots & 0 \
0 & lambda & 1 & cdots & 0\
vdots & vdots & vdots & ddots & vdots \
0 & 0 & lambda & cdots & 1
\ 0 & 0 & 0 & cdots & lambda
end{array}right]
$$
So the answer to your question is that you can always choose the basis so that you get $1$'s on the diagonal above the main diagonal for a Jordan block.
answered Dec 23 '18 at 2:20
DisintegratingByPartsDisintegratingByParts
59.2k42580
edited Dec 24 '18 at 16:54
answered Dec 23 '18 at 2:20
DisintegratingByPartsDisintegratingByParts
59.2k42580
answered Dec 23 '18 at 2:20
DisintegratingByPartsDisintegratingByParts
59.2k42580
answered Dec 23 '18 at 2:20
DisintegratingByPartsDisintegratingByParts
59.2k42580
59.2k42580
$begingroup$
Hi DisintegratingByParts, i like what you have done here. When you talk about "Then A−λI has the following matrix representation with respect to this basis: " and you get that matrix that has the ones in the SuperDiagonal, i need to try to look at this more carefully, it does not jump out immediately to me. I will try to look at this carefully. Thanks for what you have provided, it looks really interesting.
$endgroup$
– Palu
Dec 23 '18 at 22:03
$begingroup$
Hi DisintegratingByParts, I am having difficulty with the algebra manipulation of the e basis vectors. Is e1, e2, e3 etc going to be the column vectors of the Matrix, then would i factor out the A−λI with power out. Hope you can help with this.
$endgroup$
– Palu
Dec 24 '18 at 0:43
$begingroup$
By a change of basis, the k-th basis element becomes $e_k$. Jordan form is achieved by a change of basis
$endgroup$
– DisintegratingByParts
Dec 24 '18 at 2:49
$begingroup$
So when you use $e_k$ i guess you mean they are the standard basis of $R^n$. For example (1,0,0) , (0,1,0) and (0,0,1) for $R^3$.
$endgroup$
– Palu
Dec 24 '18 at 16:07
add a comment |
$begingroup$
Hi DisintegratingByParts, i like what you have done here. When you talk about "Then A−λI has the following matrix representation with respect to this basis: " and you get that matrix that has the ones in the SuperDiagonal, i need to try to look at this more carefully, it does not jump out immediately to me. I will try to look at this carefully. Thanks for what you have provided, it looks really interesting.
$endgroup$
– Palu
Dec 23 '18 at 22:03
$begingroup$
Hi DisintegratingByParts, I am having difficulty with the algebra manipulation of the e basis vectors. Is e1, e2, e3 etc going to be the column vectors of the Matrix, then would i factor out the A−λI with power out. Hope you can help with this.
$endgroup$
– Palu
Dec 24 '18 at 0:43
$begingroup$
By a change of basis, the k-th basis element becomes $e_k$. Jordan form is achieved by a change of basis
$endgroup$
– DisintegratingByParts
Dec 24 '18 at 2:49
$begingroup$
So when you use $e_k$ i guess you mean they are the standard basis of $R^n$. For example (1,0,0) , (0,1,0) and (0,0,1) for $R^3$.
$endgroup$
– Palu
Dec 24 '18 at 16:07
$begingroup$
Hi DisintegratingByParts, i like what you have done here. When you talk about "Then A−λI has the following matrix representation with respect to this basis: " and you get that matrix that has the ones in the SuperDiagonal, i need to try to look at this more carefully, it does not jump out immediately to me. I will try to look at this carefully. Thanks for what you have provided, it looks really interesting.
$endgroup$
– Palu
Dec 23 '18 at 22:03
$begingroup$
Hi DisintegratingByParts, i like what you have done here. When you talk about "Then A−λI has the following matrix representation with respect to this basis: " and you get that matrix that has the ones in the SuperDiagonal, i need to try to look at this more carefully, it does not jump out immediately to me. I will try to look at this carefully. Thanks for what you have provided, it looks really interesting.
$endgroup$
– Palu
Dec 23 '18 at 22:03
$begingroup$
Hi DisintegratingByParts, I am having difficulty with the algebra manipulation of the e basis vectors. Is e1, e2, e3 etc going to be the column vectors of the Matrix, then would i factor out the A−λI with power out. Hope you can help with this.
$endgroup$
– Palu
Dec 24 '18 at 0:43
$begingroup$
Hi DisintegratingByParts, I am having difficulty with the algebra manipulation of the e basis vectors. Is e1, e2, e3 etc going to be the column vectors of the Matrix, then would i factor out the A−λI with power out. Hope you can help with this.
$endgroup$
– Palu
Dec 24 '18 at 0:43
$begingroup$
By a change of basis, the k-th basis element becomes $e_k$. Jordan form is achieved by a change of basis
$endgroup$
– DisintegratingByParts
Dec 24 '18 at 2:49
$begingroup$
By a change of basis, the k-th basis element becomes $e_k$. Jordan form is achieved by a change of basis
$endgroup$
– DisintegratingByParts
Dec 24 '18 at 2:49
$begingroup$
So when you use $e_k$ i guess you mean they are the standard basis of $R^n$. For example (1,0,0) , (0,1,0) and (0,0,1) for $R^3$.
$endgroup$
– Palu
Dec 24 '18 at 16:07
$begingroup$
So when you use $e_k$ i guess you mean they are the standard basis of $R^n$. For example (1,0,0) , (0,1,0) and (0,0,1) for $R^3$.
$endgroup$
– Palu
Dec 24 '18 at 16:07
add a comment |
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"I have looked all over": Have you looked at a proof that every (complex) matrix has a Jordan form?
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– David C. Ullrich
Dec 22 '18 at 16:03
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Hi David, i have seen things about complex matrix has Jordan form, but when they discuss it they assume the Ones on the Super Diagonal as though this is a fundamental axiom without need of explanation.
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– Palu
Dec 22 '18 at 19:57
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I just added a link to a pdf article I want to illustrate, how they explain things, where they just assume it to be a fact. See above.
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– Palu
Dec 22 '18 at 20:00