Finding the Jordan Form of a transformation defined by $T(X)=AX$ when $A,X in M_4times 4(mathbb C)$
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Given $A=left(matrix0&1&0&0\1&0&0&0\0&0&1&0\0&0&0&-1right)$ and define $T(X)=AX$. when $A,X in M_4times 4(mathbb C)$
Find the Jordan Form of T
I found that the minimal polynomial is $(t-1)(t+1)$ therefore $T$ is diagonalizable. However, I'm not sure how to find the amount of each eigenvalue to put in the diagonal without explicity finding the $16times16$ representative matrix of T.
linear-algebra linear-transformations jordan-normal-form
asked Aug 14 at 11:31
Jason
1,12711327
 |Â
show 2 more comments
up vote
1
down vote
favorite
Given $A=left(matrix0&1&0&0\1&0&0&0\0&0&1&0\0&0&0&-1right)$ and define $T(X)=AX$. when $A,X in M_4times 4(mathbb C)$
Find the Jordan Form of T
I found that the minimal polynomial is $(t-1)(t+1)$ therefore $T$ is diagonalizable. However, I'm not sure how to find the amount of each eigenvalue to put in the diagonal without explicity finding the $16times16$ representative matrix of T.
linear-algebra linear-transformations jordan-normal-form
asked Aug 14 at 11:31
Jason
1,12711327
X is also a 4×4 matrix or 4×1?
– dmtri
Aug 14 at 11:37
By the $16times 16$ comment, I guess it should be a $4times 4$ matrix. It is also in the title. But you are right, the reader should not guess, and it should have been repeated.
– A. Pongrácz
Aug 14 at 11:39
Edited to repeat it in the body
– Jason
Aug 14 at 11:56
This might help: $(T-I)X=0Longleftrightarrow (A-I)X=0$
– Lozenges
Aug 14 at 12:00
@Lozenges I thought about that but it seemed it might lead to a lot of equations. Might be worth exploring though, I could be wrong.
– Jason
Aug 14 at 12:02
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given $A=left(matrix0&1&0&0\1&0&0&0\0&0&1&0\0&0&0&-1right)$ and define $T(X)=AX$. when $A,X in M_4times 4(mathbb C)$
Find the Jordan Form of T
I found that the minimal polynomial is $(t-1)(t+1)$ therefore $T$ is diagonalizable. However, I'm not sure how to find the amount of each eigenvalue to put in the diagonal without explicity finding the $16times16$ representative matrix of T.
linear-algebra linear-transformations jordan-normal-form
asked Aug 14 at 11:31
Jason
1,12711327
Given $A=left(matrix0&1&0&0\1&0&0&0\0&0&1&0\0&0&0&-1right)$ and define $T(X)=AX$. when $A,X in M_4times 4(mathbb C)$
Find the Jordan Form of T
I found that the minimal polynomial is $(t-1)(t+1)$ therefore $T$ is diagonalizable. However, I'm not sure how to find the amount of each eigenvalue to put in the diagonal without explicity finding the $16times16$ representative matrix of T.
linear-algebra linear-transformations jordan-normal-form
asked Aug 14 at 11:31
Jason
1,12711327
edited Aug 14 at 11:55
asked Aug 14 at 11:31
Jason
1,12711327
asked Aug 14 at 11:31
Jason
1,12711327
asked Aug 14 at 11:31
Jason
1,12711327
1,12711327
X is also a 4×4 matrix or 4×1?
– dmtri
Aug 14 at 11:37
By the $16times 16$ comment, I guess it should be a $4times 4$ matrix. It is also in the title. But you are right, the reader should not guess, and it should have been repeated.
– A. Pongrácz
Aug 14 at 11:39
Edited to repeat it in the body
– Jason
Aug 14 at 11:56
This might help: $(T-I)X=0Longleftrightarrow (A-I)X=0$
– Lozenges
Aug 14 at 12:00
@Lozenges I thought about that but it seemed it might lead to a lot of equations. Might be worth exploring though, I could be wrong.
– Jason
Aug 14 at 12:02
 |Â
show 2 more comments
X is also a 4×4 matrix or 4×1?
– dmtri
Aug 14 at 11:37
By the $16times 16$ comment, I guess it should be a $4times 4$ matrix. It is also in the title. But you are right, the reader should not guess, and it should have been repeated.
– A. Pongrácz
Aug 14 at 11:39
Edited to repeat it in the body
– Jason
Aug 14 at 11:56
This might help: $(T-I)X=0Longleftrightarrow (A-I)X=0$
– Lozenges
Aug 14 at 12:00
@Lozenges I thought about that but it seemed it might lead to a lot of equations. Might be worth exploring though, I could be wrong.
– Jason
Aug 14 at 12:02
X is also a 4×4 matrix or 4×1?
– dmtri
Aug 14 at 11:37
X is also a 4×4 matrix or 4×1?
– dmtri
Aug 14 at 11:37
By the $16times 16$ comment, I guess it should be a $4times 4$ matrix. It is also in the title. But you are right, the reader should not guess, and it should have been repeated.
– A. Pongrácz
Aug 14 at 11:39
By the $16times 16$ comment, I guess it should be a $4times 4$ matrix. It is also in the title. But you are right, the reader should not guess, and it should have been repeated.
– A. Pongrácz
Aug 14 at 11:39
Edited to repeat it in the body
– Jason
Aug 14 at 11:56
Edited to repeat it in the body
– Jason
Aug 14 at 11:56
This might help: $(T-I)X=0Longleftrightarrow (A-I)X=0$
– Lozenges
Aug 14 at 12:00
This might help: $(T-I)X=0Longleftrightarrow (A-I)X=0$
– Lozenges
Aug 14 at 12:00
@Lozenges I thought about that but it seemed it might lead to a lot of equations. Might be worth exploring though, I could be wrong.
– Jason
Aug 14 at 12:02
@Lozenges I thought about that but it seemed it might lead to a lot of equations. Might be worth exploring though, I could be wrong.
– Jason
Aug 14 at 12:02
 |Â
show 2 more comments
1 Answer
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up vote
2
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I identify a $4times 4$ matrix by a vector of length $16$ by reading it from top to bottom, from left to right. So the first four coordinates are from the first column, etc.
Then $T$ is diagonal in the last $8$ coordinates: identical on coordinates $9-12$, and negative of the identity on $13-16$, so that part is covered.
Furthermore, $T$ switches the first four coordinates by the second four as a product of four transpositions. You can view these transpositions separately. The matrix of a transposition of two coordinates is $beginpmatrix 0 & 1 \ 1 & 0 endpmatrix$, whose Jordan normal form is $beginpmatrix 1 & 0 \ 0 & -1 endpmatrix$.
So the Jordan normal form of $T$ is diagonal with eigenvalues:
$$1,-1,1,-1,1,-1,1,-1,1,1,1,1,-1,-1,-1,-1$$
answered Aug 14 at 11:46
A. Pongrácz
3,682624
1
Nice thinking about the transformation gets you most of the way and the transposition trick is nice. Thanks
– Jason
Aug 14 at 12:04
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I identify a $4times 4$ matrix by a vector of length $16$ by reading it from top to bottom, from left to right. So the first four coordinates are from the first column, etc.
Then $T$ is diagonal in the last $8$ coordinates: identical on coordinates $9-12$, and negative of the identity on $13-16$, so that part is covered.
Furthermore, $T$ switches the first four coordinates by the second four as a product of four transpositions. You can view these transpositions separately. The matrix of a transposition of two coordinates is $beginpmatrix 0 & 1 \ 1 & 0 endpmatrix$, whose Jordan normal form is $beginpmatrix 1 & 0 \ 0 & -1 endpmatrix$.
So the Jordan normal form of $T$ is diagonal with eigenvalues:
$$1,-1,1,-1,1,-1,1,-1,1,1,1,1,-1,-1,-1,-1$$
answered Aug 14 at 11:46
A. Pongrácz
3,682624
1
Nice thinking about the transformation gets you most of the way and the transposition trick is nice. Thanks
– Jason
Aug 14 at 12:04
add a comment |Â
up vote
2
down vote
accepted
I identify a $4times 4$ matrix by a vector of length $16$ by reading it from top to bottom, from left to right. So the first four coordinates are from the first column, etc.
Then $T$ is diagonal in the last $8$ coordinates: identical on coordinates $9-12$, and negative of the identity on $13-16$, so that part is covered.
Furthermore, $T$ switches the first four coordinates by the second four as a product of four transpositions. You can view these transpositions separately. The matrix of a transposition of two coordinates is $beginpmatrix 0 & 1 \ 1 & 0 endpmatrix$, whose Jordan normal form is $beginpmatrix 1 & 0 \ 0 & -1 endpmatrix$.
So the Jordan normal form of $T$ is diagonal with eigenvalues:
$$1,-1,1,-1,1,-1,1,-1,1,1,1,1,-1,-1,-1,-1$$
answered Aug 14 at 11:46
A. Pongrácz
3,682624
1
Nice thinking about the transformation gets you most of the way and the transposition trick is nice. Thanks
– Jason
Aug 14 at 12:04
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I identify a $4times 4$ matrix by a vector of length $16$ by reading it from top to bottom, from left to right. So the first four coordinates are from the first column, etc.
Then $T$ is diagonal in the last $8$ coordinates: identical on coordinates $9-12$, and negative of the identity on $13-16$, so that part is covered.
Furthermore, $T$ switches the first four coordinates by the second four as a product of four transpositions. You can view these transpositions separately. The matrix of a transposition of two coordinates is $beginpmatrix 0 & 1 \ 1 & 0 endpmatrix$, whose Jordan normal form is $beginpmatrix 1 & 0 \ 0 & -1 endpmatrix$.
So the Jordan normal form of $T$ is diagonal with eigenvalues:
$$1,-1,1,-1,1,-1,1,-1,1,1,1,1,-1,-1,-1,-1$$
answered Aug 14 at 11:46
A. Pongrácz
3,682624
I identify a $4times 4$ matrix by a vector of length $16$ by reading it from top to bottom, from left to right. So the first four coordinates are from the first column, etc.
Then $T$ is diagonal in the last $8$ coordinates: identical on coordinates $9-12$, and negative of the identity on $13-16$, so that part is covered.
Furthermore, $T$ switches the first four coordinates by the second four as a product of four transpositions. You can view these transpositions separately. The matrix of a transposition of two coordinates is $beginpmatrix 0 & 1 \ 1 & 0 endpmatrix$, whose Jordan normal form is $beginpmatrix 1 & 0 \ 0 & -1 endpmatrix$.
So the Jordan normal form of $T$ is diagonal with eigenvalues:
$$1,-1,1,-1,1,-1,1,-1,1,1,1,1,-1,-1,-1,-1$$
answered Aug 14 at 11:46
A. Pongrácz
3,682624
answered Aug 14 at 11:46
A. Pongrácz
3,682624
answered Aug 14 at 11:46
A. Pongrácz
3,682624
answered Aug 14 at 11:46
A. Pongrácz
3,682624
3,682624
1
Nice thinking about the transformation gets you most of the way and the transposition trick is nice. Thanks
– Jason
Aug 14 at 12:04
add a comment |Â
1
Nice thinking about the transformation gets you most of the way and the transposition trick is nice. Thanks
– Jason
Aug 14 at 12:04
1
1
Nice thinking about the transformation gets you most of the way and the transposition trick is nice. Thanks
– Jason
Aug 14 at 12:04
Nice thinking about the transformation gets you most of the way and the transposition trick is nice. Thanks
– Jason
Aug 14 at 12:04
add a comment |Â
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X is also a 4×4 matrix or 4×1?
– dmtri
Aug 14 at 11:37
By the $16times 16$ comment, I guess it should be a $4times 4$ matrix. It is also in the title. But you are right, the reader should not guess, and it should have been repeated.
– A. Pongrácz
Aug 14 at 11:39
Edited to repeat it in the body
– Jason
Aug 14 at 11:56
This might help: $(T-I)X=0Longleftrightarrow (A-I)X=0$
– Lozenges
Aug 14 at 12:00
@Lozenges I thought about that but it seemed it might lead to a lot of equations. Might be worth exploring though, I could be wrong.
– Jason
Aug 14 at 12:02