Mixing
Does A transpose = - A transpose?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
The question I am currently struggling with is
Does there exist an $ n×n $ real matrix$ A $ such that
$ tr(A)= 0 $ and $ A^2+A^T= I $?
For some reason, the solution starts with
$ I= A^2−A^T $
instead of $ I = A^2 + A^T $ as stated in the problem.
which leads me to think that $ A^T = -A^T $
Is this true?
The solution I am looking at is this.
https://i.stack.imgur.com/w1Q9D.png
linear-algebra transpose
add a comment |Â
up vote
0
down vote
favorite
The question I am currently struggling with is
Does there exist an $ n×n $ real matrix$ A $ such that
$ tr(A)= 0 $ and $ A^2+A^T= I $?
For some reason, the solution starts with
$ I= A^2−A^T $
instead of $ I = A^2 + A^T $ as stated in the problem.
which leads me to think that $ A^T = -A^T $
Is this true?
The solution I am looking at is this.
https://i.stack.imgur.com/w1Q9D.png
linear-algebra transpose
Since the solution states the matrix cannot exist, the question whether this non-existing matrix satisfies $A^T = - A^T$ is somewhat weird...
– Vincent
Jul 20 at 11:14
I expect that there was a typo in the question and they intended $I = A^2 - A^T$
– Vincent
Jul 20 at 11:15
Something else: the only matrix satisfying $-X = X$ is the null-matrix, so if $A^T = -A^T$ then $A^T= 0$ and hence $A = 0$. But then we check that $A^2 + A^T = 0$ and not $I$. In other words, from the conditions in the question it follows that $A^T$ is NOT equal to $-A^T$.
– Vincent
Jul 20 at 11:20
Thank you very much for the comments.
– user449415
Jul 21 at 3:10
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The question I am currently struggling with is
Does there exist an $ n×n $ real matrix$ A $ such that
$ tr(A)= 0 $ and $ A^2+A^T= I $?
For some reason, the solution starts with
$ I= A^2−A^T $
instead of $ I = A^2 + A^T $ as stated in the problem.
which leads me to think that $ A^T = -A^T $
Is this true?
The solution I am looking at is this.
https://i.stack.imgur.com/w1Q9D.png
linear-algebra transpose
The question I am currently struggling with is
Does there exist an $ n×n $ real matrix$ A $ such that
$ tr(A)= 0 $ and $ A^2+A^T= I $?
For some reason, the solution starts with
$ I= A^2−A^T $
instead of $ I = A^2 + A^T $ as stated in the problem.
which leads me to think that $ A^T = -A^T $
Is this true?
The solution I am looking at is this.
https://i.stack.imgur.com/w1Q9D.png
linear-algebra transpose
edited Jul 20 at 11:09
asked Jul 20 at 10:52
user449415
Since the solution states the matrix cannot exist, the question whether this non-existing matrix satisfies $A^T = - A^T$ is somewhat weird...
– Vincent
Jul 20 at 11:14
I expect that there was a typo in the question and they intended $I = A^2 - A^T$
– Vincent
Jul 20 at 11:15
Something else: the only matrix satisfying $-X = X$ is the null-matrix, so if $A^T = -A^T$ then $A^T= 0$ and hence $A = 0$. But then we check that $A^2 + A^T = 0$ and not $I$. In other words, from the conditions in the question it follows that $A^T$ is NOT equal to $-A^T$.
– Vincent
Jul 20 at 11:20
Thank you very much for the comments.
– user449415
Jul 21 at 3:10
add a comment |Â
Since the solution states the matrix cannot exist, the question whether this non-existing matrix satisfies $A^T = - A^T$ is somewhat weird...
– Vincent
Jul 20 at 11:14
I expect that there was a typo in the question and they intended $I = A^2 - A^T$
– Vincent
Jul 20 at 11:15
Something else: the only matrix satisfying $-X = X$ is the null-matrix, so if $A^T = -A^T$ then $A^T= 0$ and hence $A = 0$. But then we check that $A^2 + A^T = 0$ and not $I$. In other words, from the conditions in the question it follows that $A^T$ is NOT equal to $-A^T$.
– Vincent
Jul 20 at 11:20
Thank you very much for the comments.
– user449415
Jul 21 at 3:10
Since the solution states the matrix cannot exist, the question whether this non-existing matrix satisfies $A^T = - A^T$ is somewhat weird...
– Vincent
Jul 20 at 11:14
Since the solution states the matrix cannot exist, the question whether this non-existing matrix satisfies $A^T = - A^T$ is somewhat weird...
– Vincent
Jul 20 at 11:14
I expect that there was a typo in the question and they intended $I = A^2 - A^T$
– Vincent
Jul 20 at 11:15
I expect that there was a typo in the question and they intended $I = A^2 - A^T$
– Vincent
Jul 20 at 11:15
Something else: the only matrix satisfying $-X = X$ is the null-matrix, so if $A^T = -A^T$ then $A^T= 0$ and hence $A = 0$. But then we check that $A^2 + A^T = 0$ and not $I$. In other words, from the conditions in the question it follows that $A^T$ is NOT equal to $-A^T$.
– Vincent
Jul 20 at 11:20
Something else: the only matrix satisfying $-X = X$ is the null-matrix, so if $A^T = -A^T$ then $A^T= 0$ and hence $A = 0$. But then we check that $A^2 + A^T = 0$ and not $I$. In other words, from the conditions in the question it follows that $A^T$ is NOT equal to $-A^T$.
– Vincent
Jul 20 at 11:20
Thank you very much for the comments.
– user449415
Jul 21 at 3:10
Thank you very much for the comments.
– user449415
Jul 21 at 3:10
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
Well, no, that is a mistake from the solution makers. However, if you use $I=A^2+A^top$ where they write $I=A^2-A^top$, the argument doesn't change that much.
Edit: if the question makers intended to write $I=A^2 - A^top$, then note that for any solution $A$, we could define $B=-A$ and we would get $I=B^2 + B^top$. The other way around works as well. Thus, it doesn't matter what the exact sign is, the problems are equivalent, and the argument is similar.
answered Jul 20 at 11:14
Stan Tendijck
1,277110
Thank you for your answer !
– user449415
Jul 21 at 3:10
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
Well, no, that is a mistake from the solution makers. However, if you use $I=A^2+A^top$ where they write $I=A^2-A^top$, the argument doesn't change that much.
Edit: if the question makers intended to write $I=A^2 - A^top$, then note that for any solution $A$, we could define $B=-A$ and we would get $I=B^2 + B^top$. The other way around works as well. Thus, it doesn't matter what the exact sign is, the problems are equivalent, and the argument is similar.
answered Jul 20 at 11:14
Stan Tendijck
1,277110
Thank you for your answer !
– user449415
Jul 21 at 3:10
add a comment |Â
up vote
2
down vote
Well, no, that is a mistake from the solution makers. However, if you use $I=A^2+A^top$ where they write $I=A^2-A^top$, the argument doesn't change that much.
Edit: if the question makers intended to write $I=A^2 - A^top$, then note that for any solution $A$, we could define $B=-A$ and we would get $I=B^2 + B^top$. The other way around works as well. Thus, it doesn't matter what the exact sign is, the problems are equivalent, and the argument is similar.
answered Jul 20 at 11:14
Stan Tendijck
1,277110
Thank you for your answer !
– user449415
Jul 21 at 3:10
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Well, no, that is a mistake from the solution makers. However, if you use $I=A^2+A^top$ where they write $I=A^2-A^top$, the argument doesn't change that much.
Edit: if the question makers intended to write $I=A^2 - A^top$, then note that for any solution $A$, we could define $B=-A$ and we would get $I=B^2 + B^top$. The other way around works as well. Thus, it doesn't matter what the exact sign is, the problems are equivalent, and the argument is similar.
answered Jul 20 at 11:14
Stan Tendijck
1,277110
Well, no, that is a mistake from the solution makers. However, if you use $I=A^2+A^top$ where they write $I=A^2-A^top$, the argument doesn't change that much.
Edit: if the question makers intended to write $I=A^2 - A^top$, then note that for any solution $A$, we could define $B=-A$ and we would get $I=B^2 + B^top$. The other way around works as well. Thus, it doesn't matter what the exact sign is, the problems are equivalent, and the argument is similar.
answered Jul 20 at 11:14
Stan Tendijck
1,277110
edited Jul 20 at 11:27
answered Jul 20 at 11:14
Stan Tendijck
1,277110
answered Jul 20 at 11:14
Stan Tendijck
1,277110
answered Jul 20 at 11:14
Stan Tendijck
1,277110
1,277110
Thank you for your answer !
– user449415
Jul 21 at 3:10
add a comment |Â
Thank you for your answer !
– user449415
Jul 21 at 3:10
Thank you for your answer !
– user449415
Jul 21 at 3:10
Thank you for your answer !
– user449415
Jul 21 at 3:10
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857517%2fdoes-a-transpose-a-transpose%23new-answer', 'question_page');
);
Post as a guest
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
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
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
Since the solution states the matrix cannot exist, the question whether this non-existing matrix satisfies $A^T = - A^T$ is somewhat weird...
– Vincent
Jul 20 at 11:14
I expect that there was a typo in the question and they intended $I = A^2 - A^T$
– Vincent
Jul 20 at 11:15
Something else: the only matrix satisfying $-X = X$ is the null-matrix, so if $A^T = -A^T$ then $A^T= 0$ and hence $A = 0$. But then we check that $A^2 + A^T = 0$ and not $I$. In other words, from the conditions in the question it follows that $A^T$ is NOT equal to $-A^T$.
– Vincent
Jul 20 at 11:20
Thank you very much for the comments.
– user449415
Jul 21 at 3:10