Mixing
Skew-Symmetric vs Symmetric
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
If $A$ is a symmetric $n × n$ matrix and $B$ is a skew symmetric $n × n$ matrix, which of the following are true?
(a) $ABA$ is symmetric
(b) $ABA$ is skew-symmetric
(c) $AB^2A$ is symmetric
(d) $AB^2A$ is skew-symmetric
I know that b and d holds true.
I am unsure of A and C
However, for a, how does multiplying ABA preserve symmetry, but squaring B preserves symmetry as well?
linear-algebra matrices
edited Jul 22 at 11:00
Chinnapparaj R
1,569315
asked Jul 22 at 10:56
FireMeUP
456
add a comment |Â
up vote
0
down vote
favorite
If $A$ is a symmetric $n × n$ matrix and $B$ is a skew symmetric $n × n$ matrix, which of the following are true?
(a) $ABA$ is symmetric
(b) $ABA$ is skew-symmetric
(c) $AB^2A$ is symmetric
(d) $AB^2A$ is skew-symmetric
I know that b and d holds true.
I am unsure of A and C
However, for a, how does multiplying ABA preserve symmetry, but squaring B preserves symmetry as well?
linear-algebra matrices
edited Jul 22 at 11:00
Chinnapparaj R
1,569315
asked Jul 22 at 10:56
FireMeUP
456
Do you know that $(AB)^t=B^tA^t$?
– Javi
Jul 22 at 10:59
@Javi right. But I don't understand how the composition of matrices can effect whether or not it is symmetric or skew-symmetric
– FireMeUP
Jul 22 at 10:59
I'll post an answer
– Javi
Jul 22 at 11:02
@Javi thank you for your clarifications
– FireMeUP
Jul 22 at 11:02
@FireMeUP: Use this MathJax for your future posts!
– Chinnapparaj R
Jul 22 at 11:11
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $A$ is a symmetric $n × n$ matrix and $B$ is a skew symmetric $n × n$ matrix, which of the following are true?
(a) $ABA$ is symmetric
(b) $ABA$ is skew-symmetric
(c) $AB^2A$ is symmetric
(d) $AB^2A$ is skew-symmetric
I know that b and d holds true.
I am unsure of A and C
However, for a, how does multiplying ABA preserve symmetry, but squaring B preserves symmetry as well?
linear-algebra matrices
edited Jul 22 at 11:00
Chinnapparaj R
1,569315
asked Jul 22 at 10:56
FireMeUP
456
If $A$ is a symmetric $n × n$ matrix and $B$ is a skew symmetric $n × n$ matrix, which of the following are true?
(a) $ABA$ is symmetric
(b) $ABA$ is skew-symmetric
(c) $AB^2A$ is symmetric
(d) $AB^2A$ is skew-symmetric
I know that b and d holds true.
I am unsure of A and C
However, for a, how does multiplying ABA preserve symmetry, but squaring B preserves symmetry as well?
linear-algebra matrices
edited Jul 22 at 11:00
Chinnapparaj R
1,569315
asked Jul 22 at 10:56
FireMeUP
456
edited Jul 22 at 11:00
Chinnapparaj R
1,569315
edited Jul 22 at 11:00
Chinnapparaj R
1,569315
edited Jul 22 at 11:00
Chinnapparaj R
1,569315
1,569315
asked Jul 22 at 10:56
FireMeUP
456
asked Jul 22 at 10:56
FireMeUP
456
asked Jul 22 at 10:56
FireMeUP
456
456
Do you know that $(AB)^t=B^tA^t$?
– Javi
Jul 22 at 10:59
@Javi right. But I don't understand how the composition of matrices can effect whether or not it is symmetric or skew-symmetric
– FireMeUP
Jul 22 at 10:59
I'll post an answer
– Javi
Jul 22 at 11:02
@Javi thank you for your clarifications
– FireMeUP
Jul 22 at 11:02
@FireMeUP: Use this MathJax for your future posts!
– Chinnapparaj R
Jul 22 at 11:11
add a comment |Â
Do you know that $(AB)^t=B^tA^t$?
– Javi
Jul 22 at 10:59
@Javi right. But I don't understand how the composition of matrices can effect whether or not it is symmetric or skew-symmetric
– FireMeUP
Jul 22 at 10:59
I'll post an answer
– Javi
Jul 22 at 11:02
@Javi thank you for your clarifications
– FireMeUP
Jul 22 at 11:02
@FireMeUP: Use this MathJax for your future posts!
– Chinnapparaj R
Jul 22 at 11:11
Do you know that $(AB)^t=B^tA^t$?
– Javi
Jul 22 at 10:59
Do you know that $(AB)^t=B^tA^t$?
– Javi
Jul 22 at 10:59
@Javi right. But I don't understand how the composition of matrices can effect whether or not it is symmetric or skew-symmetric
– FireMeUP
Jul 22 at 10:59
@Javi right. But I don't understand how the composition of matrices can effect whether or not it is symmetric or skew-symmetric
– FireMeUP
Jul 22 at 10:59
I'll post an answer
– Javi
Jul 22 at 11:02
I'll post an answer
– Javi
Jul 22 at 11:02
@Javi thank you for your clarifications
– FireMeUP
Jul 22 at 11:02
@Javi thank you for your clarifications
– FireMeUP
Jul 22 at 11:02
@FireMeUP: Use this MathJax for your future posts!
– Chinnapparaj R
Jul 22 at 11:11
@FireMeUP: Use this MathJax for your future posts!
– Chinnapparaj R
Jul 22 at 11:11
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Use the property $(AB)^t=B^tA^t$ to compute the transpose of each matrix and the fact that $A$ is symmetric and $B$ is skew-symmetric. For example, $(ABA)^t=A^tB^tA^t=A(-B)A=-(ABA)$. Then $ABA$ is skew-symmetric (and not symmetric in general).
answered Jul 22 at 11:02
Javi
2,1631725
how does this change for B^2. Do we have to compute B^2t?
– FireMeUP
Jul 22 at 11:03
Yes, but that's just $B^tB^t=(B^t)^2$.
– Javi
Jul 22 at 11:05
ahhh I I see. So in the case for B^2. I have that AB^2A is symmetric, but not skew symmetric since using the transpose condition you have give me, we reduce back to (A^TBB*A^T) which cannot be negative
– FireMeUP
Jul 22 at 11:08
Yeah, that's it :)
– Javi
Jul 22 at 11:08
1
thank you for your help
– FireMeUP
Jul 22 at 11:11
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
Use the property $(AB)^t=B^tA^t$ to compute the transpose of each matrix and the fact that $A$ is symmetric and $B$ is skew-symmetric. For example, $(ABA)^t=A^tB^tA^t=A(-B)A=-(ABA)$. Then $ABA$ is skew-symmetric (and not symmetric in general).
answered Jul 22 at 11:02
Javi
2,1631725
how does this change for B^2. Do we have to compute B^2t?
– FireMeUP
Jul 22 at 11:03
Yes, but that's just $B^tB^t=(B^t)^2$.
– Javi
Jul 22 at 11:05
ahhh I I see. So in the case for B^2. I have that AB^2A is symmetric, but not skew symmetric since using the transpose condition you have give me, we reduce back to (A^TBB*A^T) which cannot be negative
– FireMeUP
Jul 22 at 11:08
Yeah, that's it :)
– Javi
Jul 22 at 11:08
1
thank you for your help
– FireMeUP
Jul 22 at 11:11
add a comment |Â
up vote
2
down vote
accepted
Use the property $(AB)^t=B^tA^t$ to compute the transpose of each matrix and the fact that $A$ is symmetric and $B$ is skew-symmetric. For example, $(ABA)^t=A^tB^tA^t=A(-B)A=-(ABA)$. Then $ABA$ is skew-symmetric (and not symmetric in general).
answered Jul 22 at 11:02
Javi
2,1631725
how does this change for B^2. Do we have to compute B^2t?
– FireMeUP
Jul 22 at 11:03
Yes, but that's just $B^tB^t=(B^t)^2$.
– Javi
Jul 22 at 11:05
ahhh I I see. So in the case for B^2. I have that AB^2A is symmetric, but not skew symmetric since using the transpose condition you have give me, we reduce back to (A^TBB*A^T) which cannot be negative
– FireMeUP
Jul 22 at 11:08
Yeah, that's it :)
– Javi
Jul 22 at 11:08
1
thank you for your help
– FireMeUP
Jul 22 at 11:11
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Use the property $(AB)^t=B^tA^t$ to compute the transpose of each matrix and the fact that $A$ is symmetric and $B$ is skew-symmetric. For example, $(ABA)^t=A^tB^tA^t=A(-B)A=-(ABA)$. Then $ABA$ is skew-symmetric (and not symmetric in general).
answered Jul 22 at 11:02
Javi
2,1631725
Use the property $(AB)^t=B^tA^t$ to compute the transpose of each matrix and the fact that $A$ is symmetric and $B$ is skew-symmetric. For example, $(ABA)^t=A^tB^tA^t=A(-B)A=-(ABA)$. Then $ABA$ is skew-symmetric (and not symmetric in general).
answered Jul 22 at 11:02
Javi
2,1631725
edited Jul 22 at 11:05
answered Jul 22 at 11:02
Javi
2,1631725
answered Jul 22 at 11:02
Javi
2,1631725
answered Jul 22 at 11:02
Javi
2,1631725
2,1631725
how does this change for B^2. Do we have to compute B^2t?
– FireMeUP
Jul 22 at 11:03
Yes, but that's just $B^tB^t=(B^t)^2$.
– Javi
Jul 22 at 11:05
ahhh I I see. So in the case for B^2. I have that AB^2A is symmetric, but not skew symmetric since using the transpose condition you have give me, we reduce back to (A^TBB*A^T) which cannot be negative
– FireMeUP
Jul 22 at 11:08
Yeah, that's it :)
– Javi
Jul 22 at 11:08
1
thank you for your help
– FireMeUP
Jul 22 at 11:11
add a comment |Â
how does this change for B^2. Do we have to compute B^2t?
– FireMeUP
Jul 22 at 11:03
Yes, but that's just $B^tB^t=(B^t)^2$.
– Javi
Jul 22 at 11:05
ahhh I I see. So in the case for B^2. I have that AB^2A is symmetric, but not skew symmetric since using the transpose condition you have give me, we reduce back to (A^TBB*A^T) which cannot be negative
– FireMeUP
Jul 22 at 11:08
Yeah, that's it :)
– Javi
Jul 22 at 11:08
1
thank you for your help
– FireMeUP
Jul 22 at 11:11
how does this change for B^2. Do we have to compute B^2t?
– FireMeUP
Jul 22 at 11:03
how does this change for B^2. Do we have to compute B^2t?
– FireMeUP
Jul 22 at 11:03
Yes, but that's just $B^tB^t=(B^t)^2$.
– Javi
Jul 22 at 11:05
Yes, but that's just $B^tB^t=(B^t)^2$.
– Javi
Jul 22 at 11:05
ahhh I I see. So in the case for B^2. I have that AB^2A is symmetric, but not skew symmetric since using the transpose condition you have give me, we reduce back to (A^TBB*A^T) which cannot be negative
– FireMeUP
Jul 22 at 11:08
ahhh I I see. So in the case for B^2. I have that AB^2A is symmetric, but not skew symmetric since using the transpose condition you have give me, we reduce back to (A^TBB*A^T) which cannot be negative
– FireMeUP
Jul 22 at 11:08
Yeah, that's it :)
– Javi
Jul 22 at 11:08
Yeah, that's it :)
– Javi
Jul 22 at 11:08
1
1
thank you for your help
– FireMeUP
Jul 22 at 11:11
thank you for your help
– FireMeUP
Jul 22 at 11:11
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%2f2859283%2fskew-symmetric-vs-symmetric%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
Do you know that $(AB)^t=B^tA^t$?
– Javi
Jul 22 at 10:59
@Javi right. But I don't understand how the composition of matrices can effect whether or not it is symmetric or skew-symmetric
– FireMeUP
Jul 22 at 10:59
I'll post an answer
– Javi
Jul 22 at 11:02
@Javi thank you for your clarifications
– FireMeUP
Jul 22 at 11:02
@FireMeUP: Use this MathJax for your future posts!
– Chinnapparaj R
Jul 22 at 11:11