Mixing
In $mathbbC^2$, show that $langle x,y rangle$ = $xAy^*$ is an inner product
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
In $mathbbC^2$, show that $langle x,y rangle$ = $xAy^*$ is an inner product, where
$A = beginpmatrix
1 & i \
-i & 2 \
endpmatrix$
Property (c) of an inner product on V states that $langle barx,bary rangle$ = $langle y,x rangle$
For part (c), I found the following answer online
Property (c) requires a little more work: First note that since $xAy^*$ is a $1 times 1$ matrix, it is equal to its transpose. So $xAy^* = (xAy^*)^t = bar y A^t x^t$. Now we have
$$
overlinelangle x,y rangle =
overlinexAy^* =
overlinebar y A^t x^t =
yA^*x^* = yAx^* = langle y,x rangle
$$
since $A$ is self-adjoint.
(image)
For the equality, $(xAy*)^t = baryA^tx^t$, why are we able to switch $bary$ and $x^t$? Is it simply because we're dealing with a $1$x$1$ matrix?
linear-algebra
edited Jul 23 at 21:34
Omnomnomnom
121k784170
asked Jul 23 at 20:35
K.M
480312
 |Â
show 2 more comments
up vote
0
down vote
favorite
In $mathbbC^2$, show that $langle x,y rangle$ = $xAy^*$ is an inner product, where
$A = beginpmatrix
1 & i \
-i & 2 \
endpmatrix$
Property (c) of an inner product on V states that $langle barx,bary rangle$ = $langle y,x rangle$
For part (c), I found the following answer online
Property (c) requires a little more work: First note that since $xAy^*$ is a $1 times 1$ matrix, it is equal to its transpose. So $xAy^* = (xAy^*)^t = bar y A^t x^t$. Now we have
$$
overlinelangle x,y rangle =
overlinexAy^* =
overlinebar y A^t x^t =
yA^*x^* = yAx^* = langle y,x rangle
$$
since $A$ is self-adjoint.
(image)
For the equality, $(xAy*)^t = baryA^tx^t$, why are we able to switch $bary$ and $x^t$? Is it simply because we're dealing with a $1$x$1$ matrix?
linear-algebra
edited Jul 23 at 21:34
Omnomnomnom
121k784170
asked Jul 23 at 20:35
K.M
480312
Yes, I know what * means. I still don't see why that allows me to switch.
– K.M
Jul 23 at 20:40
1
Honestly, if you don't understand how transpose and conjugate work, you should definitely not be looking up questions like this online. You are only hurting yourself in the long run. read your notes and textbook.
– AnonymousCoward
Jul 23 at 20:40
I recommend looking at the references you are using for your class and reading about those two topics.
– AnonymousCoward
Jul 23 at 20:40
I'm pretty sure it's something elementary, but going through an entire book to look for one fact would take up too much time.
– K.M
Jul 23 at 20:43
I`m not sure if I understood the question right, but using $(AB)^t=B^tA^t$, we get : $(xAy^*)^t=bar y^t^tA^tx^t=bar yA^tx^t $
– Sar
Jul 23 at 20:55
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In $mathbbC^2$, show that $langle x,y rangle$ = $xAy^*$ is an inner product, where
$A = beginpmatrix
1 & i \
-i & 2 \
endpmatrix$
Property (c) of an inner product on V states that $langle barx,bary rangle$ = $langle y,x rangle$
For part (c), I found the following answer online
Property (c) requires a little more work: First note that since $xAy^*$ is a $1 times 1$ matrix, it is equal to its transpose. So $xAy^* = (xAy^*)^t = bar y A^t x^t$. Now we have
$$
overlinelangle x,y rangle =
overlinexAy^* =
overlinebar y A^t x^t =
yA^*x^* = yAx^* = langle y,x rangle
$$
since $A$ is self-adjoint.
(image)
For the equality, $(xAy*)^t = baryA^tx^t$, why are we able to switch $bary$ and $x^t$? Is it simply because we're dealing with a $1$x$1$ matrix?
linear-algebra
edited Jul 23 at 21:34
Omnomnomnom
121k784170
asked Jul 23 at 20:35
K.M
480312
In $mathbbC^2$, show that $langle x,y rangle$ = $xAy^*$ is an inner product, where
$A = beginpmatrix
1 & i \
-i & 2 \
endpmatrix$
Property (c) of an inner product on V states that $langle barx,bary rangle$ = $langle y,x rangle$
For part (c), I found the following answer online
Property (c) requires a little more work: First note that since $xAy^*$ is a $1 times 1$ matrix, it is equal to its transpose. So $xAy^* = (xAy^*)^t = bar y A^t x^t$. Now we have
$$
overlinelangle x,y rangle =
overlinexAy^* =
overlinebar y A^t x^t =
yA^*x^* = yAx^* = langle y,x rangle
$$
since $A$ is self-adjoint.
(image)
For the equality, $(xAy*)^t = baryA^tx^t$, why are we able to switch $bary$ and $x^t$? Is it simply because we're dealing with a $1$x$1$ matrix?
linear-algebra
edited Jul 23 at 21:34
Omnomnomnom
121k784170
asked Jul 23 at 20:35
K.M
480312
edited Jul 23 at 21:34
Omnomnomnom
121k784170
edited Jul 23 at 21:34
Omnomnomnom
121k784170
edited Jul 23 at 21:34
Omnomnomnom
121k784170
121k784170
asked Jul 23 at 20:35
K.M
480312
asked Jul 23 at 20:35
K.M
480312
asked Jul 23 at 20:35
K.M
480312
480312
Yes, I know what * means. I still don't see why that allows me to switch.
– K.M
Jul 23 at 20:40
1
Honestly, if you don't understand how transpose and conjugate work, you should definitely not be looking up questions like this online. You are only hurting yourself in the long run. read your notes and textbook.
– AnonymousCoward
Jul 23 at 20:40
I recommend looking at the references you are using for your class and reading about those two topics.
– AnonymousCoward
Jul 23 at 20:40
I'm pretty sure it's something elementary, but going through an entire book to look for one fact would take up too much time.
– K.M
Jul 23 at 20:43
I`m not sure if I understood the question right, but using $(AB)^t=B^tA^t$, we get : $(xAy^*)^t=bar y^t^tA^tx^t=bar yA^tx^t $
– Sar
Jul 23 at 20:55
 |Â
show 2 more comments
Yes, I know what * means. I still don't see why that allows me to switch.
– K.M
Jul 23 at 20:40
1
Honestly, if you don't understand how transpose and conjugate work, you should definitely not be looking up questions like this online. You are only hurting yourself in the long run. read your notes and textbook.
– AnonymousCoward
Jul 23 at 20:40
I recommend looking at the references you are using for your class and reading about those two topics.
– AnonymousCoward
Jul 23 at 20:40
I'm pretty sure it's something elementary, but going through an entire book to look for one fact would take up too much time.
– K.M
Jul 23 at 20:43
I`m not sure if I understood the question right, but using $(AB)^t=B^tA^t$, we get : $(xAy^*)^t=bar y^t^tA^tx^t=bar yA^tx^t $
– Sar
Jul 23 at 20:55
Yes, I know what * means. I still don't see why that allows me to switch.
– K.M
Jul 23 at 20:40
Yes, I know what * means. I still don't see why that allows me to switch.
– K.M
Jul 23 at 20:40
1
1
Honestly, if you don't understand how transpose and conjugate work, you should definitely not be looking up questions like this online. You are only hurting yourself in the long run. read your notes and textbook.
– AnonymousCoward
Jul 23 at 20:40
Honestly, if you don't understand how transpose and conjugate work, you should definitely not be looking up questions like this online. You are only hurting yourself in the long run. read your notes and textbook.
– AnonymousCoward
Jul 23 at 20:40
I recommend looking at the references you are using for your class and reading about those two topics.
– AnonymousCoward
Jul 23 at 20:40
I recommend looking at the references you are using for your class and reading about those two topics.
– AnonymousCoward
Jul 23 at 20:40
I'm pretty sure it's something elementary, but going through an entire book to look for one fact would take up too much time.
– K.M
Jul 23 at 20:43
I'm pretty sure it's something elementary, but going through an entire book to look for one fact would take up too much time.
– K.M
Jul 23 at 20:43
I`m not sure if I understood the question right, but using $(AB)^t=B^tA^t$, we get : $(xAy^*)^t=bar y^t^tA^tx^t=bar yA^tx^t $
– Sar
Jul 23 at 20:55
I`m not sure if I understood the question right, but using $(AB)^t=B^tA^t$, we get : $(xAy^*)^t=bar y^t^tA^tx^t=bar yA^tx^t $
– Sar
Jul 23 at 20:55
 |Â
show 2 more comments
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f2860758%2fin-mathbbc2-show-that-langle-x-y-rangle-xay-is-an-inner-product%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
Yes, I know what * means. I still don't see why that allows me to switch.
– K.M
Jul 23 at 20:40
1
Honestly, if you don't understand how transpose and conjugate work, you should definitely not be looking up questions like this online. You are only hurting yourself in the long run. read your notes and textbook.
– AnonymousCoward
Jul 23 at 20:40
I recommend looking at the references you are using for your class and reading about those two topics.
– AnonymousCoward
Jul 23 at 20:40
I'm pretty sure it's something elementary, but going through an entire book to look for one fact would take up too much time.
– K.M
Jul 23 at 20:43
I`m not sure if I understood the question right, but using $(AB)^t=B^tA^t$, we get : $(xAy^*)^t=bar y^t^tA^tx^t=bar yA^tx^t $
– Sar
Jul 23 at 20:55