Mixing
Product of a skew hermitian and real matrix [closed]
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Let A and B be two 2×2 matrix with A being skew-hermitian and AB=B .show that B=0
please help need hint
matrices
asked Jul 30 at 17:30
Shrimon Mukherjee
155
closed as off-topic by Xander Henderson, Leucippus, Jyrki Lahtonen, Taroccoesbrocco, Shailesh Jul 31 at 9:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Leucippus, Jyrki Lahtonen, Taroccoesbrocco, Shailesh
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up vote
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down vote
favorite
Let A and B be two 2×2 matrix with A being skew-hermitian and AB=B .show that B=0
please help need hint
matrices
asked Jul 30 at 17:30
Shrimon Mukherjee
155
closed as off-topic by Xander Henderson, Leucippus, Jyrki Lahtonen, Taroccoesbrocco, Shailesh Jul 31 at 9:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Leucippus, Jyrki Lahtonen, Taroccoesbrocco, Shailesh
Your title says that $B$ is real, but your question body doesn't. Is $B$ assumed to be real?
– Arthur
Jul 30 at 17:34
Yes. B IS REAL matrix
– Shrimon Mukherjee
Jul 30 at 17:39
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let A and B be two 2×2 matrix with A being skew-hermitian and AB=B .show that B=0
please help need hint
matrices
asked Jul 30 at 17:30
Shrimon Mukherjee
155
Let A and B be two 2×2 matrix with A being skew-hermitian and AB=B .show that B=0
please help need hint
matrices
asked Jul 30 at 17:30
Shrimon Mukherjee
155
asked Jul 30 at 17:30
Shrimon Mukherjee
155
asked Jul 30 at 17:30
Shrimon Mukherjee
155
asked Jul 30 at 17:30
Shrimon Mukherjee
155
155
closed as off-topic by Xander Henderson, Leucippus, Jyrki Lahtonen, Taroccoesbrocco, Shailesh Jul 31 at 9:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Leucippus, Jyrki Lahtonen, Taroccoesbrocco, Shailesh
closed as off-topic by Xander Henderson, Leucippus, Jyrki Lahtonen, Taroccoesbrocco, Shailesh Jul 31 at 9:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Leucippus, Jyrki Lahtonen, Taroccoesbrocco, Shailesh
Your title says that $B$ is real, but your question body doesn't. Is $B$ assumed to be real?
– Arthur
Jul 30 at 17:34
Yes. B IS REAL matrix
– Shrimon Mukherjee
Jul 30 at 17:39
add a comment |Â
Your title says that $B$ is real, but your question body doesn't. Is $B$ assumed to be real?
– Arthur
Jul 30 at 17:34
Yes. B IS REAL matrix
– Shrimon Mukherjee
Jul 30 at 17:39
Your title says that $B$ is real, but your question body doesn't. Is $B$ assumed to be real?
– Arthur
Jul 30 at 17:34
Your title says that $B$ is real, but your question body doesn't. Is $B$ assumed to be real?
– Arthur
Jul 30 at 17:34
Yes. B IS REAL matrix
– Shrimon Mukherjee
Jul 30 at 17:39
Yes. B IS REAL matrix
– Shrimon Mukherjee
Jul 30 at 17:39
add a comment |Â
2 Answers
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Try rewriting the equation as $(A-I)B = 0.$ What piece of information about $A-I$ would let you conclude that $B=0$? Since $A-I$ is just a $2times 2$ matrix, you can check directly that this condition on $A-I$ holds.
answered Jul 30 at 18:27
William Swartworth
398210
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We are given that $A^dagger=-A$ and that $B$ is real. The first condition implies that $a_ij=-overlinea_ji.$ On the main diagonal of $A,$ we must have that $a_ii=-overlinea_ii,$ or, if $a_ii=a+bi,$ then $a+bi=-(a-bi),$ or $a+bi=-a+bi,$ and hence $a=0.$ That is, the main diagonals must have purely imaginary entries. Now suppose $a_12=c+di.$ Then $a_21=-overlinea_12=-(c-di)=-c+di.$ Hence, we may characterize $A$ as follows:
$$ A=left[beginmatrixai &c+di\ -c+di &biendmatrixright].$$
Now, for $B$ real, we simply write
$$B=left[beginmatrixe &f\ g &h endmatrixright]. $$
Here, $a, b, c, d, e, f, g, hinmathbbR.$ Multiplying out yields
$$AB=left[beginmatrixcg+(ae+dg)i&ch+(af+dh)i\ -ce+(de+bg)i &-cf+(df+bh)iendmatrixright]=left[beginmatrixe &f\ g &h endmatrixright]. $$
From this, we may conclude that
beginalign*
cg&=e \
ch&=f \
-ce&=g \
-cf&=h.
endalign*
I'm just equating the real parts of all four entries.
A couple of substitutions yields
beginalign*
-c^2g&=g implies 0=g(1+c^2)\
-c^2h&=h implies 0=h(1+c^2).
endalign*
This can only happen if $g=h=0,$ from which it follows that $e=f=0,$ and you're done.
answered Jul 30 at 18:37
Adrian Keister
3,51821533
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Try rewriting the equation as $(A-I)B = 0.$ What piece of information about $A-I$ would let you conclude that $B=0$? Since $A-I$ is just a $2times 2$ matrix, you can check directly that this condition on $A-I$ holds.
answered Jul 30 at 18:27
William Swartworth
398210
add a comment |Â
up vote
0
down vote
Try rewriting the equation as $(A-I)B = 0.$ What piece of information about $A-I$ would let you conclude that $B=0$? Since $A-I$ is just a $2times 2$ matrix, you can check directly that this condition on $A-I$ holds.
answered Jul 30 at 18:27
William Swartworth
398210
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Try rewriting the equation as $(A-I)B = 0.$ What piece of information about $A-I$ would let you conclude that $B=0$? Since $A-I$ is just a $2times 2$ matrix, you can check directly that this condition on $A-I$ holds.
answered Jul 30 at 18:27
William Swartworth
398210
Try rewriting the equation as $(A-I)B = 0.$ What piece of information about $A-I$ would let you conclude that $B=0$? Since $A-I$ is just a $2times 2$ matrix, you can check directly that this condition on $A-I$ holds.
answered Jul 30 at 18:27
William Swartworth
398210
answered Jul 30 at 18:27
William Swartworth
398210
answered Jul 30 at 18:27
William Swartworth
398210
answered Jul 30 at 18:27
William Swartworth
398210
398210
add a comment |Â
add a comment |Â
up vote
0
down vote
We are given that $A^dagger=-A$ and that $B$ is real. The first condition implies that $a_ij=-overlinea_ji.$ On the main diagonal of $A,$ we must have that $a_ii=-overlinea_ii,$ or, if $a_ii=a+bi,$ then $a+bi=-(a-bi),$ or $a+bi=-a+bi,$ and hence $a=0.$ That is, the main diagonals must have purely imaginary entries. Now suppose $a_12=c+di.$ Then $a_21=-overlinea_12=-(c-di)=-c+di.$ Hence, we may characterize $A$ as follows:
$$ A=left[beginmatrixai &c+di\ -c+di &biendmatrixright].$$
Now, for $B$ real, we simply write
$$B=left[beginmatrixe &f\ g &h endmatrixright]. $$
Here, $a, b, c, d, e, f, g, hinmathbbR.$ Multiplying out yields
$$AB=left[beginmatrixcg+(ae+dg)i&ch+(af+dh)i\ -ce+(de+bg)i &-cf+(df+bh)iendmatrixright]=left[beginmatrixe &f\ g &h endmatrixright]. $$
From this, we may conclude that
beginalign*
cg&=e \
ch&=f \
-ce&=g \
-cf&=h.
endalign*
I'm just equating the real parts of all four entries.
A couple of substitutions yields
beginalign*
-c^2g&=g implies 0=g(1+c^2)\
-c^2h&=h implies 0=h(1+c^2).
endalign*
This can only happen if $g=h=0,$ from which it follows that $e=f=0,$ and you're done.
answered Jul 30 at 18:37
Adrian Keister
3,51821533
add a comment |Â
up vote
0
down vote
We are given that $A^dagger=-A$ and that $B$ is real. The first condition implies that $a_ij=-overlinea_ji.$ On the main diagonal of $A,$ we must have that $a_ii=-overlinea_ii,$ or, if $a_ii=a+bi,$ then $a+bi=-(a-bi),$ or $a+bi=-a+bi,$ and hence $a=0.$ That is, the main diagonals must have purely imaginary entries. Now suppose $a_12=c+di.$ Then $a_21=-overlinea_12=-(c-di)=-c+di.$ Hence, we may characterize $A$ as follows:
$$ A=left[beginmatrixai &c+di\ -c+di &biendmatrixright].$$
Now, for $B$ real, we simply write
$$B=left[beginmatrixe &f\ g &h endmatrixright]. $$
Here, $a, b, c, d, e, f, g, hinmathbbR.$ Multiplying out yields
$$AB=left[beginmatrixcg+(ae+dg)i&ch+(af+dh)i\ -ce+(de+bg)i &-cf+(df+bh)iendmatrixright]=left[beginmatrixe &f\ g &h endmatrixright]. $$
From this, we may conclude that
beginalign*
cg&=e \
ch&=f \
-ce&=g \
-cf&=h.
endalign*
I'm just equating the real parts of all four entries.
A couple of substitutions yields
beginalign*
-c^2g&=g implies 0=g(1+c^2)\
-c^2h&=h implies 0=h(1+c^2).
endalign*
This can only happen if $g=h=0,$ from which it follows that $e=f=0,$ and you're done.
answered Jul 30 at 18:37
Adrian Keister
3,51821533
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We are given that $A^dagger=-A$ and that $B$ is real. The first condition implies that $a_ij=-overlinea_ji.$ On the main diagonal of $A,$ we must have that $a_ii=-overlinea_ii,$ or, if $a_ii=a+bi,$ then $a+bi=-(a-bi),$ or $a+bi=-a+bi,$ and hence $a=0.$ That is, the main diagonals must have purely imaginary entries. Now suppose $a_12=c+di.$ Then $a_21=-overlinea_12=-(c-di)=-c+di.$ Hence, we may characterize $A$ as follows:
$$ A=left[beginmatrixai &c+di\ -c+di &biendmatrixright].$$
Now, for $B$ real, we simply write
$$B=left[beginmatrixe &f\ g &h endmatrixright]. $$
Here, $a, b, c, d, e, f, g, hinmathbbR.$ Multiplying out yields
$$AB=left[beginmatrixcg+(ae+dg)i&ch+(af+dh)i\ -ce+(de+bg)i &-cf+(df+bh)iendmatrixright]=left[beginmatrixe &f\ g &h endmatrixright]. $$
From this, we may conclude that
beginalign*
cg&=e \
ch&=f \
-ce&=g \
-cf&=h.
endalign*
I'm just equating the real parts of all four entries.
A couple of substitutions yields
beginalign*
-c^2g&=g implies 0=g(1+c^2)\
-c^2h&=h implies 0=h(1+c^2).
endalign*
This can only happen if $g=h=0,$ from which it follows that $e=f=0,$ and you're done.
answered Jul 30 at 18:37
Adrian Keister
3,51821533
We are given that $A^dagger=-A$ and that $B$ is real. The first condition implies that $a_ij=-overlinea_ji.$ On the main diagonal of $A,$ we must have that $a_ii=-overlinea_ii,$ or, if $a_ii=a+bi,$ then $a+bi=-(a-bi),$ or $a+bi=-a+bi,$ and hence $a=0.$ That is, the main diagonals must have purely imaginary entries. Now suppose $a_12=c+di.$ Then $a_21=-overlinea_12=-(c-di)=-c+di.$ Hence, we may characterize $A$ as follows:
$$ A=left[beginmatrixai &c+di\ -c+di &biendmatrixright].$$
Now, for $B$ real, we simply write
$$B=left[beginmatrixe &f\ g &h endmatrixright]. $$
Here, $a, b, c, d, e, f, g, hinmathbbR.$ Multiplying out yields
$$AB=left[beginmatrixcg+(ae+dg)i&ch+(af+dh)i\ -ce+(de+bg)i &-cf+(df+bh)iendmatrixright]=left[beginmatrixe &f\ g &h endmatrixright]. $$
From this, we may conclude that
beginalign*
cg&=e \
ch&=f \
-ce&=g \
-cf&=h.
endalign*
I'm just equating the real parts of all four entries.
A couple of substitutions yields
beginalign*
-c^2g&=g implies 0=g(1+c^2)\
-c^2h&=h implies 0=h(1+c^2).
endalign*
This can only happen if $g=h=0,$ from which it follows that $e=f=0,$ and you're done.
answered Jul 30 at 18:37
Adrian Keister
3,51821533
answered Jul 30 at 18:37
Adrian Keister
3,51821533
answered Jul 30 at 18:37
Adrian Keister
3,51821533
answered Jul 30 at 18:37
Adrian Keister
3,51821533
3,51821533
add a comment |Â
add a comment |Â
Your title says that $B$ is real, but your question body doesn't. Is $B$ assumed to be real?
– Arthur
Jul 30 at 17:34
Yes. B IS REAL matrix
– Shrimon Mukherjee
Jul 30 at 17:39