Commutative property of linear operators
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I know that if we have to cascade linear system we can substitute the place of each system and get the same result.
We can simply proof this property by writing the output of each system as convolution of impulse response and combine integrals.
But my question is about linear operators.
If :
$$ L1: V rightarrow V$$
$$ L2: V rightarrow V$$
$$L1[aX+bY]=aL1[X]+bL1[Y]$$
$$L2[aX+bY]=aL2[X]+bL2[Y]$$
Does the below equality hold? if yes what is the proof?
$$L2[L1[X]]=L1[L2[X]]$$
linear-transformations
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up vote
0
down vote
favorite
I know that if we have to cascade linear system we can substitute the place of each system and get the same result.
We can simply proof this property by writing the output of each system as convolution of impulse response and combine integrals.
But my question is about linear operators.
If :
$$ L1: V rightarrow V$$
$$ L2: V rightarrow V$$
$$L1[aX+bY]=aL1[X]+bL1[Y]$$
$$L2[aX+bY]=aL2[X]+bL2[Y]$$
Does the below equality hold? if yes what is the proof?
$$L2[L1[X]]=L1[L2[X]]$$
linear-transformations
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know that if we have to cascade linear system we can substitute the place of each system and get the same result.
We can simply proof this property by writing the output of each system as convolution of impulse response and combine integrals.
But my question is about linear operators.
If :
$$ L1: V rightarrow V$$
$$ L2: V rightarrow V$$
$$L1[aX+bY]=aL1[X]+bL1[Y]$$
$$L2[aX+bY]=aL2[X]+bL2[Y]$$
Does the below equality hold? if yes what is the proof?
$$L2[L1[X]]=L1[L2[X]]$$
linear-transformations
I know that if we have to cascade linear system we can substitute the place of each system and get the same result.
We can simply proof this property by writing the output of each system as convolution of impulse response and combine integrals.
But my question is about linear operators.
If :
$$ L1: V rightarrow V$$
$$ L2: V rightarrow V$$
$$L1[aX+bY]=aL1[X]+bL1[Y]$$
$$L2[aX+bY]=aL2[X]+bL2[Y]$$
Does the below equality hold? if yes what is the proof?
$$L2[L1[X]]=L1[L2[X]]$$
linear-transformations
linear-transformations
edited Nov 15 at 19:02
asked Nov 15 at 18:54
alireza
73
73
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