Mixing
Combinations of linear mappings
$begingroup$
What I know: the composition of two linear transformations, say $(Mcirc L)(x)$, has the geometric interpretation of successively applying two transformations to $x$, e.g. first a rotation by $L$, then a shear by $M$.
My question is this: What is the geometric interpretation of
$(L+M)(x)$
$(tL)(x)$, where $t$ is some real number?
I would suppose we are simply achieving a new transformation, but I do not know how we this would be different from that defined by $(Mcirc L)(x)$.
linear-algebra
asked Jan 24 at 20:24
Julia KimJulia Kim
174
$endgroup$
add a comment |
$begingroup$
What I know: the composition of two linear transformations, say $(Mcirc L)(x)$, has the geometric interpretation of successively applying two transformations to $x$, e.g. first a rotation by $L$, then a shear by $M$.
My question is this: What is the geometric interpretation of
$(L+M)(x)$
$(tL)(x)$, where $t$ is some real number?
I would suppose we are simply achieving a new transformation, but I do not know how we this would be different from that defined by $(Mcirc L)(x)$.
linear-algebra
asked Jan 24 at 20:24
Julia KimJulia Kim
174
$endgroup$
$begingroup$
Well, if you took $frac{1}{2}(L+M)(x)$, the result would be the average of the two transformations.
$endgroup$
– Morgan Rodgers
Jan 24 at 20:40
add a comment |
$begingroup$
What I know: the composition of two linear transformations, say $(Mcirc L)(x)$, has the geometric interpretation of successively applying two transformations to $x$, e.g. first a rotation by $L$, then a shear by $M$.
My question is this: What is the geometric interpretation of
$(L+M)(x)$
$(tL)(x)$, where $t$ is some real number?
I would suppose we are simply achieving a new transformation, but I do not know how we this would be different from that defined by $(Mcirc L)(x)$.
linear-algebra
asked Jan 24 at 20:24
Julia KimJulia Kim
174
$endgroup$
What I know: the composition of two linear transformations, say $(Mcirc L)(x)$, has the geometric interpretation of successively applying two transformations to $x$, e.g. first a rotation by $L$, then a shear by $M$.
My question is this: What is the geometric interpretation of
$(L+M)(x)$
$(tL)(x)$, where $t$ is some real number?
I would suppose we are simply achieving a new transformation, but I do not know how we this would be different from that defined by $(Mcirc L)(x)$.
linear-algebra
linear-algebra
asked Jan 24 at 20:24
Julia KimJulia Kim
174
asked Jan 24 at 20:24
Julia KimJulia Kim
174
asked Jan 24 at 20:24
Julia KimJulia Kim
174
asked Jan 24 at 20:24
Julia KimJulia Kim
174
asked Jan 24 at 20:24
Julia KimJulia Kim
174
174
$begingroup$
Well, if you took $frac{1}{2}(L+M)(x)$, the result would be the average of the two transformations.
$endgroup$
– Morgan Rodgers
Jan 24 at 20:40
add a comment |
$begingroup$
Well, if you took $frac{1}{2}(L+M)(x)$, the result would be the average of the two transformations.
$endgroup$
– Morgan Rodgers
Jan 24 at 20:40
$begingroup$
Well, if you took $frac{1}{2}(L+M)(x)$, the result would be the average of the two transformations.
$endgroup$
– Morgan Rodgers
Jan 24 at 20:40
$begingroup$
Well, if you took $frac{1}{2}(L+M)(x)$, the result would be the average of the two transformations.
$endgroup$
– Morgan Rodgers
Jan 24 at 20:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In some sense, 2 is again of the form $(tI_ncirc L)(x)$, and you may regard this as first applying L, followed by the action of the linear map $tI_n$, and $tI_n$ uniformly dilates all vectors by t away from the origin. (order you perform these is immaterial in this case)
For 1, $(L+M)(x)= L(x)+M(x)$ holds, and provided that they have the same domain/codomain, then $L+M$ is just another linear transformation. You can simply think of it as adding the images under M and L together.
answered Jan 24 at 20:39
AlexandrosAlexandros
9421412
$endgroup$
add a comment |
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1 Answer
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$begingroup$
In some sense, 2 is again of the form $(tI_ncirc L)(x)$, and you may regard this as first applying L, followed by the action of the linear map $tI_n$, and $tI_n$ uniformly dilates all vectors by t away from the origin. (order you perform these is immaterial in this case)
For 1, $(L+M)(x)= L(x)+M(x)$ holds, and provided that they have the same domain/codomain, then $L+M$ is just another linear transformation. You can simply think of it as adding the images under M and L together.
answered Jan 24 at 20:39
AlexandrosAlexandros
9421412
$endgroup$
add a comment |
$begingroup$
In some sense, 2 is again of the form $(tI_ncirc L)(x)$, and you may regard this as first applying L, followed by the action of the linear map $tI_n$, and $tI_n$ uniformly dilates all vectors by t away from the origin. (order you perform these is immaterial in this case)
For 1, $(L+M)(x)= L(x)+M(x)$ holds, and provided that they have the same domain/codomain, then $L+M$ is just another linear transformation. You can simply think of it as adding the images under M and L together.
answered Jan 24 at 20:39
AlexandrosAlexandros
9421412
$endgroup$
add a comment |
$begingroup$
In some sense, 2 is again of the form $(tI_ncirc L)(x)$, and you may regard this as first applying L, followed by the action of the linear map $tI_n$, and $tI_n$ uniformly dilates all vectors by t away from the origin. (order you perform these is immaterial in this case)
For 1, $(L+M)(x)= L(x)+M(x)$ holds, and provided that they have the same domain/codomain, then $L+M$ is just another linear transformation. You can simply think of it as adding the images under M and L together.
answered Jan 24 at 20:39
AlexandrosAlexandros
9421412
$endgroup$
In some sense, 2 is again of the form $(tI_ncirc L)(x)$, and you may regard this as first applying L, followed by the action of the linear map $tI_n$, and $tI_n$ uniformly dilates all vectors by t away from the origin. (order you perform these is immaterial in this case)
For 1, $(L+M)(x)= L(x)+M(x)$ holds, and provided that they have the same domain/codomain, then $L+M$ is just another linear transformation. You can simply think of it as adding the images under M and L together.
answered Jan 24 at 20:39
AlexandrosAlexandros
9421412
answered Jan 24 at 20:39
AlexandrosAlexandros
9421412
answered Jan 24 at 20:39
AlexandrosAlexandros
9421412
answered Jan 24 at 20:39
AlexandrosAlexandros
9421412
9421412
add a comment |
add a comment |
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$begingroup$
Well, if you took $frac{1}{2}(L+M)(x)$, the result would be the average of the two transformations.
$endgroup$
– Morgan Rodgers
Jan 24 at 20:40