Mixing
Composition of linear transformation and domain and codomain
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I am trying to do a problem regarding composition of linear transformation from my book. I want to know if my reasoning is correct.
$$T_1(x,y,z)=(2x+3y-z,x-y+z,2x+9z,3y-10z)$$
$$T_2(x_1,x_2,x_3,x_4)=(9x_1 - 3x_4,2x_2 + 4x_3 - 2x_4,2x_3 + 2x_4)$$
Q1. State domain and codomain of $T_2circ T_1$ and $T_1 circ T_2$.
Since for $T_1$ we are going from $mathbb R^3to mathbb R^4$ and for $T_2$ from $mathbb R^4to mathbb R^3$
,domain of $T_2 circ T_1$ will be $mathbb R^3$ and Codomain will also be $mathbb R^3$
Domain of $T_1 circ T_2$ will be $mathbb R^4$ and codomain will also be $mathbb R^4$.
Q2 Why are $T_2 circ T_1$ and $T_1 circ T_2$ well defined?
$T_2 circ T_1$ is well defined because the range of $T_1$ is $mathbb R^4$ and $mathbb R^4$ lies in the domain of $T_2$.
Similarly for $T_1 circ T_2$, the range of $T_2$ is $mathbb R^3$ and lies in the domain of $T_1$ which is $mathbb R^3$.
Is my reasoning correct? Any help will be appreciated.
linear-algebra
edited Jan 30 at 10:45
kelvin hong 方
83919
asked Jan 30 at 10:34
SchniederSchnieder
62
$endgroup$
add a comment |
$begingroup$
I am trying to do a problem regarding composition of linear transformation from my book. I want to know if my reasoning is correct.
$$T_1(x,y,z)=(2x+3y-z,x-y+z,2x+9z,3y-10z)$$
$$T_2(x_1,x_2,x_3,x_4)=(9x_1 - 3x_4,2x_2 + 4x_3 - 2x_4,2x_3 + 2x_4)$$
Q1. State domain and codomain of $T_2circ T_1$ and $T_1 circ T_2$.
Since for $T_1$ we are going from $mathbb R^3to mathbb R^4$ and for $T_2$ from $mathbb R^4to mathbb R^3$
,domain of $T_2 circ T_1$ will be $mathbb R^3$ and Codomain will also be $mathbb R^3$
Domain of $T_1 circ T_2$ will be $mathbb R^4$ and codomain will also be $mathbb R^4$.
Q2 Why are $T_2 circ T_1$ and $T_1 circ T_2$ well defined?
$T_2 circ T_1$ is well defined because the range of $T_1$ is $mathbb R^4$ and $mathbb R^4$ lies in the domain of $T_2$.
Similarly for $T_1 circ T_2$, the range of $T_2$ is $mathbb R^3$ and lies in the domain of $T_1$ which is $mathbb R^3$.
Is my reasoning correct? Any help will be appreciated.
linear-algebra
edited Jan 30 at 10:45
kelvin hong 方
83919
asked Jan 30 at 10:34
SchniederSchnieder
62
$endgroup$
1
$begingroup$
You could improve the readability of your question by using MathJax to type the equations.
$endgroup$
– Matti P.
Jan 30 at 10:36
add a comment |
$begingroup$
I am trying to do a problem regarding composition of linear transformation from my book. I want to know if my reasoning is correct.
$$T_1(x,y,z)=(2x+3y-z,x-y+z,2x+9z,3y-10z)$$
$$T_2(x_1,x_2,x_3,x_4)=(9x_1 - 3x_4,2x_2 + 4x_3 - 2x_4,2x_3 + 2x_4)$$
Q1. State domain and codomain of $T_2circ T_1$ and $T_1 circ T_2$.
Since for $T_1$ we are going from $mathbb R^3to mathbb R^4$ and for $T_2$ from $mathbb R^4to mathbb R^3$
,domain of $T_2 circ T_1$ will be $mathbb R^3$ and Codomain will also be $mathbb R^3$
Domain of $T_1 circ T_2$ will be $mathbb R^4$ and codomain will also be $mathbb R^4$.
Q2 Why are $T_2 circ T_1$ and $T_1 circ T_2$ well defined?
$T_2 circ T_1$ is well defined because the range of $T_1$ is $mathbb R^4$ and $mathbb R^4$ lies in the domain of $T_2$.
Similarly for $T_1 circ T_2$, the range of $T_2$ is $mathbb R^3$ and lies in the domain of $T_1$ which is $mathbb R^3$.
Is my reasoning correct? Any help will be appreciated.
linear-algebra
edited Jan 30 at 10:45
kelvin hong 方
83919
asked Jan 30 at 10:34
SchniederSchnieder
62
$endgroup$
I am trying to do a problem regarding composition of linear transformation from my book. I want to know if my reasoning is correct.
$$T_1(x,y,z)=(2x+3y-z,x-y+z,2x+9z,3y-10z)$$
$$T_2(x_1,x_2,x_3,x_4)=(9x_1 - 3x_4,2x_2 + 4x_3 - 2x_4,2x_3 + 2x_4)$$
Q1. State domain and codomain of $T_2circ T_1$ and $T_1 circ T_2$.
Since for $T_1$ we are going from $mathbb R^3to mathbb R^4$ and for $T_2$ from $mathbb R^4to mathbb R^3$
,domain of $T_2 circ T_1$ will be $mathbb R^3$ and Codomain will also be $mathbb R^3$
Domain of $T_1 circ T_2$ will be $mathbb R^4$ and codomain will also be $mathbb R^4$.
Q2 Why are $T_2 circ T_1$ and $T_1 circ T_2$ well defined?
$T_2 circ T_1$ is well defined because the range of $T_1$ is $mathbb R^4$ and $mathbb R^4$ lies in the domain of $T_2$.
Similarly for $T_1 circ T_2$, the range of $T_2$ is $mathbb R^3$ and lies in the domain of $T_1$ which is $mathbb R^3$.
Is my reasoning correct? Any help will be appreciated.
linear-algebra
linear-algebra
edited Jan 30 at 10:45
kelvin hong 方
83919
asked Jan 30 at 10:34
SchniederSchnieder
62
edited Jan 30 at 10:45
kelvin hong 方
83919
asked Jan 30 at 10:34
SchniederSchnieder
62
edited Jan 30 at 10:45
kelvin hong 方
83919
edited Jan 30 at 10:45
kelvin hong 方
83919
edited Jan 30 at 10:45
kelvin hong 方
83919
83919
asked Jan 30 at 10:34
SchniederSchnieder
62
asked Jan 30 at 10:34
SchniederSchnieder
62
asked Jan 30 at 10:34
SchniederSchnieder
62
62
1
$begingroup$
You could improve the readability of your question by using MathJax to type the equations.
$endgroup$
– Matti P.
Jan 30 at 10:36
add a comment |
1
$begingroup$
You could improve the readability of your question by using MathJax to type the equations.
$endgroup$
– Matti P.
Jan 30 at 10:36
1
1
$begingroup$
You could improve the readability of your question by using MathJax to type the equations.
$endgroup$
– Matti P.
Jan 30 at 10:36
$begingroup$
You could improve the readability of your question by using MathJax to type the equations.
$endgroup$
– Matti P.
Jan 30 at 10:36
add a comment |
1 Answer
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$begingroup$
Your answer to Q1 is correct!
For Q2, first note that the range (image) of $T_1$ cannot be $mathbb{R} ^4$, since $dimoperatorname{Ran} T_1 leqdim operatorname{Dom} T_1 = 3$.
The range of $T_1$ consists only of those vectors that are of the form $v = T_1 w$ for some $winmathbb{R}^3$.
To show that $T_2circ T_1$ is well-defined, it suffices to see that the range (or just the codomain, since the range is a subset of the codomain) of $T_1$ is a subset of the domain of $T_2$. This means that $T_2$ can be applied to any output of $T_1$. But you have already seen that the $operatorname{Dom} T_2 = operatorname{Codom} T_1$.
The same reasoning goes through for $T_1circ T_2$ being well-defined. Again, domain of $T_1$ and codomain of $T_2$ coïncide.
answered Jan 30 at 11:38
ffffforallffffforall
36028
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$begingroup$
Your answer to Q1 is correct!
For Q2, first note that the range (image) of $T_1$ cannot be $mathbb{R} ^4$, since $dimoperatorname{Ran} T_1 leqdim operatorname{Dom} T_1 = 3$.
The range of $T_1$ consists only of those vectors that are of the form $v = T_1 w$ for some $winmathbb{R}^3$.
To show that $T_2circ T_1$ is well-defined, it suffices to see that the range (or just the codomain, since the range is a subset of the codomain) of $T_1$ is a subset of the domain of $T_2$. This means that $T_2$ can be applied to any output of $T_1$. But you have already seen that the $operatorname{Dom} T_2 = operatorname{Codom} T_1$.
The same reasoning goes through for $T_1circ T_2$ being well-defined. Again, domain of $T_1$ and codomain of $T_2$ coïncide.
answered Jan 30 at 11:38
ffffforallffffforall
36028
$endgroup$
add a comment |
$begingroup$
Your answer to Q1 is correct!
For Q2, first note that the range (image) of $T_1$ cannot be $mathbb{R} ^4$, since $dimoperatorname{Ran} T_1 leqdim operatorname{Dom} T_1 = 3$.
The range of $T_1$ consists only of those vectors that are of the form $v = T_1 w$ for some $winmathbb{R}^3$.
To show that $T_2circ T_1$ is well-defined, it suffices to see that the range (or just the codomain, since the range is a subset of the codomain) of $T_1$ is a subset of the domain of $T_2$. This means that $T_2$ can be applied to any output of $T_1$. But you have already seen that the $operatorname{Dom} T_2 = operatorname{Codom} T_1$.
The same reasoning goes through for $T_1circ T_2$ being well-defined. Again, domain of $T_1$ and codomain of $T_2$ coïncide.
answered Jan 30 at 11:38
ffffforallffffforall
36028
$endgroup$
add a comment |
$begingroup$
Your answer to Q1 is correct!
For Q2, first note that the range (image) of $T_1$ cannot be $mathbb{R} ^4$, since $dimoperatorname{Ran} T_1 leqdim operatorname{Dom} T_1 = 3$.
The range of $T_1$ consists only of those vectors that are of the form $v = T_1 w$ for some $winmathbb{R}^3$.
To show that $T_2circ T_1$ is well-defined, it suffices to see that the range (or just the codomain, since the range is a subset of the codomain) of $T_1$ is a subset of the domain of $T_2$. This means that $T_2$ can be applied to any output of $T_1$. But you have already seen that the $operatorname{Dom} T_2 = operatorname{Codom} T_1$.
The same reasoning goes through for $T_1circ T_2$ being well-defined. Again, domain of $T_1$ and codomain of $T_2$ coïncide.
answered Jan 30 at 11:38
ffffforallffffforall
36028
$endgroup$
Your answer to Q1 is correct!
For Q2, first note that the range (image) of $T_1$ cannot be $mathbb{R} ^4$, since $dimoperatorname{Ran} T_1 leqdim operatorname{Dom} T_1 = 3$.
The range of $T_1$ consists only of those vectors that are of the form $v = T_1 w$ for some $winmathbb{R}^3$.
To show that $T_2circ T_1$ is well-defined, it suffices to see that the range (or just the codomain, since the range is a subset of the codomain) of $T_1$ is a subset of the domain of $T_2$. This means that $T_2$ can be applied to any output of $T_1$. But you have already seen that the $operatorname{Dom} T_2 = operatorname{Codom} T_1$.
The same reasoning goes through for $T_1circ T_2$ being well-defined. Again, domain of $T_1$ and codomain of $T_2$ coïncide.
answered Jan 30 at 11:38
ffffforallffffforall
36028
answered Jan 30 at 11:38
ffffforallffffforall
36028
answered Jan 30 at 11:38
ffffforallffffforall
36028
answered Jan 30 at 11:38
ffffforallffffforall
36028
36028
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You could improve the readability of your question by using MathJax to type the equations.
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– Matti P.
Jan 30 at 10:36