Mixing
Linear transformation from $R^3$ to $R^2$.
$begingroup$
Find the matrix of the linear transformation $Tcolon {Bbb R}^3 to {Bbb R}^2$ such that
$T(1,1,1) = (1,1)$, $T(1,2,3) = (1,2)$, $T(1,2,4) = (1,4)$.
So far, I have only dealt with transformations in the same R. Any help?
linear-algebra matrices linear-transformations
edited Jan 1 at 20:52
A.Γ.
22.6k32656
asked Mar 29 '17 at 21:08
AmaCAmaC
14410
$endgroup$
add a comment |
$begingroup$
Find the matrix of the linear transformation $Tcolon {Bbb R}^3 to {Bbb R}^2$ such that
$T(1,1,1) = (1,1)$, $T(1,2,3) = (1,2)$, $T(1,2,4) = (1,4)$.
So far, I have only dealt with transformations in the same R. Any help?
linear-algebra matrices linear-transformations
edited Jan 1 at 20:52
A.Γ.
22.6k32656
asked Mar 29 '17 at 21:08
AmaCAmaC
14410
$endgroup$
$begingroup$
The idea remains the same once you fix a basis of $mathbb{R}^3$ and $mathbb{R}^2$. The only difference you'll see is that earlier you would be getting square matrices. Here you'll get a $2times 3$ matrix. Thus in general $T:mathbb{R}^nto mathbb{R}^m$ yields a $m times n$ matrix. Can you see why?
$endgroup$
– Swapnil Tripathi
Mar 29 '17 at 21:17
add a comment |
$begingroup$
Find the matrix of the linear transformation $Tcolon {Bbb R}^3 to {Bbb R}^2$ such that
$T(1,1,1) = (1,1)$, $T(1,2,3) = (1,2)$, $T(1,2,4) = (1,4)$.
So far, I have only dealt with transformations in the same R. Any help?
linear-algebra matrices linear-transformations
edited Jan 1 at 20:52
A.Γ.
22.6k32656
asked Mar 29 '17 at 21:08
AmaCAmaC
14410
$endgroup$
Find the matrix of the linear transformation $Tcolon {Bbb R}^3 to {Bbb R}^2$ such that
$T(1,1,1) = (1,1)$, $T(1,2,3) = (1,2)$, $T(1,2,4) = (1,4)$.
So far, I have only dealt with transformations in the same R. Any help?
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
edited Jan 1 at 20:52
A.Γ.
22.6k32656
asked Mar 29 '17 at 21:08
AmaCAmaC
14410
edited Jan 1 at 20:52
A.Γ.
22.6k32656
asked Mar 29 '17 at 21:08
AmaCAmaC
14410
edited Jan 1 at 20:52
A.Γ.
22.6k32656
edited Jan 1 at 20:52
A.Γ.
22.6k32656
edited Jan 1 at 20:52
A.Γ.
22.6k32656
22.6k32656
asked Mar 29 '17 at 21:08
AmaCAmaC
14410
asked Mar 29 '17 at 21:08
AmaCAmaC
14410
asked Mar 29 '17 at 21:08
AmaCAmaC
14410
14410
$begingroup$
The idea remains the same once you fix a basis of $mathbb{R}^3$ and $mathbb{R}^2$. The only difference you'll see is that earlier you would be getting square matrices. Here you'll get a $2times 3$ matrix. Thus in general $T:mathbb{R}^nto mathbb{R}^m$ yields a $m times n$ matrix. Can you see why?
$endgroup$
– Swapnil Tripathi
Mar 29 '17 at 21:17
add a comment |
$begingroup$
The idea remains the same once you fix a basis of $mathbb{R}^3$ and $mathbb{R}^2$. The only difference you'll see is that earlier you would be getting square matrices. Here you'll get a $2times 3$ matrix. Thus in general $T:mathbb{R}^nto mathbb{R}^m$ yields a $m times n$ matrix. Can you see why?
$endgroup$
– Swapnil Tripathi
Mar 29 '17 at 21:17
$begingroup$
The idea remains the same once you fix a basis of $mathbb{R}^3$ and $mathbb{R}^2$. The only difference you'll see is that earlier you would be getting square matrices. Here you'll get a $2times 3$ matrix. Thus in general $T:mathbb{R}^nto mathbb{R}^m$ yields a $m times n$ matrix. Can you see why?
$endgroup$
– Swapnil Tripathi
Mar 29 '17 at 21:17
$begingroup$
The idea remains the same once you fix a basis of $mathbb{R}^3$ and $mathbb{R}^2$. The only difference you'll see is that earlier you would be getting square matrices. Here you'll get a $2times 3$ matrix. Thus in general $T:mathbb{R}^nto mathbb{R}^m$ yields a $m times n$ matrix. Can you see why?
$endgroup$
– Swapnil Tripathi
Mar 29 '17 at 21:17
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Group your 3 constraints into a single one:
$$tag{1}T.underbrace{begin{pmatrix}1&1&1\1&2&2\1&3&4end{pmatrix}}_{M}=underbrace{begin{pmatrix}1&1&1\1&2&4end{pmatrix}}_{N}$$
(where the point means matrix product).
(1) is equivalent to $T=N.M^{-1},$ which is a $2 times 3$ matrix.
Up to you for the last calculations.
You should find $begin{pmatrix}1& 0&0\2&-3&2end{pmatrix}.$
answered Mar 29 '17 at 21:15
Jean MarieJean Marie
28.8k41949
$endgroup$
$begingroup$
Why compute $M^{-1}$ ? We can simply use elementary linear combinaisons for get $T(e_3)$, and then it is easy to conclude.
$endgroup$
– user171326
Mar 29 '17 at 21:26
$begingroup$
"Because it gives the simplest explanation" (and the easiest way to give the computation to be done to a computer), I would say.
$endgroup$
– Jean Marie
Mar 29 '17 at 21:32
$begingroup$
I have added the result one must find.
$endgroup$
– Jean Marie
Mar 29 '17 at 22:53
add a comment |
$begingroup$
Hint : $T(0,0,1) = (0,2)$ so the last column of the matrix is $begin{pmatrix} 0 \ 2 end{pmatrix}$.
Do you see how to find it ? Do you see how to continue ?
$endgroup$
add a comment |
$begingroup$
Hint:
Try finding $a$, $b$, $c$, $d$, $e$, $f$ such that:
$$
left(
begin{array}{ccc}
a & b & c \
d & e & f \
end{array}
right)
left(
begin{array}{c}
1\
1\
1
end{array}
right)
=
left(
begin{array}{c}
1\
1
end{array}
right)
$$
What other matrix equations can you form?
answered Mar 29 '17 at 21:54
D. BritoD. Brito
358111
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Group your 3 constraints into a single one:
$$tag{1}T.underbrace{begin{pmatrix}1&1&1\1&2&2\1&3&4end{pmatrix}}_{M}=underbrace{begin{pmatrix}1&1&1\1&2&4end{pmatrix}}_{N}$$
(where the point means matrix product).
(1) is equivalent to $T=N.M^{-1},$ which is a $2 times 3$ matrix.
Up to you for the last calculations.
You should find $begin{pmatrix}1& 0&0\2&-3&2end{pmatrix}.$
answered Mar 29 '17 at 21:15
Jean MarieJean Marie
28.8k41949
$endgroup$
$begingroup$
Why compute $M^{-1}$ ? We can simply use elementary linear combinaisons for get $T(e_3)$, and then it is easy to conclude.
$endgroup$
– user171326
Mar 29 '17 at 21:26
$begingroup$
"Because it gives the simplest explanation" (and the easiest way to give the computation to be done to a computer), I would say.
$endgroup$
– Jean Marie
Mar 29 '17 at 21:32
$begingroup$
I have added the result one must find.
$endgroup$
– Jean Marie
Mar 29 '17 at 22:53
add a comment |
$begingroup$
Group your 3 constraints into a single one:
$$tag{1}T.underbrace{begin{pmatrix}1&1&1\1&2&2\1&3&4end{pmatrix}}_{M}=underbrace{begin{pmatrix}1&1&1\1&2&4end{pmatrix}}_{N}$$
(where the point means matrix product).
(1) is equivalent to $T=N.M^{-1},$ which is a $2 times 3$ matrix.
Up to you for the last calculations.
You should find $begin{pmatrix}1& 0&0\2&-3&2end{pmatrix}.$
answered Mar 29 '17 at 21:15
Jean MarieJean Marie
28.8k41949
$endgroup$
$begingroup$
Why compute $M^{-1}$ ? We can simply use elementary linear combinaisons for get $T(e_3)$, and then it is easy to conclude.
$endgroup$
– user171326
Mar 29 '17 at 21:26
$begingroup$
"Because it gives the simplest explanation" (and the easiest way to give the computation to be done to a computer), I would say.
$endgroup$
– Jean Marie
Mar 29 '17 at 21:32
$begingroup$
I have added the result one must find.
$endgroup$
– Jean Marie
Mar 29 '17 at 22:53
add a comment |
$begingroup$
Group your 3 constraints into a single one:
$$tag{1}T.underbrace{begin{pmatrix}1&1&1\1&2&2\1&3&4end{pmatrix}}_{M}=underbrace{begin{pmatrix}1&1&1\1&2&4end{pmatrix}}_{N}$$
(where the point means matrix product).
(1) is equivalent to $T=N.M^{-1},$ which is a $2 times 3$ matrix.
Up to you for the last calculations.
You should find $begin{pmatrix}1& 0&0\2&-3&2end{pmatrix}.$
answered Mar 29 '17 at 21:15
Jean MarieJean Marie
28.8k41949
$endgroup$
Group your 3 constraints into a single one:
$$tag{1}T.underbrace{begin{pmatrix}1&1&1\1&2&2\1&3&4end{pmatrix}}_{M}=underbrace{begin{pmatrix}1&1&1\1&2&4end{pmatrix}}_{N}$$
(where the point means matrix product).
(1) is equivalent to $T=N.M^{-1},$ which is a $2 times 3$ matrix.
Up to you for the last calculations.
You should find $begin{pmatrix}1& 0&0\2&-3&2end{pmatrix}.$
answered Mar 29 '17 at 21:15
Jean MarieJean Marie
28.8k41949
edited Mar 29 '17 at 22:52
answered Mar 29 '17 at 21:15
Jean MarieJean Marie
28.8k41949
answered Mar 29 '17 at 21:15
Jean MarieJean Marie
28.8k41949
answered Mar 29 '17 at 21:15
Jean MarieJean Marie
28.8k41949
28.8k41949
$begingroup$
Why compute $M^{-1}$ ? We can simply use elementary linear combinaisons for get $T(e_3)$, and then it is easy to conclude.
$endgroup$
– user171326
Mar 29 '17 at 21:26
$begingroup$
"Because it gives the simplest explanation" (and the easiest way to give the computation to be done to a computer), I would say.
$endgroup$
– Jean Marie
Mar 29 '17 at 21:32
$begingroup$
I have added the result one must find.
$endgroup$
– Jean Marie
Mar 29 '17 at 22:53
add a comment |
$begingroup$
Why compute $M^{-1}$ ? We can simply use elementary linear combinaisons for get $T(e_3)$, and then it is easy to conclude.
$endgroup$
– user171326
Mar 29 '17 at 21:26
$begingroup$
"Because it gives the simplest explanation" (and the easiest way to give the computation to be done to a computer), I would say.
$endgroup$
– Jean Marie
Mar 29 '17 at 21:32
$begingroup$
I have added the result one must find.
$endgroup$
– Jean Marie
Mar 29 '17 at 22:53
$begingroup$
Why compute $M^{-1}$ ? We can simply use elementary linear combinaisons for get $T(e_3)$, and then it is easy to conclude.
$endgroup$
– user171326
Mar 29 '17 at 21:26
$begingroup$
Why compute $M^{-1}$ ? We can simply use elementary linear combinaisons for get $T(e_3)$, and then it is easy to conclude.
$endgroup$
– user171326
Mar 29 '17 at 21:26
$begingroup$
"Because it gives the simplest explanation" (and the easiest way to give the computation to be done to a computer), I would say.
$endgroup$
– Jean Marie
Mar 29 '17 at 21:32
$begingroup$
"Because it gives the simplest explanation" (and the easiest way to give the computation to be done to a computer), I would say.
$endgroup$
– Jean Marie
Mar 29 '17 at 21:32
$begingroup$
I have added the result one must find.
$endgroup$
– Jean Marie
Mar 29 '17 at 22:53
$begingroup$
I have added the result one must find.
$endgroup$
– Jean Marie
Mar 29 '17 at 22:53
add a comment |
$begingroup$
Hint : $T(0,0,1) = (0,2)$ so the last column of the matrix is $begin{pmatrix} 0 \ 2 end{pmatrix}$.
Do you see how to find it ? Do you see how to continue ?
$endgroup$
add a comment |
$begingroup$
Hint : $T(0,0,1) = (0,2)$ so the last column of the matrix is $begin{pmatrix} 0 \ 2 end{pmatrix}$.
Do you see how to find it ? Do you see how to continue ?
$endgroup$
add a comment |
$begingroup$
Hint : $T(0,0,1) = (0,2)$ so the last column of the matrix is $begin{pmatrix} 0 \ 2 end{pmatrix}$.
Do you see how to find it ? Do you see how to continue ?
$endgroup$
Hint : $T(0,0,1) = (0,2)$ so the last column of the matrix is $begin{pmatrix} 0 \ 2 end{pmatrix}$.
Do you see how to find it ? Do you see how to continue ?
answered Mar 29 '17 at 21:12
user171326
add a comment |
add a comment |
$begingroup$
Hint:
Try finding $a$, $b$, $c$, $d$, $e$, $f$ such that:
$$
left(
begin{array}{ccc}
a & b & c \
d & e & f \
end{array}
right)
left(
begin{array}{c}
1\
1\
1
end{array}
right)
=
left(
begin{array}{c}
1\
1
end{array}
right)
$$
What other matrix equations can you form?
answered Mar 29 '17 at 21:54
D. BritoD. Brito
358111
$endgroup$
add a comment |
$begingroup$
Hint:
Try finding $a$, $b$, $c$, $d$, $e$, $f$ such that:
$$
left(
begin{array}{ccc}
a & b & c \
d & e & f \
end{array}
right)
left(
begin{array}{c}
1\
1\
1
end{array}
right)
=
left(
begin{array}{c}
1\
1
end{array}
right)
$$
What other matrix equations can you form?
answered Mar 29 '17 at 21:54
D. BritoD. Brito
358111
$endgroup$
add a comment |
$begingroup$
Hint:
Try finding $a$, $b$, $c$, $d$, $e$, $f$ such that:
$$
left(
begin{array}{ccc}
a & b & c \
d & e & f \
end{array}
right)
left(
begin{array}{c}
1\
1\
1
end{array}
right)
=
left(
begin{array}{c}
1\
1
end{array}
right)
$$
What other matrix equations can you form?
answered Mar 29 '17 at 21:54
D. BritoD. Brito
358111
$endgroup$
Hint:
Try finding $a$, $b$, $c$, $d$, $e$, $f$ such that:
$$
left(
begin{array}{ccc}
a & b & c \
d & e & f \
end{array}
right)
left(
begin{array}{c}
1\
1\
1
end{array}
right)
=
left(
begin{array}{c}
1\
1
end{array}
right)
$$
What other matrix equations can you form?
answered Mar 29 '17 at 21:54
D. BritoD. Brito
358111
answered Mar 29 '17 at 21:54
D. BritoD. Brito
358111
answered Mar 29 '17 at 21:54
D. BritoD. Brito
358111
answered Mar 29 '17 at 21:54
D. BritoD. Brito
358111
358111
add a comment |
add a comment |
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$begingroup$
The idea remains the same once you fix a basis of $mathbb{R}^3$ and $mathbb{R}^2$. The only difference you'll see is that earlier you would be getting square matrices. Here you'll get a $2times 3$ matrix. Thus in general $T:mathbb{R}^nto mathbb{R}^m$ yields a $m times n$ matrix. Can you see why?
$endgroup$
– Swapnil Tripathi
Mar 29 '17 at 21:17