Let $L: mathbb{R}^2rightarrow mathbb{R}^2$ be a linear operator such that $L((1,2)^T)=(-2,3)^T$ and...
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Let $L: mathbb{R}^2rightarrow mathbb{R}^2$ be a linear operator. If $L((1,2)^T)=(-2,3)^T$ and $L((1,-1)^T)=(5,2)^T$, find the value of $L((7,5)^T).$
Is there a way to solve these kinds of problems? I only know if $L(alpha v_1+beta v_2)=alpha L(v_1)+beta L(v_2)$, then the vector space is said to be a linear transformation.
linear-algebra linear-transformations
edited Nov 18 at 9:08
Yadati Kiran
1,243417
asked Nov 18 at 8:10
Shadow Z
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Let $L: mathbb{R}^2rightarrow mathbb{R}^2$ be a linear operator. If $L((1,2)^T)=(-2,3)^T$ and $L((1,-1)^T)=(5,2)^T$, find the value of $L((7,5)^T).$
Is there a way to solve these kinds of problems? I only know if $L(alpha v_1+beta v_2)=alpha L(v_1)+beta L(v_2)$, then the vector space is said to be a linear transformation.
linear-algebra linear-transformations
edited Nov 18 at 9:08
Yadati Kiran
1,243417
asked Nov 18 at 8:10
Shadow Z
73
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $L: mathbb{R}^2rightarrow mathbb{R}^2$ be a linear operator. If $L((1,2)^T)=(-2,3)^T$ and $L((1,-1)^T)=(5,2)^T$, find the value of $L((7,5)^T).$
Is there a way to solve these kinds of problems? I only know if $L(alpha v_1+beta v_2)=alpha L(v_1)+beta L(v_2)$, then the vector space is said to be a linear transformation.
linear-algebra linear-transformations
edited Nov 18 at 9:08
Yadati Kiran
1,243417
asked Nov 18 at 8:10
Shadow Z
73
Let $L: mathbb{R}^2rightarrow mathbb{R}^2$ be a linear operator. If $L((1,2)^T)=(-2,3)^T$ and $L((1,-1)^T)=(5,2)^T$, find the value of $L((7,5)^T).$
Is there a way to solve these kinds of problems? I only know if $L(alpha v_1+beta v_2)=alpha L(v_1)+beta L(v_2)$, then the vector space is said to be a linear transformation.
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Nov 18 at 9:08
Yadati Kiran
1,243417
asked Nov 18 at 8:10
Shadow Z
73
edited Nov 18 at 9:08
Yadati Kiran
1,243417
asked Nov 18 at 8:10
Shadow Z
73
edited Nov 18 at 9:08
Yadati Kiran
1,243417
edited Nov 18 at 9:08
Yadati Kiran
1,243417
edited Nov 18 at 9:08
Yadati Kiran
1,243417
1,243417
asked Nov 18 at 8:10
Shadow Z
73
asked Nov 18 at 8:10
Shadow Z
73
asked Nov 18 at 8:10
Shadow Z
73
73
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Let your transformation matrix $L=begin{bmatrix}a&b\c&dend{bmatrix}$. Then $L((1,2)^T)=(-2,3)^T$ and $L((1,-1)^T)=(5,2)^T$ give $$begin{align}&a+2b=-2 &c+2d=3\&a-b=5 &c-b=2.end{align}$$ Solving we get $L=begin{bmatrix}dfrac{8}{3}&dfrac{-7}{3}\dfrac{7}{3}&dfrac{1}{3}end{bmatrix}$. $:$So $L((7,5)^T)=begin{bmatrix}7\18end{bmatrix}$.
answered Nov 18 at 9:22
Yadati Kiran
1,243417
Thanks! But why let the transformation matrix L=(a b c d). Doesn't R2 mean the row or column vectors whose has 2 dimensions? I thought R2 means R 2×1(2 rows,1 column)
– Shadow Z
Nov 18 at 10:35
@ShadowZ: If we have a linear transformation from $R^n$ ton $R^m$, the corresponding matrix associated with the transformation will have dimension $mtimes n$
– Yadati Kiran
Nov 18 at 10:37
Okay!Thank you!
– Shadow Z
Nov 18 at 10:41
Okay!Thank you!
– Shadow Z
Nov 18 at 10:41
add a comment |
up vote
0
down vote
Right, so you have to find $alpha$ and $beta$ with
$$ alpha(1,2)+beta(1,-1)=(7,5)$$
answered Nov 18 at 8:14
Empy2
33.2k12261
1
I figured α equals 4 and β equals 3, then I write L(7,5)^T=4*(-2,3)^T+3*(5,2)^T and I get the answer is (7,18)^T, is it correct?
– Shadow Z
Nov 18 at 8:40
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Let your transformation matrix $L=begin{bmatrix}a&b\c&dend{bmatrix}$. Then $L((1,2)^T)=(-2,3)^T$ and $L((1,-1)^T)=(5,2)^T$ give $$begin{align}&a+2b=-2 &c+2d=3\&a-b=5 &c-b=2.end{align}$$ Solving we get $L=begin{bmatrix}dfrac{8}{3}&dfrac{-7}{3}\dfrac{7}{3}&dfrac{1}{3}end{bmatrix}$. $:$So $L((7,5)^T)=begin{bmatrix}7\18end{bmatrix}$.
answered Nov 18 at 9:22
Yadati Kiran
1,243417
Thanks! But why let the transformation matrix L=(a b c d). Doesn't R2 mean the row or column vectors whose has 2 dimensions? I thought R2 means R 2×1(2 rows,1 column)
– Shadow Z
Nov 18 at 10:35
@ShadowZ: If we have a linear transformation from $R^n$ ton $R^m$, the corresponding matrix associated with the transformation will have dimension $mtimes n$
– Yadati Kiran
Nov 18 at 10:37
Okay!Thank you!
– Shadow Z
Nov 18 at 10:41
Okay!Thank you!
– Shadow Z
Nov 18 at 10:41
add a comment |
up vote
0
down vote
accepted
Let your transformation matrix $L=begin{bmatrix}a&b\c&dend{bmatrix}$. Then $L((1,2)^T)=(-2,3)^T$ and $L((1,-1)^T)=(5,2)^T$ give $$begin{align}&a+2b=-2 &c+2d=3\&a-b=5 &c-b=2.end{align}$$ Solving we get $L=begin{bmatrix}dfrac{8}{3}&dfrac{-7}{3}\dfrac{7}{3}&dfrac{1}{3}end{bmatrix}$. $:$So $L((7,5)^T)=begin{bmatrix}7\18end{bmatrix}$.
answered Nov 18 at 9:22
Yadati Kiran
1,243417
Thanks! But why let the transformation matrix L=(a b c d). Doesn't R2 mean the row or column vectors whose has 2 dimensions? I thought R2 means R 2×1(2 rows,1 column)
– Shadow Z
Nov 18 at 10:35
@ShadowZ: If we have a linear transformation from $R^n$ ton $R^m$, the corresponding matrix associated with the transformation will have dimension $mtimes n$
– Yadati Kiran
Nov 18 at 10:37
Okay!Thank you!
– Shadow Z
Nov 18 at 10:41
Okay!Thank you!
– Shadow Z
Nov 18 at 10:41
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Let your transformation matrix $L=begin{bmatrix}a&b\c&dend{bmatrix}$. Then $L((1,2)^T)=(-2,3)^T$ and $L((1,-1)^T)=(5,2)^T$ give $$begin{align}&a+2b=-2 &c+2d=3\&a-b=5 &c-b=2.end{align}$$ Solving we get $L=begin{bmatrix}dfrac{8}{3}&dfrac{-7}{3}\dfrac{7}{3}&dfrac{1}{3}end{bmatrix}$. $:$So $L((7,5)^T)=begin{bmatrix}7\18end{bmatrix}$.
answered Nov 18 at 9:22
Yadati Kiran
1,243417
Let your transformation matrix $L=begin{bmatrix}a&b\c&dend{bmatrix}$. Then $L((1,2)^T)=(-2,3)^T$ and $L((1,-1)^T)=(5,2)^T$ give $$begin{align}&a+2b=-2 &c+2d=3\&a-b=5 &c-b=2.end{align}$$ Solving we get $L=begin{bmatrix}dfrac{8}{3}&dfrac{-7}{3}\dfrac{7}{3}&dfrac{1}{3}end{bmatrix}$. $:$So $L((7,5)^T)=begin{bmatrix}7\18end{bmatrix}$.
answered Nov 18 at 9:22
Yadati Kiran
1,243417
answered Nov 18 at 9:22
Yadati Kiran
1,243417
answered Nov 18 at 9:22
Yadati Kiran
1,243417
answered Nov 18 at 9:22
Yadati Kiran
1,243417
1,243417
Thanks! But why let the transformation matrix L=(a b c d). Doesn't R2 mean the row or column vectors whose has 2 dimensions? I thought R2 means R 2×1(2 rows,1 column)
– Shadow Z
Nov 18 at 10:35
@ShadowZ: If we have a linear transformation from $R^n$ ton $R^m$, the corresponding matrix associated with the transformation will have dimension $mtimes n$
– Yadati Kiran
Nov 18 at 10:37
Okay!Thank you!
– Shadow Z
Nov 18 at 10:41
Okay!Thank you!
– Shadow Z
Nov 18 at 10:41
add a comment |
Thanks! But why let the transformation matrix L=(a b c d). Doesn't R2 mean the row or column vectors whose has 2 dimensions? I thought R2 means R 2×1(2 rows,1 column)
– Shadow Z
Nov 18 at 10:35
@ShadowZ: If we have a linear transformation from $R^n$ ton $R^m$, the corresponding matrix associated with the transformation will have dimension $mtimes n$
– Yadati Kiran
Nov 18 at 10:37
Okay!Thank you!
– Shadow Z
Nov 18 at 10:41
Okay!Thank you!
– Shadow Z
Nov 18 at 10:41
Thanks! But why let the transformation matrix L=(a b c d). Doesn't R2 mean the row or column vectors whose has 2 dimensions? I thought R2 means R 2×1(2 rows,1 column)
– Shadow Z
Nov 18 at 10:35
Thanks! But why let the transformation matrix L=(a b c d). Doesn't R2 mean the row or column vectors whose has 2 dimensions? I thought R2 means R 2×1(2 rows,1 column)
– Shadow Z
Nov 18 at 10:35
@ShadowZ: If we have a linear transformation from $R^n$ ton $R^m$, the corresponding matrix associated with the transformation will have dimension $mtimes n$
– Yadati Kiran
Nov 18 at 10:37
@ShadowZ: If we have a linear transformation from $R^n$ ton $R^m$, the corresponding matrix associated with the transformation will have dimension $mtimes n$
– Yadati Kiran
Nov 18 at 10:37
Okay!Thank you!
– Shadow Z
Nov 18 at 10:41
Okay!Thank you!
– Shadow Z
Nov 18 at 10:41
Okay!Thank you!
– Shadow Z
Nov 18 at 10:41
Okay!Thank you!
– Shadow Z
Nov 18 at 10:41
add a comment |
up vote
0
down vote
Right, so you have to find $alpha$ and $beta$ with
$$ alpha(1,2)+beta(1,-1)=(7,5)$$
answered Nov 18 at 8:14
Empy2
33.2k12261
1
I figured α equals 4 and β equals 3, then I write L(7,5)^T=4*(-2,3)^T+3*(5,2)^T and I get the answer is (7,18)^T, is it correct?
– Shadow Z
Nov 18 at 8:40
add a comment |
up vote
0
down vote
Right, so you have to find $alpha$ and $beta$ with
$$ alpha(1,2)+beta(1,-1)=(7,5)$$
answered Nov 18 at 8:14
Empy2
33.2k12261
1
I figured α equals 4 and β equals 3, then I write L(7,5)^T=4*(-2,3)^T+3*(5,2)^T and I get the answer is (7,18)^T, is it correct?
– Shadow Z
Nov 18 at 8:40
add a comment |
up vote
0
down vote
up vote
0
down vote
Right, so you have to find $alpha$ and $beta$ with
$$ alpha(1,2)+beta(1,-1)=(7,5)$$
answered Nov 18 at 8:14
Empy2
33.2k12261
Right, so you have to find $alpha$ and $beta$ with
$$ alpha(1,2)+beta(1,-1)=(7,5)$$
answered Nov 18 at 8:14
Empy2
33.2k12261
answered Nov 18 at 8:14
Empy2
33.2k12261
answered Nov 18 at 8:14
Empy2
33.2k12261
answered Nov 18 at 8:14
Empy2
33.2k12261
33.2k12261
1
I figured α equals 4 and β equals 3, then I write L(7,5)^T=4*(-2,3)^T+3*(5,2)^T and I get the answer is (7,18)^T, is it correct?
– Shadow Z
Nov 18 at 8:40
add a comment |
1
I figured α equals 4 and β equals 3, then I write L(7,5)^T=4*(-2,3)^T+3*(5,2)^T and I get the answer is (7,18)^T, is it correct?
– Shadow Z
Nov 18 at 8:40
1
1
I figured α equals 4 and β equals 3, then I write L(7,5)^T=4*(-2,3)^T+3*(5,2)^T and I get the answer is (7,18)^T, is it correct?
– Shadow Z
Nov 18 at 8:40
I figured α equals 4 and β equals 3, then I write L(7,5)^T=4*(-2,3)^T+3*(5,2)^T and I get the answer is (7,18)^T, is it correct?
– Shadow Z
Nov 18 at 8:40
add a comment |
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