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Concept of linear mappings are confusing me
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$begingroup$
I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.
Thanks!
linear-algebra
asked 1 hour ago
mingming
4306
$endgroup$
add a comment |
$begingroup$
I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.
Thanks!
linear-algebra
asked 1 hour ago
mingming
4306
$endgroup$
1
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
1 hour ago
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
1 hour ago
add a comment |
$begingroup$
I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.
Thanks!
linear-algebra
asked 1 hour ago
mingming
4306
$endgroup$
I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.
Thanks!
linear-algebra
linear-algebra
asked 1 hour ago
mingming
4306
asked 1 hour ago
mingming
4306
asked 1 hour ago
mingming
4306
asked 1 hour ago
mingming
4306
asked 1 hour ago
mingming
4306
4306
1
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
1 hour ago
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
1 hour ago
add a comment |
1
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
1 hour ago
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
1 hour ago
1
1
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
1 hour ago
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
1 hour ago
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
1 hour ago
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For the moment don't think about multiplication and matrices.
You can imagine starting from a vector $(x,y,z)$ in $mathbb{R}^3$ and mapping it to a vector in $mathbb{R}^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$
Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
begin{bmatrix}
2 & 0 & 1 \
3 & 4 & 0
end{bmatrix}
begin{bmatrix}
1 \
2 \
3
end{bmatrix}
=
begin{bmatrix}
5\
11
end{bmatrix}.
$$
You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.
answered 1 hour ago
Ethan BolkerEthan Bolker
45.8k553120
$endgroup$
add a comment |
$begingroup$
A linear mapping has the property that it maps subspaces to subspaces.
So it will map a line to a line or ${0}$, a plane to a plane, a line, or ${0}$, and so on.
By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.
answered 1 hour ago
rschwiebrschwieb
108k12103253
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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votes
$begingroup$
For the moment don't think about multiplication and matrices.
You can imagine starting from a vector $(x,y,z)$ in $mathbb{R}^3$ and mapping it to a vector in $mathbb{R}^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$
Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
begin{bmatrix}
2 & 0 & 1 \
3 & 4 & 0
end{bmatrix}
begin{bmatrix}
1 \
2 \
3
end{bmatrix}
=
begin{bmatrix}
5\
11
end{bmatrix}.
$$
You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.
answered 1 hour ago
Ethan BolkerEthan Bolker
45.8k553120
$endgroup$
add a comment |
$begingroup$
For the moment don't think about multiplication and matrices.
You can imagine starting from a vector $(x,y,z)$ in $mathbb{R}^3$ and mapping it to a vector in $mathbb{R}^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$
Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
begin{bmatrix}
2 & 0 & 1 \
3 & 4 & 0
end{bmatrix}
begin{bmatrix}
1 \
2 \
3
end{bmatrix}
=
begin{bmatrix}
5\
11
end{bmatrix}.
$$
You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.
answered 1 hour ago
Ethan BolkerEthan Bolker
45.8k553120
$endgroup$
add a comment |
$begingroup$
For the moment don't think about multiplication and matrices.
You can imagine starting from a vector $(x,y,z)$ in $mathbb{R}^3$ and mapping it to a vector in $mathbb{R}^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$
Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
begin{bmatrix}
2 & 0 & 1 \
3 & 4 & 0
end{bmatrix}
begin{bmatrix}
1 \
2 \
3
end{bmatrix}
=
begin{bmatrix}
5\
11
end{bmatrix}.
$$
You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.
answered 1 hour ago
Ethan BolkerEthan Bolker
45.8k553120
$endgroup$
For the moment don't think about multiplication and matrices.
You can imagine starting from a vector $(x,y,z)$ in $mathbb{R}^3$ and mapping it to a vector in $mathbb{R}^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$
Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
begin{bmatrix}
2 & 0 & 1 \
3 & 4 & 0
end{bmatrix}
begin{bmatrix}
1 \
2 \
3
end{bmatrix}
=
begin{bmatrix}
5\
11
end{bmatrix}.
$$
You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.
answered 1 hour ago
Ethan BolkerEthan Bolker
45.8k553120
answered 1 hour ago
Ethan BolkerEthan Bolker
45.8k553120
answered 1 hour ago
Ethan BolkerEthan Bolker
45.8k553120
answered 1 hour ago
Ethan BolkerEthan Bolker
45.8k553120
45.8k553120
add a comment |
add a comment |
$begingroup$
A linear mapping has the property that it maps subspaces to subspaces.
So it will map a line to a line or ${0}$, a plane to a plane, a line, or ${0}$, and so on.
By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.
answered 1 hour ago
rschwiebrschwieb
108k12103253
$endgroup$
add a comment |
$begingroup$
A linear mapping has the property that it maps subspaces to subspaces.
So it will map a line to a line or ${0}$, a plane to a plane, a line, or ${0}$, and so on.
By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.
answered 1 hour ago
rschwiebrschwieb
108k12103253
$endgroup$
add a comment |
$begingroup$
A linear mapping has the property that it maps subspaces to subspaces.
So it will map a line to a line or ${0}$, a plane to a plane, a line, or ${0}$, and so on.
By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.
answered 1 hour ago
rschwiebrschwieb
108k12103253
$endgroup$
A linear mapping has the property that it maps subspaces to subspaces.
So it will map a line to a line or ${0}$, a plane to a plane, a line, or ${0}$, and so on.
By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.
answered 1 hour ago
rschwiebrschwieb
108k12103253
answered 1 hour ago
rschwiebrschwieb
108k12103253
answered 1 hour ago
rschwiebrschwieb
108k12103253
answered 1 hour ago
rschwiebrschwieb
108k12103253
108k12103253
add a comment |
add a comment |
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1
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
1 hour ago
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
1 hour ago