Mixing
Matrix associated with a projection mapping
$begingroup$
From S.L Linear Algebra:
Find the matrix associated with the following linear maps. The vectors
are written horizontally with a transpose sign for typographical
reasons.
(a) $F:mathbb{R}^4 rightarrow mathbb{R}^2$ given by $Fleft ((x_1, x_2, x_3, x_4)^T right)=(x_1, x_2)^T$ (the projection)
Solution Attempt
I will use Theorem 2.1 from the book:
Let $L: K^n rightarrow K^m$ be a linear map. Then there exists a
unique matrix $A$ such that $L = L_A$.
Hence, for all $X$ we have $L(X)=AX$.
In this case, some $A$ is the matrix associated with linear map $F$ such that:
$$F(X^T)=AX^T$$
$$Fleft ((x_1, x_2, x_3, x_4)^T right)=Aleft ((x_1, x_2, x_3, x_4)^T right)$$
$$AX^T=Aleft ((x_1, x_2, x_3, x_4)^T right)=(x_1, x_2)^T$$
Solving for $A$:
$$A=left ((x_1, x_2, x_3, x_4)^T right)^{-1}(x_1, x_2)^T$$
Now this makes no sense, because $X$ must be non-degenerate square matrix in order to be invertible, but $X$ is just a vector with cardinality (dimension) $4$.
Perhaps I could invert $A$ (then I would have to show that $F$ is injective with trivial kernel), but I don't see any point of doing this.
Is $A$ truly a matrix associated with $F$? If not, how to find $A$ if it is a matrix that is truly associated with $F$?
P.S
I also found this in the book:
Let $F: mathbb{R}^3 rightarrow mathbb{R}^2$ be the projection, in
other words the mapping such that $F(x_1, x_2, x_3) = (x_1, x_2)$.
Then the matrix associated with $F$ is:
$$begin{pmatrix} 1 & 0 & 0 \ 0&1 & 0 end{pmatrix}$$
I don't know how was it exactly calculated, but the matrix above contains standard basis vectors only, which I believe must mean something.
Thank you!
linear-algebra matrices linear-transformations projection
asked Jan 6 at 14:06
ShellRoxShellRox
26928
$endgroup$
add a comment |
$begingroup$
From S.L Linear Algebra:
Find the matrix associated with the following linear maps. The vectors
are written horizontally with a transpose sign for typographical
reasons.
(a) $F:mathbb{R}^4 rightarrow mathbb{R}^2$ given by $Fleft ((x_1, x_2, x_3, x_4)^T right)=(x_1, x_2)^T$ (the projection)
Solution Attempt
I will use Theorem 2.1 from the book:
Let $L: K^n rightarrow K^m$ be a linear map. Then there exists a
unique matrix $A$ such that $L = L_A$.
Hence, for all $X$ we have $L(X)=AX$.
In this case, some $A$ is the matrix associated with linear map $F$ such that:
$$F(X^T)=AX^T$$
$$Fleft ((x_1, x_2, x_3, x_4)^T right)=Aleft ((x_1, x_2, x_3, x_4)^T right)$$
$$AX^T=Aleft ((x_1, x_2, x_3, x_4)^T right)=(x_1, x_2)^T$$
Solving for $A$:
$$A=left ((x_1, x_2, x_3, x_4)^T right)^{-1}(x_1, x_2)^T$$
Now this makes no sense, because $X$ must be non-degenerate square matrix in order to be invertible, but $X$ is just a vector with cardinality (dimension) $4$.
Perhaps I could invert $A$ (then I would have to show that $F$ is injective with trivial kernel), but I don't see any point of doing this.
Is $A$ truly a matrix associated with $F$? If not, how to find $A$ if it is a matrix that is truly associated with $F$?
P.S
I also found this in the book:
Let $F: mathbb{R}^3 rightarrow mathbb{R}^2$ be the projection, in
other words the mapping such that $F(x_1, x_2, x_3) = (x_1, x_2)$.
Then the matrix associated with $F$ is:
$$begin{pmatrix} 1 & 0 & 0 \ 0&1 & 0 end{pmatrix}$$
I don't know how was it exactly calculated, but the matrix above contains standard basis vectors only, which I believe must mean something.
Thank you!
linear-algebra matrices linear-transformations projection
asked Jan 6 at 14:06
ShellRoxShellRox
26928
$endgroup$
$begingroup$
Hint: The columns of the matrix are the images of the basis vectors.
$endgroup$
– amd
Jan 6 at 21:05
add a comment |
$begingroup$
From S.L Linear Algebra:
Find the matrix associated with the following linear maps. The vectors
are written horizontally with a transpose sign for typographical
reasons.
(a) $F:mathbb{R}^4 rightarrow mathbb{R}^2$ given by $Fleft ((x_1, x_2, x_3, x_4)^T right)=(x_1, x_2)^T$ (the projection)
Solution Attempt
I will use Theorem 2.1 from the book:
Let $L: K^n rightarrow K^m$ be a linear map. Then there exists a
unique matrix $A$ such that $L = L_A$.
Hence, for all $X$ we have $L(X)=AX$.
In this case, some $A$ is the matrix associated with linear map $F$ such that:
$$F(X^T)=AX^T$$
$$Fleft ((x_1, x_2, x_3, x_4)^T right)=Aleft ((x_1, x_2, x_3, x_4)^T right)$$
$$AX^T=Aleft ((x_1, x_2, x_3, x_4)^T right)=(x_1, x_2)^T$$
Solving for $A$:
$$A=left ((x_1, x_2, x_3, x_4)^T right)^{-1}(x_1, x_2)^T$$
Now this makes no sense, because $X$ must be non-degenerate square matrix in order to be invertible, but $X$ is just a vector with cardinality (dimension) $4$.
Perhaps I could invert $A$ (then I would have to show that $F$ is injective with trivial kernel), but I don't see any point of doing this.
Is $A$ truly a matrix associated with $F$? If not, how to find $A$ if it is a matrix that is truly associated with $F$?
P.S
I also found this in the book:
Let $F: mathbb{R}^3 rightarrow mathbb{R}^2$ be the projection, in
other words the mapping such that $F(x_1, x_2, x_3) = (x_1, x_2)$.
Then the matrix associated with $F$ is:
$$begin{pmatrix} 1 & 0 & 0 \ 0&1 & 0 end{pmatrix}$$
I don't know how was it exactly calculated, but the matrix above contains standard basis vectors only, which I believe must mean something.
Thank you!
linear-algebra matrices linear-transformations projection
asked Jan 6 at 14:06
ShellRoxShellRox
26928
$endgroup$
From S.L Linear Algebra:
Find the matrix associated with the following linear maps. The vectors
are written horizontally with a transpose sign for typographical
reasons.
(a) $F:mathbb{R}^4 rightarrow mathbb{R}^2$ given by $Fleft ((x_1, x_2, x_3, x_4)^T right)=(x_1, x_2)^T$ (the projection)
Solution Attempt
I will use Theorem 2.1 from the book:
Let $L: K^n rightarrow K^m$ be a linear map. Then there exists a
unique matrix $A$ such that $L = L_A$.
Hence, for all $X$ we have $L(X)=AX$.
In this case, some $A$ is the matrix associated with linear map $F$ such that:
$$F(X^T)=AX^T$$
$$Fleft ((x_1, x_2, x_3, x_4)^T right)=Aleft ((x_1, x_2, x_3, x_4)^T right)$$
$$AX^T=Aleft ((x_1, x_2, x_3, x_4)^T right)=(x_1, x_2)^T$$
Solving for $A$:
$$A=left ((x_1, x_2, x_3, x_4)^T right)^{-1}(x_1, x_2)^T$$
Now this makes no sense, because $X$ must be non-degenerate square matrix in order to be invertible, but $X$ is just a vector with cardinality (dimension) $4$.
Perhaps I could invert $A$ (then I would have to show that $F$ is injective with trivial kernel), but I don't see any point of doing this.
Is $A$ truly a matrix associated with $F$? If not, how to find $A$ if it is a matrix that is truly associated with $F$?
P.S
I also found this in the book:
Let $F: mathbb{R}^3 rightarrow mathbb{R}^2$ be the projection, in
other words the mapping such that $F(x_1, x_2, x_3) = (x_1, x_2)$.
Then the matrix associated with $F$ is:
$$begin{pmatrix} 1 & 0 & 0 \ 0&1 & 0 end{pmatrix}$$
I don't know how was it exactly calculated, but the matrix above contains standard basis vectors only, which I believe must mean something.
Thank you!
linear-algebra matrices linear-transformations projection
linear-algebra matrices linear-transformations projection
asked Jan 6 at 14:06
ShellRoxShellRox
26928
asked Jan 6 at 14:06
ShellRoxShellRox
26928
asked Jan 6 at 14:06
ShellRoxShellRox
26928
asked Jan 6 at 14:06
ShellRoxShellRox
26928
asked Jan 6 at 14:06
ShellRoxShellRox
26928
26928
$begingroup$
Hint: The columns of the matrix are the images of the basis vectors.
$endgroup$
– amd
Jan 6 at 21:05
add a comment |
$begingroup$
Hint: The columns of the matrix are the images of the basis vectors.
$endgroup$
– amd
Jan 6 at 21:05
$begingroup$
Hint: The columns of the matrix are the images of the basis vectors.
$endgroup$
– amd
Jan 6 at 21:05
$begingroup$
Hint: The columns of the matrix are the images of the basis vectors.
$endgroup$
– amd
Jan 6 at 21:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To begin with, the Vector Space of Matrices $mathscr{M}_{mn}(mathbb{R}^4)$ is not abelian under matrix product! It is a fundamental error to change sides of a matrix by "inverting it" in an equation.
And keep in mind: you can see obviously that $mathbb{R}^2$ can be extended so to be isomorphic to $mathbb{R}^4$. It is why they indicated that the image was a projection in fact.
Now let's calculate:
The first intuition is to write down what your matrix operations look like; to find out if the product is well-defined somehow: this will eliminate some big issues in your thoughts.
$$ F(left (
begin{matrix}
x_1 \
x_2 \
x_3 \
x_4
end{matrix}
right )
)=left (
begin{matrix}
x_1 \
x_2
end{matrix}
right )
$$
ie.
$$ left (
begin{matrix}
a_{11} & a_{12} & a_{13} & a_{14} \
a_{21} & a_{22} & a_{23} & a_{24}
end{matrix}
right )
.left (
begin{matrix}
x_1 \
x_2 \
x_3 \
x_4
end{matrix}
right )
=left (
begin{matrix}
x_1 \
x_2
end{matrix}
right )
$$
when we introduce the Matrix of elements $(a_{ij})$ as the Matrix associated to the linear transformation. We must have a $2x4$ matrix for this transformation.
When you do the calculations, you'll find:
$$ begin{cases}
a_{11}x_{1} + a_{12}x_{2} + a_{13}x_{3} + a_{14}x_{4} = x_1 \
a_{21}x_{1} + a_{22}x_{2} + a_{23}x_{3} + a_{24}x_{4} = x_2
end{cases} $$
ie.
$$ begin{cases}
(a_{11}-1)x_{1} + a_{12}x_{2} + a_{13}x_{3} + a_{14}x_{4} = 0 \
a_{21}x_{1} + (a_{22}-1)x_{2} + a_{23}x_{3} + a_{24}x_{4} = 0
end{cases} $$
But as we're looking for a basis, we know it must be linearly independent (within row vctors). So that, we know, every coefficients are zero, with $(a_{11}-1)=0$ and $(a_{22}-1)=0$ so that we have $a_{11}=1$ and $a_{22}=1$, over row vectors.
Thus, we conclude that one basis for this linear transformation in $mathbb{R}^4$ is:
$$
left (
begin{matrix}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0
end{matrix}
right )$$
answered Jan 6 at 17:44
freehumoristfreehumorist
173112
$endgroup$
$begingroup$
Wow that makes a lot of sense. Thank you! My lack of familiarity with matrices is the primary reason for my ignorance. Also, you have mentioned that linear transformation is an isomorphism, that means that it is bijective and hence kernel under $F$ is trivial, correct?
$endgroup$
– ShellRox
Jan 6 at 20:59
1
$begingroup$
I'm glad you see more clearly. But attention: I didn't say that every linear transformation is isomorphism. Indeed, here it is surjective because we have a projection of a vector of $mathbb{R}^4$ into $mathbb{R}^2$. I wanted you to see the reverse: We could have extended the image by null vectors to have an isomorphism. So that you can visualize the fact that matrix associated needn't to be square, but only well-defined.
$endgroup$
– freehumorist
Jan 6 at 21:05
1
$begingroup$
Yes, of course. I just wanted to further discover properties of only linear transformations that are defined by projection. Thank you for the help!
$endgroup$
– ShellRox
Jan 6 at 21:10
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
oldest
votes
$begingroup$
To begin with, the Vector Space of Matrices $mathscr{M}_{mn}(mathbb{R}^4)$ is not abelian under matrix product! It is a fundamental error to change sides of a matrix by "inverting it" in an equation.
And keep in mind: you can see obviously that $mathbb{R}^2$ can be extended so to be isomorphic to $mathbb{R}^4$. It is why they indicated that the image was a projection in fact.
Now let's calculate:
The first intuition is to write down what your matrix operations look like; to find out if the product is well-defined somehow: this will eliminate some big issues in your thoughts.
$$ F(left (
begin{matrix}
x_1 \
x_2 \
x_3 \
x_4
end{matrix}
right )
)=left (
begin{matrix}
x_1 \
x_2
end{matrix}
right )
$$
ie.
$$ left (
begin{matrix}
a_{11} & a_{12} & a_{13} & a_{14} \
a_{21} & a_{22} & a_{23} & a_{24}
end{matrix}
right )
.left (
begin{matrix}
x_1 \
x_2 \
x_3 \
x_4
end{matrix}
right )
=left (
begin{matrix}
x_1 \
x_2
end{matrix}
right )
$$
when we introduce the Matrix of elements $(a_{ij})$ as the Matrix associated to the linear transformation. We must have a $2x4$ matrix for this transformation.
When you do the calculations, you'll find:
$$ begin{cases}
a_{11}x_{1} + a_{12}x_{2} + a_{13}x_{3} + a_{14}x_{4} = x_1 \
a_{21}x_{1} + a_{22}x_{2} + a_{23}x_{3} + a_{24}x_{4} = x_2
end{cases} $$
ie.
$$ begin{cases}
(a_{11}-1)x_{1} + a_{12}x_{2} + a_{13}x_{3} + a_{14}x_{4} = 0 \
a_{21}x_{1} + (a_{22}-1)x_{2} + a_{23}x_{3} + a_{24}x_{4} = 0
end{cases} $$
But as we're looking for a basis, we know it must be linearly independent (within row vctors). So that, we know, every coefficients are zero, with $(a_{11}-1)=0$ and $(a_{22}-1)=0$ so that we have $a_{11}=1$ and $a_{22}=1$, over row vectors.
Thus, we conclude that one basis for this linear transformation in $mathbb{R}^4$ is:
$$
left (
begin{matrix}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0
end{matrix}
right )$$
answered Jan 6 at 17:44
freehumoristfreehumorist
173112
$endgroup$
$begingroup$
Wow that makes a lot of sense. Thank you! My lack of familiarity with matrices is the primary reason for my ignorance. Also, you have mentioned that linear transformation is an isomorphism, that means that it is bijective and hence kernel under $F$ is trivial, correct?
$endgroup$
– ShellRox
Jan 6 at 20:59
1
$begingroup$
I'm glad you see more clearly. But attention: I didn't say that every linear transformation is isomorphism. Indeed, here it is surjective because we have a projection of a vector of $mathbb{R}^4$ into $mathbb{R}^2$. I wanted you to see the reverse: We could have extended the image by null vectors to have an isomorphism. So that you can visualize the fact that matrix associated needn't to be square, but only well-defined.
$endgroup$
– freehumorist
Jan 6 at 21:05
1
$begingroup$
Yes, of course. I just wanted to further discover properties of only linear transformations that are defined by projection. Thank you for the help!
$endgroup$
– ShellRox
Jan 6 at 21:10
add a comment |
$begingroup$
To begin with, the Vector Space of Matrices $mathscr{M}_{mn}(mathbb{R}^4)$ is not abelian under matrix product! It is a fundamental error to change sides of a matrix by "inverting it" in an equation.
And keep in mind: you can see obviously that $mathbb{R}^2$ can be extended so to be isomorphic to $mathbb{R}^4$. It is why they indicated that the image was a projection in fact.
Now let's calculate:
The first intuition is to write down what your matrix operations look like; to find out if the product is well-defined somehow: this will eliminate some big issues in your thoughts.
$$ F(left (
begin{matrix}
x_1 \
x_2 \
x_3 \
x_4
end{matrix}
right )
)=left (
begin{matrix}
x_1 \
x_2
end{matrix}
right )
$$
ie.
$$ left (
begin{matrix}
a_{11} & a_{12} & a_{13} & a_{14} \
a_{21} & a_{22} & a_{23} & a_{24}
end{matrix}
right )
.left (
begin{matrix}
x_1 \
x_2 \
x_3 \
x_4
end{matrix}
right )
=left (
begin{matrix}
x_1 \
x_2
end{matrix}
right )
$$
when we introduce the Matrix of elements $(a_{ij})$ as the Matrix associated to the linear transformation. We must have a $2x4$ matrix for this transformation.
When you do the calculations, you'll find:
$$ begin{cases}
a_{11}x_{1} + a_{12}x_{2} + a_{13}x_{3} + a_{14}x_{4} = x_1 \
a_{21}x_{1} + a_{22}x_{2} + a_{23}x_{3} + a_{24}x_{4} = x_2
end{cases} $$
ie.
$$ begin{cases}
(a_{11}-1)x_{1} + a_{12}x_{2} + a_{13}x_{3} + a_{14}x_{4} = 0 \
a_{21}x_{1} + (a_{22}-1)x_{2} + a_{23}x_{3} + a_{24}x_{4} = 0
end{cases} $$
But as we're looking for a basis, we know it must be linearly independent (within row vctors). So that, we know, every coefficients are zero, with $(a_{11}-1)=0$ and $(a_{22}-1)=0$ so that we have $a_{11}=1$ and $a_{22}=1$, over row vectors.
Thus, we conclude that one basis for this linear transformation in $mathbb{R}^4$ is:
$$
left (
begin{matrix}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0
end{matrix}
right )$$
answered Jan 6 at 17:44
freehumoristfreehumorist
173112
$endgroup$
$begingroup$
Wow that makes a lot of sense. Thank you! My lack of familiarity with matrices is the primary reason for my ignorance. Also, you have mentioned that linear transformation is an isomorphism, that means that it is bijective and hence kernel under $F$ is trivial, correct?
$endgroup$
– ShellRox
Jan 6 at 20:59
1
$begingroup$
I'm glad you see more clearly. But attention: I didn't say that every linear transformation is isomorphism. Indeed, here it is surjective because we have a projection of a vector of $mathbb{R}^4$ into $mathbb{R}^2$. I wanted you to see the reverse: We could have extended the image by null vectors to have an isomorphism. So that you can visualize the fact that matrix associated needn't to be square, but only well-defined.
$endgroup$
– freehumorist
Jan 6 at 21:05
1
$begingroup$
Yes, of course. I just wanted to further discover properties of only linear transformations that are defined by projection. Thank you for the help!
$endgroup$
– ShellRox
Jan 6 at 21:10
add a comment |
$begingroup$
To begin with, the Vector Space of Matrices $mathscr{M}_{mn}(mathbb{R}^4)$ is not abelian under matrix product! It is a fundamental error to change sides of a matrix by "inverting it" in an equation.
And keep in mind: you can see obviously that $mathbb{R}^2$ can be extended so to be isomorphic to $mathbb{R}^4$. It is why they indicated that the image was a projection in fact.
Now let's calculate:
The first intuition is to write down what your matrix operations look like; to find out if the product is well-defined somehow: this will eliminate some big issues in your thoughts.
$$ F(left (
begin{matrix}
x_1 \
x_2 \
x_3 \
x_4
end{matrix}
right )
)=left (
begin{matrix}
x_1 \
x_2
end{matrix}
right )
$$
ie.
$$ left (
begin{matrix}
a_{11} & a_{12} & a_{13} & a_{14} \
a_{21} & a_{22} & a_{23} & a_{24}
end{matrix}
right )
.left (
begin{matrix}
x_1 \
x_2 \
x_3 \
x_4
end{matrix}
right )
=left (
begin{matrix}
x_1 \
x_2
end{matrix}
right )
$$
when we introduce the Matrix of elements $(a_{ij})$ as the Matrix associated to the linear transformation. We must have a $2x4$ matrix for this transformation.
When you do the calculations, you'll find:
$$ begin{cases}
a_{11}x_{1} + a_{12}x_{2} + a_{13}x_{3} + a_{14}x_{4} = x_1 \
a_{21}x_{1} + a_{22}x_{2} + a_{23}x_{3} + a_{24}x_{4} = x_2
end{cases} $$
ie.
$$ begin{cases}
(a_{11}-1)x_{1} + a_{12}x_{2} + a_{13}x_{3} + a_{14}x_{4} = 0 \
a_{21}x_{1} + (a_{22}-1)x_{2} + a_{23}x_{3} + a_{24}x_{4} = 0
end{cases} $$
But as we're looking for a basis, we know it must be linearly independent (within row vctors). So that, we know, every coefficients are zero, with $(a_{11}-1)=0$ and $(a_{22}-1)=0$ so that we have $a_{11}=1$ and $a_{22}=1$, over row vectors.
Thus, we conclude that one basis for this linear transformation in $mathbb{R}^4$ is:
$$
left (
begin{matrix}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0
end{matrix}
right )$$
answered Jan 6 at 17:44
freehumoristfreehumorist
173112
$endgroup$
To begin with, the Vector Space of Matrices $mathscr{M}_{mn}(mathbb{R}^4)$ is not abelian under matrix product! It is a fundamental error to change sides of a matrix by "inverting it" in an equation.
And keep in mind: you can see obviously that $mathbb{R}^2$ can be extended so to be isomorphic to $mathbb{R}^4$. It is why they indicated that the image was a projection in fact.
Now let's calculate:
The first intuition is to write down what your matrix operations look like; to find out if the product is well-defined somehow: this will eliminate some big issues in your thoughts.
$$ F(left (
begin{matrix}
x_1 \
x_2 \
x_3 \
x_4
end{matrix}
right )
)=left (
begin{matrix}
x_1 \
x_2
end{matrix}
right )
$$
ie.
$$ left (
begin{matrix}
a_{11} & a_{12} & a_{13} & a_{14} \
a_{21} & a_{22} & a_{23} & a_{24}
end{matrix}
right )
.left (
begin{matrix}
x_1 \
x_2 \
x_3 \
x_4
end{matrix}
right )
=left (
begin{matrix}
x_1 \
x_2
end{matrix}
right )
$$
when we introduce the Matrix of elements $(a_{ij})$ as the Matrix associated to the linear transformation. We must have a $2x4$ matrix for this transformation.
When you do the calculations, you'll find:
$$ begin{cases}
a_{11}x_{1} + a_{12}x_{2} + a_{13}x_{3} + a_{14}x_{4} = x_1 \
a_{21}x_{1} + a_{22}x_{2} + a_{23}x_{3} + a_{24}x_{4} = x_2
end{cases} $$
ie.
$$ begin{cases}
(a_{11}-1)x_{1} + a_{12}x_{2} + a_{13}x_{3} + a_{14}x_{4} = 0 \
a_{21}x_{1} + (a_{22}-1)x_{2} + a_{23}x_{3} + a_{24}x_{4} = 0
end{cases} $$
But as we're looking for a basis, we know it must be linearly independent (within row vctors). So that, we know, every coefficients are zero, with $(a_{11}-1)=0$ and $(a_{22}-1)=0$ so that we have $a_{11}=1$ and $a_{22}=1$, over row vectors.
Thus, we conclude that one basis for this linear transformation in $mathbb{R}^4$ is:
$$
left (
begin{matrix}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0
end{matrix}
right )$$
answered Jan 6 at 17:44
freehumoristfreehumorist
173112
answered Jan 6 at 17:44
freehumoristfreehumorist
173112
answered Jan 6 at 17:44
freehumoristfreehumorist
173112
answered Jan 6 at 17:44
freehumoristfreehumorist
173112
173112
$begingroup$
Wow that makes a lot of sense. Thank you! My lack of familiarity with matrices is the primary reason for my ignorance. Also, you have mentioned that linear transformation is an isomorphism, that means that it is bijective and hence kernel under $F$ is trivial, correct?
$endgroup$
– ShellRox
Jan 6 at 20:59
1
$begingroup$
I'm glad you see more clearly. But attention: I didn't say that every linear transformation is isomorphism. Indeed, here it is surjective because we have a projection of a vector of $mathbb{R}^4$ into $mathbb{R}^2$. I wanted you to see the reverse: We could have extended the image by null vectors to have an isomorphism. So that you can visualize the fact that matrix associated needn't to be square, but only well-defined.
$endgroup$
– freehumorist
Jan 6 at 21:05
1
$begingroup$
Yes, of course. I just wanted to further discover properties of only linear transformations that are defined by projection. Thank you for the help!
$endgroup$
– ShellRox
Jan 6 at 21:10
add a comment |
$begingroup$
Wow that makes a lot of sense. Thank you! My lack of familiarity with matrices is the primary reason for my ignorance. Also, you have mentioned that linear transformation is an isomorphism, that means that it is bijective and hence kernel under $F$ is trivial, correct?
$endgroup$
– ShellRox
Jan 6 at 20:59
1
$begingroup$
I'm glad you see more clearly. But attention: I didn't say that every linear transformation is isomorphism. Indeed, here it is surjective because we have a projection of a vector of $mathbb{R}^4$ into $mathbb{R}^2$. I wanted you to see the reverse: We could have extended the image by null vectors to have an isomorphism. So that you can visualize the fact that matrix associated needn't to be square, but only well-defined.
$endgroup$
– freehumorist
Jan 6 at 21:05
1
$begingroup$
Yes, of course. I just wanted to further discover properties of only linear transformations that are defined by projection. Thank you for the help!
$endgroup$
– ShellRox
Jan 6 at 21:10
$begingroup$
Wow that makes a lot of sense. Thank you! My lack of familiarity with matrices is the primary reason for my ignorance. Also, you have mentioned that linear transformation is an isomorphism, that means that it is bijective and hence kernel under $F$ is trivial, correct?
$endgroup$
– ShellRox
Jan 6 at 20:59
$begingroup$
Wow that makes a lot of sense. Thank you! My lack of familiarity with matrices is the primary reason for my ignorance. Also, you have mentioned that linear transformation is an isomorphism, that means that it is bijective and hence kernel under $F$ is trivial, correct?
$endgroup$
– ShellRox
Jan 6 at 20:59
1
1
$begingroup$
I'm glad you see more clearly. But attention: I didn't say that every linear transformation is isomorphism. Indeed, here it is surjective because we have a projection of a vector of $mathbb{R}^4$ into $mathbb{R}^2$. I wanted you to see the reverse: We could have extended the image by null vectors to have an isomorphism. So that you can visualize the fact that matrix associated needn't to be square, but only well-defined.
$endgroup$
– freehumorist
Jan 6 at 21:05
$begingroup$
I'm glad you see more clearly. But attention: I didn't say that every linear transformation is isomorphism. Indeed, here it is surjective because we have a projection of a vector of $mathbb{R}^4$ into $mathbb{R}^2$. I wanted you to see the reverse: We could have extended the image by null vectors to have an isomorphism. So that you can visualize the fact that matrix associated needn't to be square, but only well-defined.
$endgroup$
– freehumorist
Jan 6 at 21:05
1
1
$begingroup$
Yes, of course. I just wanted to further discover properties of only linear transformations that are defined by projection. Thank you for the help!
$endgroup$
– ShellRox
Jan 6 at 21:10
$begingroup$
Yes, of course. I just wanted to further discover properties of only linear transformations that are defined by projection. Thank you for the help!
$endgroup$
– ShellRox
Jan 6 at 21:10
add a comment |
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$begingroup$
Hint: The columns of the matrix are the images of the basis vectors.
$endgroup$
– amd
Jan 6 at 21:05