Find a basis for the image of T, where T is a linear transformation. How do you find an image of just a...
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Doesn't it have to be "an image of V under T, where V is a subset" or "an image of v under T, where v is a vector"? What is the image of JUST a transformation?
And if you were given the matrix representation of that transformation, how might you figure out the "image of the transformation" is just from the matrix/the RREF of the matrix?
matrices linear-transformations
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
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add a comment |
$begingroup$
Doesn't it have to be "an image of V under T, where V is a subset" or "an image of v under T, where v is a vector"? What is the image of JUST a transformation?
And if you were given the matrix representation of that transformation, how might you figure out the "image of the transformation" is just from the matrix/the RREF of the matrix?
matrices linear-transformations
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
$endgroup$
$begingroup$
$V$ is tacitly assumed to be the entire space on which $T$ is defined.
$endgroup$
– amd
Dec 16 '18 at 0:11
add a comment |
$begingroup$
Doesn't it have to be "an image of V under T, where V is a subset" or "an image of v under T, where v is a vector"? What is the image of JUST a transformation?
And if you were given the matrix representation of that transformation, how might you figure out the "image of the transformation" is just from the matrix/the RREF of the matrix?
matrices linear-transformations
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
$endgroup$
Doesn't it have to be "an image of V under T, where V is a subset" or "an image of v under T, where v is a vector"? What is the image of JUST a transformation?
And if you were given the matrix representation of that transformation, how might you figure out the "image of the transformation" is just from the matrix/the RREF of the matrix?
matrices linear-transformations
matrices linear-transformations
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
1257
$begingroup$
$V$ is tacitly assumed to be the entire space on which $T$ is defined.
$endgroup$
– amd
Dec 16 '18 at 0:11
add a comment |
$begingroup$
$V$ is tacitly assumed to be the entire space on which $T$ is defined.
$endgroup$
– amd
Dec 16 '18 at 0:11
$begingroup$
$V$ is tacitly assumed to be the entire space on which $T$ is defined.
$endgroup$
– amd
Dec 16 '18 at 0:11
$begingroup$
$V$ is tacitly assumed to be the entire space on which $T$ is defined.
$endgroup$
– amd
Dec 16 '18 at 0:11
add a comment |
2 Answers
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The image of a linear transformation $T : V to W$ (or indeed any function) is a synonym for the range of the transformation. It's the set
$$operatorname{Im} T = operatorname{Range} T = { Tv : v in V }.$$
Given the standard matrix $mathcal{M}(T)$ of $T$ (when $T$ maps between Euclidean spaces), the range is the columnspace of $mathcal{M}(T)$, meaning the span of the columns of $mathcal{M}(T)$. This cannot be deduced purely from the reduced row echelon form of the matrix, as elementary row operations do not preserve the columnspace.
However, if you take note of which columns in the RREF contain pivots (leading $1$s), the corresponding columns of the standard matrix will form a basis for the columnspace of $mathcal{M}(T)$, and hence the image of $T$. Alternatively, row reducing $mathcal{M}(T)^top$ will also yield a basis for the image.
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.6k12152
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add a comment |
$begingroup$
The image of a linear transformation is actually the span of the columns of the matrix rel a basis (see this answer of mine).
Thus, you just need a basis for the column space. To do that, you can read off which columns form a basis for the column space of the RREF. The corresponding columns in the original matrix are a basis (for the original column space, and hence the image).
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.7k3827
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add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
The image of a linear transformation $T : V to W$ (or indeed any function) is a synonym for the range of the transformation. It's the set
$$operatorname{Im} T = operatorname{Range} T = { Tv : v in V }.$$
Given the standard matrix $mathcal{M}(T)$ of $T$ (when $T$ maps between Euclidean spaces), the range is the columnspace of $mathcal{M}(T)$, meaning the span of the columns of $mathcal{M}(T)$. This cannot be deduced purely from the reduced row echelon form of the matrix, as elementary row operations do not preserve the columnspace.
However, if you take note of which columns in the RREF contain pivots (leading $1$s), the corresponding columns of the standard matrix will form a basis for the columnspace of $mathcal{M}(T)$, and hence the image of $T$. Alternatively, row reducing $mathcal{M}(T)^top$ will also yield a basis for the image.
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.6k12152
$endgroup$
add a comment |
$begingroup$
The image of a linear transformation $T : V to W$ (or indeed any function) is a synonym for the range of the transformation. It's the set
$$operatorname{Im} T = operatorname{Range} T = { Tv : v in V }.$$
Given the standard matrix $mathcal{M}(T)$ of $T$ (when $T$ maps between Euclidean spaces), the range is the columnspace of $mathcal{M}(T)$, meaning the span of the columns of $mathcal{M}(T)$. This cannot be deduced purely from the reduced row echelon form of the matrix, as elementary row operations do not preserve the columnspace.
However, if you take note of which columns in the RREF contain pivots (leading $1$s), the corresponding columns of the standard matrix will form a basis for the columnspace of $mathcal{M}(T)$, and hence the image of $T$. Alternatively, row reducing $mathcal{M}(T)^top$ will also yield a basis for the image.
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.6k12152
$endgroup$
add a comment |
$begingroup$
The image of a linear transformation $T : V to W$ (or indeed any function) is a synonym for the range of the transformation. It's the set
$$operatorname{Im} T = operatorname{Range} T = { Tv : v in V }.$$
Given the standard matrix $mathcal{M}(T)$ of $T$ (when $T$ maps between Euclidean spaces), the range is the columnspace of $mathcal{M}(T)$, meaning the span of the columns of $mathcal{M}(T)$. This cannot be deduced purely from the reduced row echelon form of the matrix, as elementary row operations do not preserve the columnspace.
However, if you take note of which columns in the RREF contain pivots (leading $1$s), the corresponding columns of the standard matrix will form a basis for the columnspace of $mathcal{M}(T)$, and hence the image of $T$. Alternatively, row reducing $mathcal{M}(T)^top$ will also yield a basis for the image.
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.6k12152
$endgroup$
The image of a linear transformation $T : V to W$ (or indeed any function) is a synonym for the range of the transformation. It's the set
$$operatorname{Im} T = operatorname{Range} T = { Tv : v in V }.$$
Given the standard matrix $mathcal{M}(T)$ of $T$ (when $T$ maps between Euclidean spaces), the range is the columnspace of $mathcal{M}(T)$, meaning the span of the columns of $mathcal{M}(T)$. This cannot be deduced purely from the reduced row echelon form of the matrix, as elementary row operations do not preserve the columnspace.
However, if you take note of which columns in the RREF contain pivots (leading $1$s), the corresponding columns of the standard matrix will form a basis for the columnspace of $mathcal{M}(T)$, and hence the image of $T$. Alternatively, row reducing $mathcal{M}(T)^top$ will also yield a basis for the image.
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.6k12152
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.6k12152
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.6k12152
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.6k12152
18.6k12152
add a comment |
add a comment |
$begingroup$
The image of a linear transformation is actually the span of the columns of the matrix rel a basis (see this answer of mine).
Thus, you just need a basis for the column space. To do that, you can read off which columns form a basis for the column space of the RREF. The corresponding columns in the original matrix are a basis (for the original column space, and hence the image).
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.7k3827
$endgroup$
add a comment |
$begingroup$
The image of a linear transformation is actually the span of the columns of the matrix rel a basis (see this answer of mine).
Thus, you just need a basis for the column space. To do that, you can read off which columns form a basis for the column space of the RREF. The corresponding columns in the original matrix are a basis (for the original column space, and hence the image).
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.7k3827
$endgroup$
add a comment |
$begingroup$
The image of a linear transformation is actually the span of the columns of the matrix rel a basis (see this answer of mine).
Thus, you just need a basis for the column space. To do that, you can read off which columns form a basis for the column space of the RREF. The corresponding columns in the original matrix are a basis (for the original column space, and hence the image).
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.7k3827
$endgroup$
The image of a linear transformation is actually the span of the columns of the matrix rel a basis (see this answer of mine).
Thus, you just need a basis for the column space. To do that, you can read off which columns form a basis for the column space of the RREF. The corresponding columns in the original matrix are a basis (for the original column space, and hence the image).
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.7k3827
edited Dec 16 '18 at 1:00
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.7k3827
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.7k3827
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.7k3827
13.7k3827
add a comment |
add a comment |
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$begingroup$
$V$ is tacitly assumed to be the entire space on which $T$ is defined.
$endgroup$
– amd
Dec 16 '18 at 0:11