Finding Range of a linear mapping
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In the following textbook example I understand how they get Null(L) but not how they get Range(L) which they say is clear to see. Can anyone elucidate the method of finding the Range of a linear mapping? Thank you!
linear-algebra linear-transformations
asked Jan 1 at 18:02
P.ythonP.ython
205
$endgroup$
add a comment |
$begingroup$
In the following textbook example I understand how they get Null(L) but not how they get Range(L) which they say is clear to see. Can anyone elucidate the method of finding the Range of a linear mapping? Thank you!
linear-algebra linear-transformations
asked Jan 1 at 18:02
P.ythonP.ython
205
$endgroup$
$begingroup$
Exactly where does the expression “clear to see” appear?
$endgroup$
– José Carlos Santos
Jan 1 at 18:04
add a comment |
$begingroup$
In the following textbook example I understand how they get Null(L) but not how they get Range(L) which they say is clear to see. Can anyone elucidate the method of finding the Range of a linear mapping? Thank you!
linear-algebra linear-transformations
asked Jan 1 at 18:02
P.ythonP.ython
205
$endgroup$
In the following textbook example I understand how they get Null(L) but not how they get Range(L) which they say is clear to see. Can anyone elucidate the method of finding the Range of a linear mapping? Thank you!
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Jan 1 at 18:02
P.ythonP.ython
205
asked Jan 1 at 18:02
P.ythonP.ython
205
edited Jan 1 at 18:10
P.ython
asked Jan 1 at 18:02
P.ythonP.ython
205
asked Jan 1 at 18:02
P.ythonP.ython
205
asked Jan 1 at 18:02
P.ythonP.ython
205
205
$begingroup$
Exactly where does the expression “clear to see” appear?
$endgroup$
– José Carlos Santos
Jan 1 at 18:04
add a comment |
$begingroup$
Exactly where does the expression “clear to see” appear?
$endgroup$
– José Carlos Santos
Jan 1 at 18:04
$begingroup$
Exactly where does the expression “clear to see” appear?
$endgroup$
– José Carlos Santos
Jan 1 at 18:04
$begingroup$
Exactly where does the expression “clear to see” appear?
$endgroup$
– José Carlos Santos
Jan 1 at 18:04
add a comment |
3 Answers
3
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The range of $L, R(L)subseteqBbb P_3$. Every element in $R(L)$ is of the form $k_1x+k_2x^3$, which is a linear combination of $x$ and $x^3$. So ${x,x^3}$ is a basis of $R(L)$, giving its dimension as $2$.
answered Jan 1 at 18:38
Shubham JohriShubham Johri
5,204718
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You know $dim M(2,2)=4$ and $dim null(L)=2$, so by the Bank nullity theorem, $dim rank(L)=2$. But all the polynomials which are image of L are of the form $alpha x + beta x^3$, so you can get all of them with a linear combination of the polynomials $x$ and $x^3$. Since $dim rank(L)=2$ and $x,x^3$ are two linearly independent polynomials, it follows that $rank(L)$ is the subspace of $P_3$ generated by $x,x^3$.
answered Jan 1 at 18:48
R. N. MarleyR. N. Marley
788
$endgroup$
add a comment |
$begingroup$
${x,x^3}$ is a linearly independent set and spans the range, as can be seen by looking at $cx+(a+b)x^3$. That's why it says "clearly".
In general the range of a linear map is the span of the columns of the matrix rel the standard basis, but we are able to take a short cut in this case.
answered Jan 1 at 19:01
Chris CusterChris Custer
14k3827
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add a comment |
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3 Answers
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active
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3 Answers
3
active
oldest
votes
active
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active
oldest
votes
$begingroup$
The range of $L, R(L)subseteqBbb P_3$. Every element in $R(L)$ is of the form $k_1x+k_2x^3$, which is a linear combination of $x$ and $x^3$. So ${x,x^3}$ is a basis of $R(L)$, giving its dimension as $2$.
answered Jan 1 at 18:38
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
$begingroup$
The range of $L, R(L)subseteqBbb P_3$. Every element in $R(L)$ is of the form $k_1x+k_2x^3$, which is a linear combination of $x$ and $x^3$. So ${x,x^3}$ is a basis of $R(L)$, giving its dimension as $2$.
answered Jan 1 at 18:38
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
$begingroup$
The range of $L, R(L)subseteqBbb P_3$. Every element in $R(L)$ is of the form $k_1x+k_2x^3$, which is a linear combination of $x$ and $x^3$. So ${x,x^3}$ is a basis of $R(L)$, giving its dimension as $2$.
answered Jan 1 at 18:38
Shubham JohriShubham Johri
5,204718
$endgroup$
The range of $L, R(L)subseteqBbb P_3$. Every element in $R(L)$ is of the form $k_1x+k_2x^3$, which is a linear combination of $x$ and $x^3$. So ${x,x^3}$ is a basis of $R(L)$, giving its dimension as $2$.
answered Jan 1 at 18:38
Shubham JohriShubham Johri
5,204718
answered Jan 1 at 18:38
Shubham JohriShubham Johri
5,204718
answered Jan 1 at 18:38
Shubham JohriShubham Johri
5,204718
answered Jan 1 at 18:38
Shubham JohriShubham Johri
5,204718
5,204718
add a comment |
add a comment |
$begingroup$
You know $dim M(2,2)=4$ and $dim null(L)=2$, so by the Bank nullity theorem, $dim rank(L)=2$. But all the polynomials which are image of L are of the form $alpha x + beta x^3$, so you can get all of them with a linear combination of the polynomials $x$ and $x^3$. Since $dim rank(L)=2$ and $x,x^3$ are two linearly independent polynomials, it follows that $rank(L)$ is the subspace of $P_3$ generated by $x,x^3$.
answered Jan 1 at 18:48
R. N. MarleyR. N. Marley
788
$endgroup$
add a comment |
$begingroup$
You know $dim M(2,2)=4$ and $dim null(L)=2$, so by the Bank nullity theorem, $dim rank(L)=2$. But all the polynomials which are image of L are of the form $alpha x + beta x^3$, so you can get all of them with a linear combination of the polynomials $x$ and $x^3$. Since $dim rank(L)=2$ and $x,x^3$ are two linearly independent polynomials, it follows that $rank(L)$ is the subspace of $P_3$ generated by $x,x^3$.
answered Jan 1 at 18:48
R. N. MarleyR. N. Marley
788
$endgroup$
add a comment |
$begingroup$
You know $dim M(2,2)=4$ and $dim null(L)=2$, so by the Bank nullity theorem, $dim rank(L)=2$. But all the polynomials which are image of L are of the form $alpha x + beta x^3$, so you can get all of them with a linear combination of the polynomials $x$ and $x^3$. Since $dim rank(L)=2$ and $x,x^3$ are two linearly independent polynomials, it follows that $rank(L)$ is the subspace of $P_3$ generated by $x,x^3$.
answered Jan 1 at 18:48
R. N. MarleyR. N. Marley
788
$endgroup$
You know $dim M(2,2)=4$ and $dim null(L)=2$, so by the Bank nullity theorem, $dim rank(L)=2$. But all the polynomials which are image of L are of the form $alpha x + beta x^3$, so you can get all of them with a linear combination of the polynomials $x$ and $x^3$. Since $dim rank(L)=2$ and $x,x^3$ are two linearly independent polynomials, it follows that $rank(L)$ is the subspace of $P_3$ generated by $x,x^3$.
answered Jan 1 at 18:48
R. N. MarleyR. N. Marley
788
answered Jan 1 at 18:48
R. N. MarleyR. N. Marley
788
answered Jan 1 at 18:48
R. N. MarleyR. N. Marley
788
answered Jan 1 at 18:48
R. N. MarleyR. N. Marley
788
788
add a comment |
add a comment |
$begingroup$
${x,x^3}$ is a linearly independent set and spans the range, as can be seen by looking at $cx+(a+b)x^3$. That's why it says "clearly".
In general the range of a linear map is the span of the columns of the matrix rel the standard basis, but we are able to take a short cut in this case.
answered Jan 1 at 19:01
Chris CusterChris Custer
14k3827
$endgroup$
add a comment |
$begingroup$
${x,x^3}$ is a linearly independent set and spans the range, as can be seen by looking at $cx+(a+b)x^3$. That's why it says "clearly".
In general the range of a linear map is the span of the columns of the matrix rel the standard basis, but we are able to take a short cut in this case.
answered Jan 1 at 19:01
Chris CusterChris Custer
14k3827
$endgroup$
add a comment |
$begingroup$
${x,x^3}$ is a linearly independent set and spans the range, as can be seen by looking at $cx+(a+b)x^3$. That's why it says "clearly".
In general the range of a linear map is the span of the columns of the matrix rel the standard basis, but we are able to take a short cut in this case.
answered Jan 1 at 19:01
Chris CusterChris Custer
14k3827
$endgroup$
${x,x^3}$ is a linearly independent set and spans the range, as can be seen by looking at $cx+(a+b)x^3$. That's why it says "clearly".
In general the range of a linear map is the span of the columns of the matrix rel the standard basis, but we are able to take a short cut in this case.
answered Jan 1 at 19:01
Chris CusterChris Custer
14k3827
edited Jan 1 at 19:10
answered Jan 1 at 19:01
Chris CusterChris Custer
14k3827
answered Jan 1 at 19:01
Chris CusterChris Custer
14k3827
answered Jan 1 at 19:01
Chris CusterChris Custer
14k3827
14k3827
add a comment |
add a comment |
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$begingroup$
Exactly where does the expression “clear to see” appear?
$endgroup$
– José Carlos Santos
Jan 1 at 18:04