Deriving a subspace through matrix multiplication [on hold]
Let x ∈ vector space V of dimension m and let W be a n x m matrix and U be a m x n matrix where m > n. Fix any U,W and consider the mapping x -> UWx. How do we prove that the range of this mapping, R = {UWx : x ∈ V} is a n dimensional subspace of V ?
linear-algebra
asked 2 days ago
SonnySonny
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put on hold as off-topic by amWhy, Paul Frost, Shailesh, KReiser, mrtaurho 2 days ago
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Let x ∈ vector space V of dimension m and let W be a n x m matrix and U be a m x n matrix where m > n. Fix any U,W and consider the mapping x -> UWx. How do we prove that the range of this mapping, R = {UWx : x ∈ V} is a n dimensional subspace of V ?
linear-algebra
asked 2 days ago
SonnySonny
1
New contributor
Sonny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by amWhy, Paul Frost, Shailesh, KReiser, mrtaurho 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, Shailesh, KReiser, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
$R$ is at-most $n$ dimensional.
– Shubham Johri
2 days ago
add a comment |
Let x ∈ vector space V of dimension m and let W be a n x m matrix and U be a m x n matrix where m > n. Fix any U,W and consider the mapping x -> UWx. How do we prove that the range of this mapping, R = {UWx : x ∈ V} is a n dimensional subspace of V ?
linear-algebra
asked 2 days ago
SonnySonny
1
New contributor
Sonny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let x ∈ vector space V of dimension m and let W be a n x m matrix and U be a m x n matrix where m > n. Fix any U,W and consider the mapping x -> UWx. How do we prove that the range of this mapping, R = {UWx : x ∈ V} is a n dimensional subspace of V ?
linear-algebra
linear-algebra
asked 2 days ago
SonnySonny
1
New contributor
Sonny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
SonnySonny
1
New contributor
Sonny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
SonnySonny
1
New contributor
Sonny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
SonnySonny
1
asked 2 days ago
SonnySonny
1
1
New contributor
Sonny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sonny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sonny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by amWhy, Paul Frost, Shailesh, KReiser, mrtaurho 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, Shailesh, KReiser, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by amWhy, Paul Frost, Shailesh, KReiser, mrtaurho 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, Shailesh, KReiser, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
$R$ is at-most $n$ dimensional.
– Shubham Johri
2 days ago
add a comment |
$R$ is at-most $n$ dimensional.
– Shubham Johri
2 days ago
$R$ is at-most $n$ dimensional.
– Shubham Johri
2 days ago
$R$ is at-most $n$ dimensional.
– Shubham Johri
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
You’ll have hard time to prove that...
Take for $U$ the zero matrix. Then $R$ is the zero subspace which dimension is also equal to zero.
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
add a comment |
Let ${v_1,v_2,v_3,...,v_m}$ be a basis of $V$. Then $$UWx=UW(a_1v_1+a_2v_2+...+a_mv_m)=a_1UWv_1+a_2UWv_2+...+a_mUWv_m$$ which suggests that $R$ is the linear span of $UWv_1,UWv_2,...,UWv_m$, and hence a linear subspace of $V$.
The dimension of $R$ is the rank of the matrix $UW$. Note that the row vectors of $UW$ are linear combinations of the row vectors of $W$, so the rank of $UW$, which is equal to the number of linearly independent rows in $UW$, cannot exceed the rank of $W$.$$text{rank}(W)lemin{m,n}=n$$Therefore, the dimension of $R$ is at-most $n$.
answered 2 days ago
Shubham JohriShubham Johri
4,469717
Could you elaborate on "The dimension of R is the rank of the matrix UW" ? Is it possible to show that the null space of the linear map UW has the dimension of m-n ?
– Sonny
2 days ago
$R$ is the image of the linear transformation having matrix $UW$. The dimension of the image of a linear transformation is the same as the rank of the matrix representing it. Now that you know $text{rank}(UW)le n$, given $UW_{mtimes m}$ represents a linear transformation from $Vto V$, you have by the rank-nullity theorem,$$text{rank}(UW)+text{nullity}(UW)=dim V=m$$which gives $text{nullity}(UW)ge m-n$.
– Shubham Johri
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You’ll have hard time to prove that...
Take for $U$ the zero matrix. Then $R$ is the zero subspace which dimension is also equal to zero.
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
add a comment |
You’ll have hard time to prove that...
Take for $U$ the zero matrix. Then $R$ is the zero subspace which dimension is also equal to zero.
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
add a comment |
You’ll have hard time to prove that...
Take for $U$ the zero matrix. Then $R$ is the zero subspace which dimension is also equal to zero.
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
You’ll have hard time to prove that...
Take for $U$ the zero matrix. Then $R$ is the zero subspace which dimension is also equal to zero.
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
25.2k21953
add a comment |
add a comment |
Let ${v_1,v_2,v_3,...,v_m}$ be a basis of $V$. Then $$UWx=UW(a_1v_1+a_2v_2+...+a_mv_m)=a_1UWv_1+a_2UWv_2+...+a_mUWv_m$$ which suggests that $R$ is the linear span of $UWv_1,UWv_2,...,UWv_m$, and hence a linear subspace of $V$.
The dimension of $R$ is the rank of the matrix $UW$. Note that the row vectors of $UW$ are linear combinations of the row vectors of $W$, so the rank of $UW$, which is equal to the number of linearly independent rows in $UW$, cannot exceed the rank of $W$.$$text{rank}(W)lemin{m,n}=n$$Therefore, the dimension of $R$ is at-most $n$.
answered 2 days ago
Shubham JohriShubham Johri
4,469717
Could you elaborate on "The dimension of R is the rank of the matrix UW" ? Is it possible to show that the null space of the linear map UW has the dimension of m-n ?
– Sonny
2 days ago
$R$ is the image of the linear transformation having matrix $UW$. The dimension of the image of a linear transformation is the same as the rank of the matrix representing it. Now that you know $text{rank}(UW)le n$, given $UW_{mtimes m}$ represents a linear transformation from $Vto V$, you have by the rank-nullity theorem,$$text{rank}(UW)+text{nullity}(UW)=dim V=m$$which gives $text{nullity}(UW)ge m-n$.
– Shubham Johri
2 days ago
add a comment |
Let ${v_1,v_2,v_3,...,v_m}$ be a basis of $V$. Then $$UWx=UW(a_1v_1+a_2v_2+...+a_mv_m)=a_1UWv_1+a_2UWv_2+...+a_mUWv_m$$ which suggests that $R$ is the linear span of $UWv_1,UWv_2,...,UWv_m$, and hence a linear subspace of $V$.
The dimension of $R$ is the rank of the matrix $UW$. Note that the row vectors of $UW$ are linear combinations of the row vectors of $W$, so the rank of $UW$, which is equal to the number of linearly independent rows in $UW$, cannot exceed the rank of $W$.$$text{rank}(W)lemin{m,n}=n$$Therefore, the dimension of $R$ is at-most $n$.
answered 2 days ago
Shubham JohriShubham Johri
4,469717
Could you elaborate on "The dimension of R is the rank of the matrix UW" ? Is it possible to show that the null space of the linear map UW has the dimension of m-n ?
– Sonny
2 days ago
$R$ is the image of the linear transformation having matrix $UW$. The dimension of the image of a linear transformation is the same as the rank of the matrix representing it. Now that you know $text{rank}(UW)le n$, given $UW_{mtimes m}$ represents a linear transformation from $Vto V$, you have by the rank-nullity theorem,$$text{rank}(UW)+text{nullity}(UW)=dim V=m$$which gives $text{nullity}(UW)ge m-n$.
– Shubham Johri
2 days ago
add a comment |
Let ${v_1,v_2,v_3,...,v_m}$ be a basis of $V$. Then $$UWx=UW(a_1v_1+a_2v_2+...+a_mv_m)=a_1UWv_1+a_2UWv_2+...+a_mUWv_m$$ which suggests that $R$ is the linear span of $UWv_1,UWv_2,...,UWv_m$, and hence a linear subspace of $V$.
The dimension of $R$ is the rank of the matrix $UW$. Note that the row vectors of $UW$ are linear combinations of the row vectors of $W$, so the rank of $UW$, which is equal to the number of linearly independent rows in $UW$, cannot exceed the rank of $W$.$$text{rank}(W)lemin{m,n}=n$$Therefore, the dimension of $R$ is at-most $n$.
answered 2 days ago
Shubham JohriShubham Johri
4,469717
Let ${v_1,v_2,v_3,...,v_m}$ be a basis of $V$. Then $$UWx=UW(a_1v_1+a_2v_2+...+a_mv_m)=a_1UWv_1+a_2UWv_2+...+a_mUWv_m$$ which suggests that $R$ is the linear span of $UWv_1,UWv_2,...,UWv_m$, and hence a linear subspace of $V$.
The dimension of $R$ is the rank of the matrix $UW$. Note that the row vectors of $UW$ are linear combinations of the row vectors of $W$, so the rank of $UW$, which is equal to the number of linearly independent rows in $UW$, cannot exceed the rank of $W$.$$text{rank}(W)lemin{m,n}=n$$Therefore, the dimension of $R$ is at-most $n$.
answered 2 days ago
Shubham JohriShubham Johri
4,469717
answered 2 days ago
Shubham JohriShubham Johri
4,469717
answered 2 days ago
Shubham JohriShubham Johri
4,469717
answered 2 days ago
Shubham JohriShubham Johri
4,469717
4,469717
Could you elaborate on "The dimension of R is the rank of the matrix UW" ? Is it possible to show that the null space of the linear map UW has the dimension of m-n ?
– Sonny
2 days ago
$R$ is the image of the linear transformation having matrix $UW$. The dimension of the image of a linear transformation is the same as the rank of the matrix representing it. Now that you know $text{rank}(UW)le n$, given $UW_{mtimes m}$ represents a linear transformation from $Vto V$, you have by the rank-nullity theorem,$$text{rank}(UW)+text{nullity}(UW)=dim V=m$$which gives $text{nullity}(UW)ge m-n$.
– Shubham Johri
2 days ago
add a comment |
Could you elaborate on "The dimension of R is the rank of the matrix UW" ? Is it possible to show that the null space of the linear map UW has the dimension of m-n ?
– Sonny
2 days ago
$R$ is the image of the linear transformation having matrix $UW$. The dimension of the image of a linear transformation is the same as the rank of the matrix representing it. Now that you know $text{rank}(UW)le n$, given $UW_{mtimes m}$ represents a linear transformation from $Vto V$, you have by the rank-nullity theorem,$$text{rank}(UW)+text{nullity}(UW)=dim V=m$$which gives $text{nullity}(UW)ge m-n$.
– Shubham Johri
2 days ago
Could you elaborate on "The dimension of R is the rank of the matrix UW" ? Is it possible to show that the null space of the linear map UW has the dimension of m-n ?
– Sonny
2 days ago
Could you elaborate on "The dimension of R is the rank of the matrix UW" ? Is it possible to show that the null space of the linear map UW has the dimension of m-n ?
– Sonny
2 days ago
$R$ is the image of the linear transformation having matrix $UW$. The dimension of the image of a linear transformation is the same as the rank of the matrix representing it. Now that you know $text{rank}(UW)le n$, given $UW_{mtimes m}$ represents a linear transformation from $Vto V$, you have by the rank-nullity theorem,$$text{rank}(UW)+text{nullity}(UW)=dim V=m$$which gives $text{nullity}(UW)ge m-n$.
– Shubham Johri
2 days ago
$R$ is the image of the linear transformation having matrix $UW$. The dimension of the image of a linear transformation is the same as the rank of the matrix representing it. Now that you know $text{rank}(UW)le n$, given $UW_{mtimes m}$ represents a linear transformation from $Vto V$, you have by the rank-nullity theorem,$$text{rank}(UW)+text{nullity}(UW)=dim V=m$$which gives $text{nullity}(UW)ge m-n$.
– Shubham Johri
2 days ago
add a comment |
$R$ is at-most $n$ dimensional.
– Shubham Johri
2 days ago