Rank of Linear Transformation Preserved
$begingroup$
Show that the linear transformation with rank $m$ on $n$-dimensional subspace $V$ can be expressed as the sum of $m$ linear transformations with rank $1$.
Since any linear transformation can be represented as a matrix product, i.e $mathbf x mapsto A mathbf x$, along with the property of rank of the matrices,
A rank-$k$ matrix can be written as the sum of $k$ rank-$1$ matrices.
This follows the conclusion.
The proof above is my attempt, and I think I might miss out something, or just have a proof for special case.
Any thought or suggestion would be appreciated.
linear-algebra proof-verification linear-transformations
asked Dec 10 '18 at 13:21
weilam06weilam06
9511
$endgroup$
add a comment |
$begingroup$
Show that the linear transformation with rank $m$ on $n$-dimensional subspace $V$ can be expressed as the sum of $m$ linear transformations with rank $1$.
Since any linear transformation can be represented as a matrix product, i.e $mathbf x mapsto A mathbf x$, along with the property of rank of the matrices,
A rank-$k$ matrix can be written as the sum of $k$ rank-$1$ matrices.
This follows the conclusion.
The proof above is my attempt, and I think I might miss out something, or just have a proof for special case.
Any thought or suggestion would be appreciated.
linear-algebra proof-verification linear-transformations
asked Dec 10 '18 at 13:21
weilam06weilam06
9511
$endgroup$
$begingroup$
Where is your proof that any rank $k$ matrix can be written as a sum of $k$ matrices of rank $1$?
$endgroup$
– Christoph
Dec 10 '18 at 13:27
$begingroup$
It is shown in the Wikipedia page for Rank (Linear Algebra). I directly used it.
$endgroup$
– weilam06
Dec 10 '18 at 13:29
$begingroup$
It is fine if you can use the result directly.
$endgroup$
– Shubham Johri
Dec 10 '18 at 13:32
$begingroup$
I don't find a proof of that statement on the wikipedia article you mentioned. But yes, when you have a proof for the statement about rank $k$ matrices, then the statement of about rank $k$ linear transformations is an immediate consequence.
$endgroup$
– Christoph
Dec 10 '18 at 13:32
add a comment |
$begingroup$
Show that the linear transformation with rank $m$ on $n$-dimensional subspace $V$ can be expressed as the sum of $m$ linear transformations with rank $1$.
Since any linear transformation can be represented as a matrix product, i.e $mathbf x mapsto A mathbf x$, along with the property of rank of the matrices,
A rank-$k$ matrix can be written as the sum of $k$ rank-$1$ matrices.
This follows the conclusion.
The proof above is my attempt, and I think I might miss out something, or just have a proof for special case.
Any thought or suggestion would be appreciated.
linear-algebra proof-verification linear-transformations
asked Dec 10 '18 at 13:21
weilam06weilam06
9511
$endgroup$
Show that the linear transformation with rank $m$ on $n$-dimensional subspace $V$ can be expressed as the sum of $m$ linear transformations with rank $1$.
Since any linear transformation can be represented as a matrix product, i.e $mathbf x mapsto A mathbf x$, along with the property of rank of the matrices,
A rank-$k$ matrix can be written as the sum of $k$ rank-$1$ matrices.
This follows the conclusion.
The proof above is my attempt, and I think I might miss out something, or just have a proof for special case.
Any thought or suggestion would be appreciated.
linear-algebra proof-verification linear-transformations
linear-algebra proof-verification linear-transformations
asked Dec 10 '18 at 13:21
weilam06weilam06
9511
asked Dec 10 '18 at 13:21
weilam06weilam06
9511
asked Dec 10 '18 at 13:21
weilam06weilam06
9511
asked Dec 10 '18 at 13:21
weilam06weilam06
9511
asked Dec 10 '18 at 13:21
weilam06weilam06
9511
9511
$begingroup$
Where is your proof that any rank $k$ matrix can be written as a sum of $k$ matrices of rank $1$?
$endgroup$
– Christoph
Dec 10 '18 at 13:27
$begingroup$
It is shown in the Wikipedia page for Rank (Linear Algebra). I directly used it.
$endgroup$
– weilam06
Dec 10 '18 at 13:29
$begingroup$
It is fine if you can use the result directly.
$endgroup$
– Shubham Johri
Dec 10 '18 at 13:32
$begingroup$
I don't find a proof of that statement on the wikipedia article you mentioned. But yes, when you have a proof for the statement about rank $k$ matrices, then the statement of about rank $k$ linear transformations is an immediate consequence.
$endgroup$
– Christoph
Dec 10 '18 at 13:32
add a comment |
$begingroup$
Where is your proof that any rank $k$ matrix can be written as a sum of $k$ matrices of rank $1$?
$endgroup$
– Christoph
Dec 10 '18 at 13:27
$begingroup$
It is shown in the Wikipedia page for Rank (Linear Algebra). I directly used it.
$endgroup$
– weilam06
Dec 10 '18 at 13:29
$begingroup$
It is fine if you can use the result directly.
$endgroup$
– Shubham Johri
Dec 10 '18 at 13:32
$begingroup$
I don't find a proof of that statement on the wikipedia article you mentioned. But yes, when you have a proof for the statement about rank $k$ matrices, then the statement of about rank $k$ linear transformations is an immediate consequence.
$endgroup$
– Christoph
Dec 10 '18 at 13:32
$begingroup$
Where is your proof that any rank $k$ matrix can be written as a sum of $k$ matrices of rank $1$?
$endgroup$
– Christoph
Dec 10 '18 at 13:27
$begingroup$
Where is your proof that any rank $k$ matrix can be written as a sum of $k$ matrices of rank $1$?
$endgroup$
– Christoph
Dec 10 '18 at 13:27
$begingroup$
It is shown in the Wikipedia page for Rank (Linear Algebra). I directly used it.
$endgroup$
– weilam06
Dec 10 '18 at 13:29
$begingroup$
It is shown in the Wikipedia page for Rank (Linear Algebra). I directly used it.
$endgroup$
– weilam06
Dec 10 '18 at 13:29
$begingroup$
It is fine if you can use the result directly.
$endgroup$
– Shubham Johri
Dec 10 '18 at 13:32
$begingroup$
It is fine if you can use the result directly.
$endgroup$
– Shubham Johri
Dec 10 '18 at 13:32
$begingroup$
I don't find a proof of that statement on the wikipedia article you mentioned. But yes, when you have a proof for the statement about rank $k$ matrices, then the statement of about rank $k$ linear transformations is an immediate consequence.
$endgroup$
– Christoph
Dec 10 '18 at 13:32
$begingroup$
I don't find a proof of that statement on the wikipedia article you mentioned. But yes, when you have a proof for the statement about rank $k$ matrices, then the statement of about rank $k$ linear transformations is an immediate consequence.
$endgroup$
– Christoph
Dec 10 '18 at 13:32
add a comment |
1 Answer
1
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oldest
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$begingroup$
Let me give a direct proof in addition to your approach reducing to the case of matrices. Let $fcolon Vto W$ be a linear map such that $dim(f(V))=m$. Pick a basis $w_1,dots,w_m$ of $f(V)$ and define projections
begin{align*}
pi_j colonquadquad f(V)&longrightarrow f(V), \
sum_i lambda_i w_i &longmapsto lambda_j w_j
end{align*}
for $j=1,dots,m$. Note that $pi_1+cdots+pi_m = operatorname{id}_{f(V)}$.
Hence, defining $f_icolon Vto W$ by $f_i(v)=pi_i(f(v))$ we have
$$
f = f_1+cdots + f_n.
$$
Each $f_i$ is of rank $1$ since $f_i(V) = langle w_irangle$.
answered Dec 10 '18 at 13:39
ChristophChristoph
11.9k1642
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add a comment |
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1 Answer
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active
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1 Answer
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oldest
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$begingroup$
Let me give a direct proof in addition to your approach reducing to the case of matrices. Let $fcolon Vto W$ be a linear map such that $dim(f(V))=m$. Pick a basis $w_1,dots,w_m$ of $f(V)$ and define projections
begin{align*}
pi_j colonquadquad f(V)&longrightarrow f(V), \
sum_i lambda_i w_i &longmapsto lambda_j w_j
end{align*}
for $j=1,dots,m$. Note that $pi_1+cdots+pi_m = operatorname{id}_{f(V)}$.
Hence, defining $f_icolon Vto W$ by $f_i(v)=pi_i(f(v))$ we have
$$
f = f_1+cdots + f_n.
$$
Each $f_i$ is of rank $1$ since $f_i(V) = langle w_irangle$.
answered Dec 10 '18 at 13:39
ChristophChristoph
11.9k1642
$endgroup$
add a comment |
$begingroup$
Let me give a direct proof in addition to your approach reducing to the case of matrices. Let $fcolon Vto W$ be a linear map such that $dim(f(V))=m$. Pick a basis $w_1,dots,w_m$ of $f(V)$ and define projections
begin{align*}
pi_j colonquadquad f(V)&longrightarrow f(V), \
sum_i lambda_i w_i &longmapsto lambda_j w_j
end{align*}
for $j=1,dots,m$. Note that $pi_1+cdots+pi_m = operatorname{id}_{f(V)}$.
Hence, defining $f_icolon Vto W$ by $f_i(v)=pi_i(f(v))$ we have
$$
f = f_1+cdots + f_n.
$$
Each $f_i$ is of rank $1$ since $f_i(V) = langle w_irangle$.
answered Dec 10 '18 at 13:39
ChristophChristoph
11.9k1642
$endgroup$
add a comment |
$begingroup$
Let me give a direct proof in addition to your approach reducing to the case of matrices. Let $fcolon Vto W$ be a linear map such that $dim(f(V))=m$. Pick a basis $w_1,dots,w_m$ of $f(V)$ and define projections
begin{align*}
pi_j colonquadquad f(V)&longrightarrow f(V), \
sum_i lambda_i w_i &longmapsto lambda_j w_j
end{align*}
for $j=1,dots,m$. Note that $pi_1+cdots+pi_m = operatorname{id}_{f(V)}$.
Hence, defining $f_icolon Vto W$ by $f_i(v)=pi_i(f(v))$ we have
$$
f = f_1+cdots + f_n.
$$
Each $f_i$ is of rank $1$ since $f_i(V) = langle w_irangle$.
answered Dec 10 '18 at 13:39
ChristophChristoph
11.9k1642
$endgroup$
Let me give a direct proof in addition to your approach reducing to the case of matrices. Let $fcolon Vto W$ be a linear map such that $dim(f(V))=m$. Pick a basis $w_1,dots,w_m$ of $f(V)$ and define projections
begin{align*}
pi_j colonquadquad f(V)&longrightarrow f(V), \
sum_i lambda_i w_i &longmapsto lambda_j w_j
end{align*}
for $j=1,dots,m$. Note that $pi_1+cdots+pi_m = operatorname{id}_{f(V)}$.
Hence, defining $f_icolon Vto W$ by $f_i(v)=pi_i(f(v))$ we have
$$
f = f_1+cdots + f_n.
$$
Each $f_i$ is of rank $1$ since $f_i(V) = langle w_irangle$.
answered Dec 10 '18 at 13:39
ChristophChristoph
11.9k1642
answered Dec 10 '18 at 13:39
ChristophChristoph
11.9k1642
answered Dec 10 '18 at 13:39
ChristophChristoph
11.9k1642
answered Dec 10 '18 at 13:39
ChristophChristoph
11.9k1642
11.9k1642
add a comment |
add a comment |
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$begingroup$
Where is your proof that any rank $k$ matrix can be written as a sum of $k$ matrices of rank $1$?
$endgroup$
– Christoph
Dec 10 '18 at 13:27
$begingroup$
It is shown in the Wikipedia page for Rank (Linear Algebra). I directly used it.
$endgroup$
– weilam06
Dec 10 '18 at 13:29
$begingroup$
It is fine if you can use the result directly.
$endgroup$
– Shubham Johri
Dec 10 '18 at 13:32
$begingroup$
I don't find a proof of that statement on the wikipedia article you mentioned. But yes, when you have a proof for the statement about rank $k$ matrices, then the statement of about rank $k$ linear transformations is an immediate consequence.
$endgroup$
– Christoph
Dec 10 '18 at 13:32