prove $dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$
$begingroup$
I'm a student and I'm studying linear algebra. in Polar Decomposition we have:
for a linear operator $T$, there exist a linear isometry $S$ that:
$$ T =Ssqrt{T^*T}$$
so if $S$ is a linear transformation then it must be $dim(operatorname{range}(T)) leq dim(operatorname{range}(sqrt{T^*T}))$.
But why?
edit: I think it must be equal, I mean:
$$dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$$
I know that $dim(operatorname{range}(T)) = dim(operatorname{range}(T^*))$ but because of the square root I cannot prove that $dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$.
linear-algebra operator-theory adjoint-operators isometry
edited Jan 6 at 23:05
egreg
186k1486208
asked Jan 6 at 17:17
Peyman mohseni kiasariPeyman mohseni kiasari
13711
$endgroup$
add a comment |
$begingroup$
I'm a student and I'm studying linear algebra. in Polar Decomposition we have:
for a linear operator $T$, there exist a linear isometry $S$ that:
$$ T =Ssqrt{T^*T}$$
so if $S$ is a linear transformation then it must be $dim(operatorname{range}(T)) leq dim(operatorname{range}(sqrt{T^*T}))$.
But why?
edit: I think it must be equal, I mean:
$$dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$$
I know that $dim(operatorname{range}(T)) = dim(operatorname{range}(T^*))$ but because of the square root I cannot prove that $dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$.
linear-algebra operator-theory adjoint-operators isometry
edited Jan 6 at 23:05
egreg
186k1486208
asked Jan 6 at 17:17
Peyman mohseni kiasariPeyman mohseni kiasari
13711
$endgroup$
$begingroup$
If $fcolon Uto V$ and $gcolon Vto W$ are linear maps, then the rank of $gcirc f$ cannot be greater than the ranks of $g$ and $f$.
$endgroup$
– egreg
Jan 6 at 17:24
add a comment |
$begingroup$
I'm a student and I'm studying linear algebra. in Polar Decomposition we have:
for a linear operator $T$, there exist a linear isometry $S$ that:
$$ T =Ssqrt{T^*T}$$
so if $S$ is a linear transformation then it must be $dim(operatorname{range}(T)) leq dim(operatorname{range}(sqrt{T^*T}))$.
But why?
edit: I think it must be equal, I mean:
$$dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$$
I know that $dim(operatorname{range}(T)) = dim(operatorname{range}(T^*))$ but because of the square root I cannot prove that $dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$.
linear-algebra operator-theory adjoint-operators isometry
edited Jan 6 at 23:05
egreg
186k1486208
asked Jan 6 at 17:17
Peyman mohseni kiasariPeyman mohseni kiasari
13711
$endgroup$
I'm a student and I'm studying linear algebra. in Polar Decomposition we have:
for a linear operator $T$, there exist a linear isometry $S$ that:
$$ T =Ssqrt{T^*T}$$
so if $S$ is a linear transformation then it must be $dim(operatorname{range}(T)) leq dim(operatorname{range}(sqrt{T^*T}))$.
But why?
edit: I think it must be equal, I mean:
$$dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$$
I know that $dim(operatorname{range}(T)) = dim(operatorname{range}(T^*))$ but because of the square root I cannot prove that $dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$.
linear-algebra operator-theory adjoint-operators isometry
linear-algebra operator-theory adjoint-operators isometry
edited Jan 6 at 23:05
egreg
186k1486208
asked Jan 6 at 17:17
Peyman mohseni kiasariPeyman mohseni kiasari
13711
edited Jan 6 at 23:05
egreg
186k1486208
asked Jan 6 at 17:17
Peyman mohseni kiasariPeyman mohseni kiasari
13711
edited Jan 6 at 23:05
egreg
186k1486208
edited Jan 6 at 23:05
egreg
186k1486208
edited Jan 6 at 23:05
egreg
186k1486208
186k1486208
asked Jan 6 at 17:17
Peyman mohseni kiasariPeyman mohseni kiasari
13711
asked Jan 6 at 17:17
Peyman mohseni kiasariPeyman mohseni kiasari
13711
asked Jan 6 at 17:17
Peyman mohseni kiasariPeyman mohseni kiasari
13711
13711
$begingroup$
If $fcolon Uto V$ and $gcolon Vto W$ are linear maps, then the rank of $gcirc f$ cannot be greater than the ranks of $g$ and $f$.
$endgroup$
– egreg
Jan 6 at 17:24
add a comment |
$begingroup$
If $fcolon Uto V$ and $gcolon Vto W$ are linear maps, then the rank of $gcirc f$ cannot be greater than the ranks of $g$ and $f$.
$endgroup$
– egreg
Jan 6 at 17:24
$begingroup$
If $fcolon Uto V$ and $gcolon Vto W$ are linear maps, then the rank of $gcirc f$ cannot be greater than the ranks of $g$ and $f$.
$endgroup$
– egreg
Jan 6 at 17:24
$begingroup$
If $fcolon Uto V$ and $gcolon Vto W$ are linear maps, then the rank of $gcirc f$ cannot be greater than the ranks of $g$ and $f$.
$endgroup$
– egreg
Jan 6 at 17:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As @egreg pointed out, for any linear maps $S,T$, the rank of $ST$ is always less than or equal to that of $T$. To see this, note that by dimension theorem $$dim text{ran} L =dim operatorname{dom} L-dimker Lledim operatorname{dom} L$$ for any linear map $L$. Now, the image of $ST$ can be seen as the image of
$$
Sbig|_{text{ran} T} :text{ran} Tto V,
$$ we can see that $dim text{ran} (ST)le dim text{ran} T$ as wanted. Moreover, equality holds if $dim ker Sbig|_{text{ran}T}=dim [ker Scaptext{ran}T]=0$. (Here, $V$ denotes the vector space where $S,T$ are defined.)
To see that $dim text{ran}T =dim text{ran}(T^*T)$, note that $Tx = 0$ if and only if $T^*Tx =0$, which is saying that $ker T = ker (T^*T)$. Now, by dimension theorem, we have
$$
dim text{ran} T = dim V - dim ker T = dim V - dim ker (T^*T) = dimtext{ran}(T^*T).
$$ (Or we can use the fact that $S$ is an isometry.)
answered Jan 6 at 22:15
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
If $Tcolon Uto V$ and $Scolon Vto W$ are linear maps between finite dimensional vector spaces, then
$$DeclareMathOperator{range}{range}
dimrange{ST}ledimrange(T)
$$
This follows from the rank-nullity theorem:
begin{align}
dim U&=dimrange(T)+dimker(T) \
dim U&=dimrange(ST)+dimker(ST)
end{align}
Therefore
$$
dimrange(ST)=dimrange(T)+dimker(T)-dimker(ST)ledimrange(T)
$$
because from $ker(T)subseteqker(ST)$ we have $dimker(T)ledimker(ST)$.
In your case you can conclude that
$$
dimrange(T)ledimrange(sqrt{T^*T})
$$
On the other hand, $S$ is an isometry, so it is invertible and
$$
sqrt{T^*T}=S^{-1}T
$$
The same argument as before implies
$$
dimrange(sqrt{T^*T})ledimrange(T)
$$
answered Jan 6 at 22:34
egregegreg
186k1486208
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As @egreg pointed out, for any linear maps $S,T$, the rank of $ST$ is always less than or equal to that of $T$. To see this, note that by dimension theorem $$dim text{ran} L =dim operatorname{dom} L-dimker Lledim operatorname{dom} L$$ for any linear map $L$. Now, the image of $ST$ can be seen as the image of
$$
Sbig|_{text{ran} T} :text{ran} Tto V,
$$ we can see that $dim text{ran} (ST)le dim text{ran} T$ as wanted. Moreover, equality holds if $dim ker Sbig|_{text{ran}T}=dim [ker Scaptext{ran}T]=0$. (Here, $V$ denotes the vector space where $S,T$ are defined.)
To see that $dim text{ran}T =dim text{ran}(T^*T)$, note that $Tx = 0$ if and only if $T^*Tx =0$, which is saying that $ker T = ker (T^*T)$. Now, by dimension theorem, we have
$$
dim text{ran} T = dim V - dim ker T = dim V - dim ker (T^*T) = dimtext{ran}(T^*T).
$$ (Or we can use the fact that $S$ is an isometry.)
answered Jan 6 at 22:15
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
As @egreg pointed out, for any linear maps $S,T$, the rank of $ST$ is always less than or equal to that of $T$. To see this, note that by dimension theorem $$dim text{ran} L =dim operatorname{dom} L-dimker Lledim operatorname{dom} L$$ for any linear map $L$. Now, the image of $ST$ can be seen as the image of
$$
Sbig|_{text{ran} T} :text{ran} Tto V,
$$ we can see that $dim text{ran} (ST)le dim text{ran} T$ as wanted. Moreover, equality holds if $dim ker Sbig|_{text{ran}T}=dim [ker Scaptext{ran}T]=0$. (Here, $V$ denotes the vector space where $S,T$ are defined.)
To see that $dim text{ran}T =dim text{ran}(T^*T)$, note that $Tx = 0$ if and only if $T^*Tx =0$, which is saying that $ker T = ker (T^*T)$. Now, by dimension theorem, we have
$$
dim text{ran} T = dim V - dim ker T = dim V - dim ker (T^*T) = dimtext{ran}(T^*T).
$$ (Or we can use the fact that $S$ is an isometry.)
answered Jan 6 at 22:15
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
As @egreg pointed out, for any linear maps $S,T$, the rank of $ST$ is always less than or equal to that of $T$. To see this, note that by dimension theorem $$dim text{ran} L =dim operatorname{dom} L-dimker Lledim operatorname{dom} L$$ for any linear map $L$. Now, the image of $ST$ can be seen as the image of
$$
Sbig|_{text{ran} T} :text{ran} Tto V,
$$ we can see that $dim text{ran} (ST)le dim text{ran} T$ as wanted. Moreover, equality holds if $dim ker Sbig|_{text{ran}T}=dim [ker Scaptext{ran}T]=0$. (Here, $V$ denotes the vector space where $S,T$ are defined.)
To see that $dim text{ran}T =dim text{ran}(T^*T)$, note that $Tx = 0$ if and only if $T^*Tx =0$, which is saying that $ker T = ker (T^*T)$. Now, by dimension theorem, we have
$$
dim text{ran} T = dim V - dim ker T = dim V - dim ker (T^*T) = dimtext{ran}(T^*T).
$$ (Or we can use the fact that $S$ is an isometry.)
answered Jan 6 at 22:15
SongSong
18.6k21651
$endgroup$
As @egreg pointed out, for any linear maps $S,T$, the rank of $ST$ is always less than or equal to that of $T$. To see this, note that by dimension theorem $$dim text{ran} L =dim operatorname{dom} L-dimker Lledim operatorname{dom} L$$ for any linear map $L$. Now, the image of $ST$ can be seen as the image of
$$
Sbig|_{text{ran} T} :text{ran} Tto V,
$$ we can see that $dim text{ran} (ST)le dim text{ran} T$ as wanted. Moreover, equality holds if $dim ker Sbig|_{text{ran}T}=dim [ker Scaptext{ran}T]=0$. (Here, $V$ denotes the vector space where $S,T$ are defined.)
To see that $dim text{ran}T =dim text{ran}(T^*T)$, note that $Tx = 0$ if and only if $T^*Tx =0$, which is saying that $ker T = ker (T^*T)$. Now, by dimension theorem, we have
$$
dim text{ran} T = dim V - dim ker T = dim V - dim ker (T^*T) = dimtext{ran}(T^*T).
$$ (Or we can use the fact that $S$ is an isometry.)
answered Jan 6 at 22:15
SongSong
18.6k21651
edited Jan 6 at 22:20
answered Jan 6 at 22:15
SongSong
18.6k21651
answered Jan 6 at 22:15
SongSong
18.6k21651
answered Jan 6 at 22:15
SongSong
18.6k21651
18.6k21651
add a comment |
add a comment |
$begingroup$
If $Tcolon Uto V$ and $Scolon Vto W$ are linear maps between finite dimensional vector spaces, then
$$DeclareMathOperator{range}{range}
dimrange{ST}ledimrange(T)
$$
This follows from the rank-nullity theorem:
begin{align}
dim U&=dimrange(T)+dimker(T) \
dim U&=dimrange(ST)+dimker(ST)
end{align}
Therefore
$$
dimrange(ST)=dimrange(T)+dimker(T)-dimker(ST)ledimrange(T)
$$
because from $ker(T)subseteqker(ST)$ we have $dimker(T)ledimker(ST)$.
In your case you can conclude that
$$
dimrange(T)ledimrange(sqrt{T^*T})
$$
On the other hand, $S$ is an isometry, so it is invertible and
$$
sqrt{T^*T}=S^{-1}T
$$
The same argument as before implies
$$
dimrange(sqrt{T^*T})ledimrange(T)
$$
answered Jan 6 at 22:34
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
If $Tcolon Uto V$ and $Scolon Vto W$ are linear maps between finite dimensional vector spaces, then
$$DeclareMathOperator{range}{range}
dimrange{ST}ledimrange(T)
$$
This follows from the rank-nullity theorem:
begin{align}
dim U&=dimrange(T)+dimker(T) \
dim U&=dimrange(ST)+dimker(ST)
end{align}
Therefore
$$
dimrange(ST)=dimrange(T)+dimker(T)-dimker(ST)ledimrange(T)
$$
because from $ker(T)subseteqker(ST)$ we have $dimker(T)ledimker(ST)$.
In your case you can conclude that
$$
dimrange(T)ledimrange(sqrt{T^*T})
$$
On the other hand, $S$ is an isometry, so it is invertible and
$$
sqrt{T^*T}=S^{-1}T
$$
The same argument as before implies
$$
dimrange(sqrt{T^*T})ledimrange(T)
$$
answered Jan 6 at 22:34
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
If $Tcolon Uto V$ and $Scolon Vto W$ are linear maps between finite dimensional vector spaces, then
$$DeclareMathOperator{range}{range}
dimrange{ST}ledimrange(T)
$$
This follows from the rank-nullity theorem:
begin{align}
dim U&=dimrange(T)+dimker(T) \
dim U&=dimrange(ST)+dimker(ST)
end{align}
Therefore
$$
dimrange(ST)=dimrange(T)+dimker(T)-dimker(ST)ledimrange(T)
$$
because from $ker(T)subseteqker(ST)$ we have $dimker(T)ledimker(ST)$.
In your case you can conclude that
$$
dimrange(T)ledimrange(sqrt{T^*T})
$$
On the other hand, $S$ is an isometry, so it is invertible and
$$
sqrt{T^*T}=S^{-1}T
$$
The same argument as before implies
$$
dimrange(sqrt{T^*T})ledimrange(T)
$$
answered Jan 6 at 22:34
egregegreg
186k1486208
$endgroup$
If $Tcolon Uto V$ and $Scolon Vto W$ are linear maps between finite dimensional vector spaces, then
$$DeclareMathOperator{range}{range}
dimrange{ST}ledimrange(T)
$$
This follows from the rank-nullity theorem:
begin{align}
dim U&=dimrange(T)+dimker(T) \
dim U&=dimrange(ST)+dimker(ST)
end{align}
Therefore
$$
dimrange(ST)=dimrange(T)+dimker(T)-dimker(ST)ledimrange(T)
$$
because from $ker(T)subseteqker(ST)$ we have $dimker(T)ledimker(ST)$.
In your case you can conclude that
$$
dimrange(T)ledimrange(sqrt{T^*T})
$$
On the other hand, $S$ is an isometry, so it is invertible and
$$
sqrt{T^*T}=S^{-1}T
$$
The same argument as before implies
$$
dimrange(sqrt{T^*T})ledimrange(T)
$$
answered Jan 6 at 22:34
egregegreg
186k1486208
answered Jan 6 at 22:34
egregegreg
186k1486208
answered Jan 6 at 22:34
egregegreg
186k1486208
answered Jan 6 at 22:34
egregegreg
186k1486208
186k1486208
add a comment |
add a comment |
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$begingroup$
If $fcolon Uto V$ and $gcolon Vto W$ are linear maps, then the rank of $gcirc f$ cannot be greater than the ranks of $g$ and $f$.
$endgroup$
– egreg
Jan 6 at 17:24