Proving $(ker{T})^{perp}subseteq operatorname{Im} T^{*}$ The 2019 Stack Overflow Developer...
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Proving $(ker{T})^{perp}subseteq operatorname{Im} T^{*}$
The 2019 Stack Overflow Developer Survey Results Are InWhy is $ker(rho-sigma)subseteqker(rho)oplusoperatorname{im}(sigma)$?Transformation and its adjointOn proving $ker(TT^*+T^*T)=ker(T^*)cap ker(T)$if $E^2=E$, then $text{Im};Esubsetleft(ker E+(ker E)^perpright)$?prove that $operatorname{Ker}T=operatorname{Im}(S)^perp$ for some transformationProblem proving: $V = ker T oplus operatorname{im}T$$ker A=(operatorname{im} A)^perp$$V = operatorname{Im} T oplus operatorname{Im}T^t oplus H$ where $T^2 = 0$ and $H = ker T cap ker T^t$Prove: $ operatorname{Ker}(T)^perp= operatorname{Im}(T^*)$$ operatorname{Ker}TR = {0}$ but $ operatorname{Ker}T not= {0}$ where $R$ is an isomorphism?
$begingroup$
Let $V$ be a finite inner product space with $Tcolon Vto V$ a linear transformation.
How can I prove that, $(ker{T})^{perp}subseteq operatorname{Im}T^{*}$ ?
Edit:
My purpose is to prove that:
$operatorname{rank}(T)=operatorname{rank}(T^{*})$
linear-algebra
edited Sep 7 '16 at 2:44
Will R
6,78231429
asked May 6 '13 at 9:47
17SI.34SA17SI.34SA
9581119
$endgroup$
add a comment |
$begingroup$
Let $V$ be a finite inner product space with $Tcolon Vto V$ a linear transformation.
How can I prove that, $(ker{T})^{perp}subseteq operatorname{Im}T^{*}$ ?
Edit:
My purpose is to prove that:
$operatorname{rank}(T)=operatorname{rank}(T^{*})$
linear-algebra
edited Sep 7 '16 at 2:44
Will R
6,78231429
asked May 6 '13 at 9:47
17SI.34SA17SI.34SA
9581119
$endgroup$
4
$begingroup$
It is easy to prove $(mbox{Im} T^*)^perp=mbox{Ker}T$ by writing $0=(Tx,y)=(x,T^*y)$ for every $y$, and every $xinmbox{Ker} T$. So $((mbox{Im} T^*)^perp)^perp=(mbox{Ker}T)^perp$. Now for $F$ a subspace in general, $(F^perp)^perp=overline{F}$. This follows from $H=overline{F}oplus overline{F}^perp=overline{F}oplus F^perp$. Finally, in finite dimension, every subspace is closed: $overline{F}=F$.
$endgroup$
– Julien
May 6 '13 at 10:36
add a comment |
$begingroup$
Let $V$ be a finite inner product space with $Tcolon Vto V$ a linear transformation.
How can I prove that, $(ker{T})^{perp}subseteq operatorname{Im}T^{*}$ ?
Edit:
My purpose is to prove that:
$operatorname{rank}(T)=operatorname{rank}(T^{*})$
linear-algebra
edited Sep 7 '16 at 2:44
Will R
6,78231429
asked May 6 '13 at 9:47
17SI.34SA17SI.34SA
9581119
$endgroup$
Let $V$ be a finite inner product space with $Tcolon Vto V$ a linear transformation.
How can I prove that, $(ker{T})^{perp}subseteq operatorname{Im}T^{*}$ ?
Edit:
My purpose is to prove that:
$operatorname{rank}(T)=operatorname{rank}(T^{*})$
linear-algebra
linear-algebra
edited Sep 7 '16 at 2:44
Will R
6,78231429
asked May 6 '13 at 9:47
17SI.34SA17SI.34SA
9581119
edited Sep 7 '16 at 2:44
Will R
6,78231429
asked May 6 '13 at 9:47
17SI.34SA17SI.34SA
9581119
edited Sep 7 '16 at 2:44
Will R
6,78231429
edited Sep 7 '16 at 2:44
Will R
6,78231429
edited Sep 7 '16 at 2:44
Will R
6,78231429
6,78231429
asked May 6 '13 at 9:47
17SI.34SA17SI.34SA
9581119
asked May 6 '13 at 9:47
17SI.34SA17SI.34SA
9581119
asked May 6 '13 at 9:47
17SI.34SA17SI.34SA
9581119
9581119
4
$begingroup$
It is easy to prove $(mbox{Im} T^*)^perp=mbox{Ker}T$ by writing $0=(Tx,y)=(x,T^*y)$ for every $y$, and every $xinmbox{Ker} T$. So $((mbox{Im} T^*)^perp)^perp=(mbox{Ker}T)^perp$. Now for $F$ a subspace in general, $(F^perp)^perp=overline{F}$. This follows from $H=overline{F}oplus overline{F}^perp=overline{F}oplus F^perp$. Finally, in finite dimension, every subspace is closed: $overline{F}=F$.
$endgroup$
– Julien
May 6 '13 at 10:36
add a comment |
4
$begingroup$
It is easy to prove $(mbox{Im} T^*)^perp=mbox{Ker}T$ by writing $0=(Tx,y)=(x,T^*y)$ for every $y$, and every $xinmbox{Ker} T$. So $((mbox{Im} T^*)^perp)^perp=(mbox{Ker}T)^perp$. Now for $F$ a subspace in general, $(F^perp)^perp=overline{F}$. This follows from $H=overline{F}oplus overline{F}^perp=overline{F}oplus F^perp$. Finally, in finite dimension, every subspace is closed: $overline{F}=F$.
$endgroup$
– Julien
May 6 '13 at 10:36
4
4
$begingroup$
It is easy to prove $(mbox{Im} T^*)^perp=mbox{Ker}T$ by writing $0=(Tx,y)=(x,T^*y)$ for every $y$, and every $xinmbox{Ker} T$. So $((mbox{Im} T^*)^perp)^perp=(mbox{Ker}T)^perp$. Now for $F$ a subspace in general, $(F^perp)^perp=overline{F}$. This follows from $H=overline{F}oplus overline{F}^perp=overline{F}oplus F^perp$. Finally, in finite dimension, every subspace is closed: $overline{F}=F$.
$endgroup$
– Julien
May 6 '13 at 10:36
$begingroup$
It is easy to prove $(mbox{Im} T^*)^perp=mbox{Ker}T$ by writing $0=(Tx,y)=(x,T^*y)$ for every $y$, and every $xinmbox{Ker} T$. So $((mbox{Im} T^*)^perp)^perp=(mbox{Ker}T)^perp$. Now for $F$ a subspace in general, $(F^perp)^perp=overline{F}$. This follows from $H=overline{F}oplus overline{F}^perp=overline{F}oplus F^perp$. Finally, in finite dimension, every subspace is closed: $overline{F}=F$.
$endgroup$
– Julien
May 6 '13 at 10:36
add a comment |
1 Answer
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$begingroup$
I thought of an alternative proof to what you originally wanted to show:
If $T = [A]$ in an orthonormal basis (exists because we're in a finite space) $[T^*] = bar{A}^t = A^*$.
$rank A^* = rank left(bar{A}^t right)$ and because the conjugate does not change the rank(tell me if you need a proof to this one as well) it's the same as:
$rank left(A^tright)$ but as you've probably proved that $forall M_{ntimes n} rank_Cleft(M right) = rank_Rleft(M right) $ so the transpose does not have any change the rank so it equals to $rank left(Aright)$.
Which therefore means that given the isomorphism of matrices and the linear transformation they describe that $operatorname{rank}(T)=operatorname{rank}(T^{*})$.
answered May 6 '13 at 10:29
ScisScis
20019
$endgroup$
add a comment |
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$begingroup$
I thought of an alternative proof to what you originally wanted to show:
If $T = [A]$ in an orthonormal basis (exists because we're in a finite space) $[T^*] = bar{A}^t = A^*$.
$rank A^* = rank left(bar{A}^t right)$ and because the conjugate does not change the rank(tell me if you need a proof to this one as well) it's the same as:
$rank left(A^tright)$ but as you've probably proved that $forall M_{ntimes n} rank_Cleft(M right) = rank_Rleft(M right) $ so the transpose does not have any change the rank so it equals to $rank left(Aright)$.
Which therefore means that given the isomorphism of matrices and the linear transformation they describe that $operatorname{rank}(T)=operatorname{rank}(T^{*})$.
answered May 6 '13 at 10:29
ScisScis
20019
$endgroup$
add a comment |
$begingroup$
I thought of an alternative proof to what you originally wanted to show:
If $T = [A]$ in an orthonormal basis (exists because we're in a finite space) $[T^*] = bar{A}^t = A^*$.
$rank A^* = rank left(bar{A}^t right)$ and because the conjugate does not change the rank(tell me if you need a proof to this one as well) it's the same as:
$rank left(A^tright)$ but as you've probably proved that $forall M_{ntimes n} rank_Cleft(M right) = rank_Rleft(M right) $ so the transpose does not have any change the rank so it equals to $rank left(Aright)$.
Which therefore means that given the isomorphism of matrices and the linear transformation they describe that $operatorname{rank}(T)=operatorname{rank}(T^{*})$.
answered May 6 '13 at 10:29
ScisScis
20019
$endgroup$
add a comment |
$begingroup$
I thought of an alternative proof to what you originally wanted to show:
If $T = [A]$ in an orthonormal basis (exists because we're in a finite space) $[T^*] = bar{A}^t = A^*$.
$rank A^* = rank left(bar{A}^t right)$ and because the conjugate does not change the rank(tell me if you need a proof to this one as well) it's the same as:
$rank left(A^tright)$ but as you've probably proved that $forall M_{ntimes n} rank_Cleft(M right) = rank_Rleft(M right) $ so the transpose does not have any change the rank so it equals to $rank left(Aright)$.
Which therefore means that given the isomorphism of matrices and the linear transformation they describe that $operatorname{rank}(T)=operatorname{rank}(T^{*})$.
answered May 6 '13 at 10:29
ScisScis
20019
$endgroup$
I thought of an alternative proof to what you originally wanted to show:
If $T = [A]$ in an orthonormal basis (exists because we're in a finite space) $[T^*] = bar{A}^t = A^*$.
$rank A^* = rank left(bar{A}^t right)$ and because the conjugate does not change the rank(tell me if you need a proof to this one as well) it's the same as:
$rank left(A^tright)$ but as you've probably proved that $forall M_{ntimes n} rank_Cleft(M right) = rank_Rleft(M right) $ so the transpose does not have any change the rank so it equals to $rank left(Aright)$.
Which therefore means that given the isomorphism of matrices and the linear transformation they describe that $operatorname{rank}(T)=operatorname{rank}(T^{*})$.
answered May 6 '13 at 10:29
ScisScis
20019
edited May 6 '13 at 11:23
answered May 6 '13 at 10:29
ScisScis
20019
answered May 6 '13 at 10:29
ScisScis
20019
answered May 6 '13 at 10:29
ScisScis
20019
20019
add a comment |
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$begingroup$
It is easy to prove $(mbox{Im} T^*)^perp=mbox{Ker}T$ by writing $0=(Tx,y)=(x,T^*y)$ for every $y$, and every $xinmbox{Ker} T$. So $((mbox{Im} T^*)^perp)^perp=(mbox{Ker}T)^perp$. Now for $F$ a subspace in general, $(F^perp)^perp=overline{F}$. This follows from $H=overline{F}oplus overline{F}^perp=overline{F}oplus F^perp$. Finally, in finite dimension, every subspace is closed: $overline{F}=F$.
$endgroup$
– Julien
May 6 '13 at 10:36