Mixing
If $TS=ST$, then $S=alpha T+beta$.
$begingroup$
Let $T=begin{pmatrix}a&b\c&dend{pmatrix}$ be a non-scalar matrix.
If $S=begin{pmatrix}e&f\g&hend{pmatrix}$ be such that $TS=ST$. Why there exists $alpha,betain mathbb{C}$ such that
$$S=alpha T+beta I;?$$
Note that $TS-ST=0$ is equivalent to
$$begin{bmatrix}bg-fc & af+bh-eb-fd\
ce+dg-ga-hc & fc-bgend{bmatrix} = begin{bmatrix}0 & 0\0 & 0end{bmatrix}$$
This implies that
$$begin{cases}
bg-fc = 0,\
af+bh-eb-fd = 0,\
ce+dg-ga-hc = 0,\
fc-bg = 0.
end{cases}$$
Since $T$ is non scalar, then $bneq 0$ or $cneq 0$ or $aneq d$. However, I cannot find $alpha$ and $beta$.
linear-algebra
asked Feb 3 at 13:05
SchülerSchüler
1,5591421
$endgroup$
add a comment |
$begingroup$
Let $T=begin{pmatrix}a&b\c&dend{pmatrix}$ be a non-scalar matrix.
If $S=begin{pmatrix}e&f\g&hend{pmatrix}$ be such that $TS=ST$. Why there exists $alpha,betain mathbb{C}$ such that
$$S=alpha T+beta I;?$$
Note that $TS-ST=0$ is equivalent to
$$begin{bmatrix}bg-fc & af+bh-eb-fd\
ce+dg-ga-hc & fc-bgend{bmatrix} = begin{bmatrix}0 & 0\0 & 0end{bmatrix}$$
This implies that
$$begin{cases}
bg-fc = 0,\
af+bh-eb-fd = 0,\
ce+dg-ga-hc = 0,\
fc-bg = 0.
end{cases}$$
Since $T$ is non scalar, then $bneq 0$ or $cneq 0$ or $aneq d$. However, I cannot find $alpha$ and $beta$.
linear-algebra
asked Feb 3 at 13:05
SchülerSchüler
1,5591421
$endgroup$
$begingroup$
Dear professor. According to Lemaa 2.3 this result works only when $text{dim}(H)=2$ (please see books.google.tn/… ) I find contradiction between this result and the provided answer. Thank you for the explanation.
$endgroup$
– Schüler
Feb 9 at 7:40
$begingroup$
I'm Ok. The problem is his comment he says:''The proof works in every vector space on any field K.''
$endgroup$
– Schüler
Feb 9 at 8:56
$begingroup$
Thank you very much for your explanation.
$endgroup$
– Schüler
Feb 9 at 9:07
add a comment |
$begingroup$
Let $T=begin{pmatrix}a&b\c&dend{pmatrix}$ be a non-scalar matrix.
If $S=begin{pmatrix}e&f\g&hend{pmatrix}$ be such that $TS=ST$. Why there exists $alpha,betain mathbb{C}$ such that
$$S=alpha T+beta I;?$$
Note that $TS-ST=0$ is equivalent to
$$begin{bmatrix}bg-fc & af+bh-eb-fd\
ce+dg-ga-hc & fc-bgend{bmatrix} = begin{bmatrix}0 & 0\0 & 0end{bmatrix}$$
This implies that
$$begin{cases}
bg-fc = 0,\
af+bh-eb-fd = 0,\
ce+dg-ga-hc = 0,\
fc-bg = 0.
end{cases}$$
Since $T$ is non scalar, then $bneq 0$ or $cneq 0$ or $aneq d$. However, I cannot find $alpha$ and $beta$.
linear-algebra
asked Feb 3 at 13:05
SchülerSchüler
1,5591421
$endgroup$
Let $T=begin{pmatrix}a&b\c&dend{pmatrix}$ be a non-scalar matrix.
If $S=begin{pmatrix}e&f\g&hend{pmatrix}$ be such that $TS=ST$. Why there exists $alpha,betain mathbb{C}$ such that
$$S=alpha T+beta I;?$$
Note that $TS-ST=0$ is equivalent to
$$begin{bmatrix}bg-fc & af+bh-eb-fd\
ce+dg-ga-hc & fc-bgend{bmatrix} = begin{bmatrix}0 & 0\0 & 0end{bmatrix}$$
This implies that
$$begin{cases}
bg-fc = 0,\
af+bh-eb-fd = 0,\
ce+dg-ga-hc = 0,\
fc-bg = 0.
end{cases}$$
Since $T$ is non scalar, then $bneq 0$ or $cneq 0$ or $aneq d$. However, I cannot find $alpha$ and $beta$.
linear-algebra
linear-algebra
asked Feb 3 at 13:05
SchülerSchüler
1,5591421
asked Feb 3 at 13:05
SchülerSchüler
1,5591421
edited Feb 3 at 13:15
Schüler
asked Feb 3 at 13:05
SchülerSchüler
1,5591421
asked Feb 3 at 13:05
SchülerSchüler
1,5591421
asked Feb 3 at 13:05
SchülerSchüler
1,5591421
1,5591421
$begingroup$
Dear professor. According to Lemaa 2.3 this result works only when $text{dim}(H)=2$ (please see books.google.tn/… ) I find contradiction between this result and the provided answer. Thank you for the explanation.
$endgroup$
– Schüler
Feb 9 at 7:40
$begingroup$
I'm Ok. The problem is his comment he says:''The proof works in every vector space on any field K.''
$endgroup$
– Schüler
Feb 9 at 8:56
$begingroup$
Thank you very much for your explanation.
$endgroup$
– Schüler
Feb 9 at 9:07
add a comment |
$begingroup$
Dear professor. According to Lemaa 2.3 this result works only when $text{dim}(H)=2$ (please see books.google.tn/… ) I find contradiction between this result and the provided answer. Thank you for the explanation.
$endgroup$
– Schüler
Feb 9 at 7:40
$begingroup$
I'm Ok. The problem is his comment he says:''The proof works in every vector space on any field K.''
$endgroup$
– Schüler
Feb 9 at 8:56
$begingroup$
Thank you very much for your explanation.
$endgroup$
– Schüler
Feb 9 at 9:07
$begingroup$
Dear professor. According to Lemaa 2.3 this result works only when $text{dim}(H)=2$ (please see books.google.tn/… ) I find contradiction between this result and the provided answer. Thank you for the explanation.
$endgroup$
– Schüler
Feb 9 at 7:40
$begingroup$
Dear professor. According to Lemaa 2.3 this result works only when $text{dim}(H)=2$ (please see books.google.tn/… ) I find contradiction between this result and the provided answer. Thank you for the explanation.
$endgroup$
– Schüler
Feb 9 at 7:40
$begingroup$
I'm Ok. The problem is his comment he says:''The proof works in every vector space on any field K.''
$endgroup$
– Schüler
Feb 9 at 8:56
$begingroup$
I'm Ok. The problem is his comment he says:''The proof works in every vector space on any field K.''
$endgroup$
– Schüler
Feb 9 at 8:56
$begingroup$
Thank you very much for your explanation.
$endgroup$
– Schüler
Feb 9 at 9:07
$begingroup$
Thank you very much for your explanation.
$endgroup$
– Schüler
Feb 9 at 9:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
EDIT. We assume that $Tin M_2(K)$, where $K$ is
a field. $C(T)={Sin M_2(K);TS=ST}$ is a vector space containing ${I,T}$, that are linearly independent; then it suffices to show that $dim(C(A))leq 2$, that is, the entries of such a matrix $S=begin{pmatrix}p&q\r&send{pmatrix}$ depend at most on $2$ parameters.
$textbf{Proof}$. There is a vector $u$ s.t. ${u,Tu}$ is a basis of $K^2$; otherwise, let $v,w$ be a basis of $K^2$; one has $Tv=av,Tw=bw,T(v+w)=c(v+w)$, that implies that $T=aI$, a contradiction.
In the basis ${u,Tu}$, $T$ becomes $begin{pmatrix}0&a\1&bend{pmatrix}$ and
$(TS)_{1,1}=(ST)_{1,1},(TS)_{2,1}=(ST)_{2,1}$ iff
$q=ar,p=-br+s$; finally, the entries of $S$ depend at most on the $2$ parameters $r,s$ and we are done.
answered Feb 3 at 15:59
loup blancloup blanc
24k21852
$endgroup$
$begingroup$
Thank you for your answer. I think you don't proceed as my attempt. Please why $T(Tu))=au+bTu$?
$endgroup$
– Schüler
Feb 4 at 16:06
$begingroup$
Please have you seen my question ? Thanks
$endgroup$
– Schüler
Feb 5 at 8:20
$begingroup$
I just read it. $a,b$ exist because $u,Tu$ is a basis. Note also that the proof works on every field $K$ and not only over $mathbb{C}$.
$endgroup$
– loup blanc
Feb 5 at 8:31
$begingroup$
I think that you used $a$ and $b$ in my question. I think also that the proof work in an arbitrary Hilbert space such that its dimension is 2. Do you agree with me?
$endgroup$
– Schüler
Feb 5 at 14:17
$begingroup$
The proof works in every vector space on any field $K$. In particular, $K$ does not need to contain the eigenvalues of $A$. Do you understand why $u$ exists ?
$endgroup$
– loup blanc
Feb 6 at 15:13
|
show 3 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
EDIT. We assume that $Tin M_2(K)$, where $K$ is
a field. $C(T)={Sin M_2(K);TS=ST}$ is a vector space containing ${I,T}$, that are linearly independent; then it suffices to show that $dim(C(A))leq 2$, that is, the entries of such a matrix $S=begin{pmatrix}p&q\r&send{pmatrix}$ depend at most on $2$ parameters.
$textbf{Proof}$. There is a vector $u$ s.t. ${u,Tu}$ is a basis of $K^2$; otherwise, let $v,w$ be a basis of $K^2$; one has $Tv=av,Tw=bw,T(v+w)=c(v+w)$, that implies that $T=aI$, a contradiction.
In the basis ${u,Tu}$, $T$ becomes $begin{pmatrix}0&a\1&bend{pmatrix}$ and
$(TS)_{1,1}=(ST)_{1,1},(TS)_{2,1}=(ST)_{2,1}$ iff
$q=ar,p=-br+s$; finally, the entries of $S$ depend at most on the $2$ parameters $r,s$ and we are done.
answered Feb 3 at 15:59
loup blancloup blanc
24k21852
$endgroup$
$begingroup$
Thank you for your answer. I think you don't proceed as my attempt. Please why $T(Tu))=au+bTu$?
$endgroup$
– Schüler
Feb 4 at 16:06
$begingroup$
Please have you seen my question ? Thanks
$endgroup$
– Schüler
Feb 5 at 8:20
$begingroup$
I just read it. $a,b$ exist because $u,Tu$ is a basis. Note also that the proof works on every field $K$ and not only over $mathbb{C}$.
$endgroup$
– loup blanc
Feb 5 at 8:31
$begingroup$
I think that you used $a$ and $b$ in my question. I think also that the proof work in an arbitrary Hilbert space such that its dimension is 2. Do you agree with me?
$endgroup$
– Schüler
Feb 5 at 14:17
$begingroup$
The proof works in every vector space on any field $K$. In particular, $K$ does not need to contain the eigenvalues of $A$. Do you understand why $u$ exists ?
$endgroup$
– loup blanc
Feb 6 at 15:13
|
show 3 more comments
$begingroup$
EDIT. We assume that $Tin M_2(K)$, where $K$ is
a field. $C(T)={Sin M_2(K);TS=ST}$ is a vector space containing ${I,T}$, that are linearly independent; then it suffices to show that $dim(C(A))leq 2$, that is, the entries of such a matrix $S=begin{pmatrix}p&q\r&send{pmatrix}$ depend at most on $2$ parameters.
$textbf{Proof}$. There is a vector $u$ s.t. ${u,Tu}$ is a basis of $K^2$; otherwise, let $v,w$ be a basis of $K^2$; one has $Tv=av,Tw=bw,T(v+w)=c(v+w)$, that implies that $T=aI$, a contradiction.
In the basis ${u,Tu}$, $T$ becomes $begin{pmatrix}0&a\1&bend{pmatrix}$ and
$(TS)_{1,1}=(ST)_{1,1},(TS)_{2,1}=(ST)_{2,1}$ iff
$q=ar,p=-br+s$; finally, the entries of $S$ depend at most on the $2$ parameters $r,s$ and we are done.
answered Feb 3 at 15:59
loup blancloup blanc
24k21852
$endgroup$
$begingroup$
Thank you for your answer. I think you don't proceed as my attempt. Please why $T(Tu))=au+bTu$?
$endgroup$
– Schüler
Feb 4 at 16:06
$begingroup$
Please have you seen my question ? Thanks
$endgroup$
– Schüler
Feb 5 at 8:20
$begingroup$
I just read it. $a,b$ exist because $u,Tu$ is a basis. Note also that the proof works on every field $K$ and not only over $mathbb{C}$.
$endgroup$
– loup blanc
Feb 5 at 8:31
$begingroup$
I think that you used $a$ and $b$ in my question. I think also that the proof work in an arbitrary Hilbert space such that its dimension is 2. Do you agree with me?
$endgroup$
– Schüler
Feb 5 at 14:17
$begingroup$
The proof works in every vector space on any field $K$. In particular, $K$ does not need to contain the eigenvalues of $A$. Do you understand why $u$ exists ?
$endgroup$
– loup blanc
Feb 6 at 15:13
|
show 3 more comments
$begingroup$
EDIT. We assume that $Tin M_2(K)$, where $K$ is
a field. $C(T)={Sin M_2(K);TS=ST}$ is a vector space containing ${I,T}$, that are linearly independent; then it suffices to show that $dim(C(A))leq 2$, that is, the entries of such a matrix $S=begin{pmatrix}p&q\r&send{pmatrix}$ depend at most on $2$ parameters.
$textbf{Proof}$. There is a vector $u$ s.t. ${u,Tu}$ is a basis of $K^2$; otherwise, let $v,w$ be a basis of $K^2$; one has $Tv=av,Tw=bw,T(v+w)=c(v+w)$, that implies that $T=aI$, a contradiction.
In the basis ${u,Tu}$, $T$ becomes $begin{pmatrix}0&a\1&bend{pmatrix}$ and
$(TS)_{1,1}=(ST)_{1,1},(TS)_{2,1}=(ST)_{2,1}$ iff
$q=ar,p=-br+s$; finally, the entries of $S$ depend at most on the $2$ parameters $r,s$ and we are done.
answered Feb 3 at 15:59
loup blancloup blanc
24k21852
$endgroup$
EDIT. We assume that $Tin M_2(K)$, where $K$ is
a field. $C(T)={Sin M_2(K);TS=ST}$ is a vector space containing ${I,T}$, that are linearly independent; then it suffices to show that $dim(C(A))leq 2$, that is, the entries of such a matrix $S=begin{pmatrix}p&q\r&send{pmatrix}$ depend at most on $2$ parameters.
$textbf{Proof}$. There is a vector $u$ s.t. ${u,Tu}$ is a basis of $K^2$; otherwise, let $v,w$ be a basis of $K^2$; one has $Tv=av,Tw=bw,T(v+w)=c(v+w)$, that implies that $T=aI$, a contradiction.
In the basis ${u,Tu}$, $T$ becomes $begin{pmatrix}0&a\1&bend{pmatrix}$ and
$(TS)_{1,1}=(ST)_{1,1},(TS)_{2,1}=(ST)_{2,1}$ iff
$q=ar,p=-br+s$; finally, the entries of $S$ depend at most on the $2$ parameters $r,s$ and we are done.
answered Feb 3 at 15:59
loup blancloup blanc
24k21852
edited Feb 9 at 15:26
answered Feb 3 at 15:59
loup blancloup blanc
24k21852
answered Feb 3 at 15:59
loup blancloup blanc
24k21852
answered Feb 3 at 15:59
loup blancloup blanc
24k21852
24k21852
$begingroup$
Thank you for your answer. I think you don't proceed as my attempt. Please why $T(Tu))=au+bTu$?
$endgroup$
– Schüler
Feb 4 at 16:06
$begingroup$
Please have you seen my question ? Thanks
$endgroup$
– Schüler
Feb 5 at 8:20
$begingroup$
I just read it. $a,b$ exist because $u,Tu$ is a basis. Note also that the proof works on every field $K$ and not only over $mathbb{C}$.
$endgroup$
– loup blanc
Feb 5 at 8:31
$begingroup$
I think that you used $a$ and $b$ in my question. I think also that the proof work in an arbitrary Hilbert space such that its dimension is 2. Do you agree with me?
$endgroup$
– Schüler
Feb 5 at 14:17
$begingroup$
The proof works in every vector space on any field $K$. In particular, $K$ does not need to contain the eigenvalues of $A$. Do you understand why $u$ exists ?
$endgroup$
– loup blanc
Feb 6 at 15:13
|
show 3 more comments
$begingroup$
Thank you for your answer. I think you don't proceed as my attempt. Please why $T(Tu))=au+bTu$?
$endgroup$
– Schüler
Feb 4 at 16:06
$begingroup$
Please have you seen my question ? Thanks
$endgroup$
– Schüler
Feb 5 at 8:20
$begingroup$
I just read it. $a,b$ exist because $u,Tu$ is a basis. Note also that the proof works on every field $K$ and not only over $mathbb{C}$.
$endgroup$
– loup blanc
Feb 5 at 8:31
$begingroup$
I think that you used $a$ and $b$ in my question. I think also that the proof work in an arbitrary Hilbert space such that its dimension is 2. Do you agree with me?
$endgroup$
– Schüler
Feb 5 at 14:17
$begingroup$
The proof works in every vector space on any field $K$. In particular, $K$ does not need to contain the eigenvalues of $A$. Do you understand why $u$ exists ?
$endgroup$
– loup blanc
Feb 6 at 15:13
$begingroup$
Thank you for your answer. I think you don't proceed as my attempt. Please why $T(Tu))=au+bTu$?
$endgroup$
– Schüler
Feb 4 at 16:06
$begingroup$
Thank you for your answer. I think you don't proceed as my attempt. Please why $T(Tu))=au+bTu$?
$endgroup$
– Schüler
Feb 4 at 16:06
$begingroup$
Please have you seen my question ? Thanks
$endgroup$
– Schüler
Feb 5 at 8:20
$begingroup$
Please have you seen my question ? Thanks
$endgroup$
– Schüler
Feb 5 at 8:20
$begingroup$
I just read it. $a,b$ exist because $u,Tu$ is a basis. Note also that the proof works on every field $K$ and not only over $mathbb{C}$.
$endgroup$
– loup blanc
Feb 5 at 8:31
$begingroup$
I just read it. $a,b$ exist because $u,Tu$ is a basis. Note also that the proof works on every field $K$ and not only over $mathbb{C}$.
$endgroup$
– loup blanc
Feb 5 at 8:31
$begingroup$
I think that you used $a$ and $b$ in my question. I think also that the proof work in an arbitrary Hilbert space such that its dimension is 2. Do you agree with me?
$endgroup$
– Schüler
Feb 5 at 14:17
$begingroup$
I think that you used $a$ and $b$ in my question. I think also that the proof work in an arbitrary Hilbert space such that its dimension is 2. Do you agree with me?
$endgroup$
– Schüler
Feb 5 at 14:17
$begingroup$
The proof works in every vector space on any field $K$. In particular, $K$ does not need to contain the eigenvalues of $A$. Do you understand why $u$ exists ?
$endgroup$
– loup blanc
Feb 6 at 15:13
$begingroup$
The proof works in every vector space on any field $K$. In particular, $K$ does not need to contain the eigenvalues of $A$. Do you understand why $u$ exists ?
$endgroup$
– loup blanc
Feb 6 at 15:13
|
show 3 more comments
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$begingroup$
Dear professor. According to Lemaa 2.3 this result works only when $text{dim}(H)=2$ (please see books.google.tn/… ) I find contradiction between this result and the provided answer. Thank you for the explanation.
$endgroup$
– Schüler
Feb 9 at 7:40
$begingroup$
I'm Ok. The problem is his comment he says:''The proof works in every vector space on any field K.''
$endgroup$
– Schüler
Feb 9 at 8:56
$begingroup$
Thank you very much for your explanation.
$endgroup$
– Schüler
Feb 9 at 9:07