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If $AB=BA$, prove that $ A=beginbmatrix a & 0 \ 0 & a endbmatrix $
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Let $A$ a $2x2$ matrix, if $AB=BA$ for every $B$ of the size $2x2$, Prove that:
$$
A=beginbmatrix a & 0 \ 0 & a endbmatrix
$$
$a in mathbbR$
My attempt:
Let $$
A=beginbmatrix a_1 & b_1 \ c_1 & d_1 endbmatrix
$$
$$B=beginbmatrix a_2 & b_2 \ c_2 & d_2 endbmatrix$$
And since $AB=BA$, then
$a_1 a_2 + b_1 c_2 = a_1a_2+b_2 c_1$
So $b_2 c_1=b_1 c_2$
And
$a_1 b_2+b_1 d_2=a_2 b_1+b_2 d_1$
$c_1 a_2+d_1 c_2=c_2 a_1+d_2 c_1$
But what can I do now ?
Thanks :)
linear-algebra
asked Jul 19 at 10:58
Dima
438214
add a comment |Â
up vote
7
down vote
favorite
Let $A$ a $2x2$ matrix, if $AB=BA$ for every $B$ of the size $2x2$, Prove that:
$$
A=beginbmatrix a & 0 \ 0 & a endbmatrix
$$
$a in mathbbR$
My attempt:
Let $$
A=beginbmatrix a_1 & b_1 \ c_1 & d_1 endbmatrix
$$
$$B=beginbmatrix a_2 & b_2 \ c_2 & d_2 endbmatrix$$
And since $AB=BA$, then
$a_1 a_2 + b_1 c_2 = a_1a_2+b_2 c_1$
So $b_2 c_1=b_1 c_2$
And
$a_1 b_2+b_1 d_2=a_2 b_1+b_2 d_1$
$c_1 a_2+d_1 c_2=c_2 a_1+d_2 c_1$
But what can I do now ?
Thanks :)
linear-algebra
asked Jul 19 at 10:58
Dima
438214
This works for $ntimes n$ matrices in general. In fact, if $D$ is a division ring, then the center of $M_n(D)$ consists of the scalar matrices $dcdot I_n$, where $I_n$ is the identity matrix and $d$ is an element of the center of $D$.
– A. Pongrácz
Jul 19 at 12:55
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Let $A$ a $2x2$ matrix, if $AB=BA$ for every $B$ of the size $2x2$, Prove that:
$$
A=beginbmatrix a & 0 \ 0 & a endbmatrix
$$
$a in mathbbR$
My attempt:
Let $$
A=beginbmatrix a_1 & b_1 \ c_1 & d_1 endbmatrix
$$
$$B=beginbmatrix a_2 & b_2 \ c_2 & d_2 endbmatrix$$
And since $AB=BA$, then
$a_1 a_2 + b_1 c_2 = a_1a_2+b_2 c_1$
So $b_2 c_1=b_1 c_2$
And
$a_1 b_2+b_1 d_2=a_2 b_1+b_2 d_1$
$c_1 a_2+d_1 c_2=c_2 a_1+d_2 c_1$
But what can I do now ?
Thanks :)
linear-algebra
asked Jul 19 at 10:58
Dima
438214
Let $A$ a $2x2$ matrix, if $AB=BA$ for every $B$ of the size $2x2$, Prove that:
$$
A=beginbmatrix a & 0 \ 0 & a endbmatrix
$$
$a in mathbbR$
My attempt:
Let $$
A=beginbmatrix a_1 & b_1 \ c_1 & d_1 endbmatrix
$$
$$B=beginbmatrix a_2 & b_2 \ c_2 & d_2 endbmatrix$$
And since $AB=BA$, then
$a_1 a_2 + b_1 c_2 = a_1a_2+b_2 c_1$
So $b_2 c_1=b_1 c_2$
And
$a_1 b_2+b_1 d_2=a_2 b_1+b_2 d_1$
$c_1 a_2+d_1 c_2=c_2 a_1+d_2 c_1$
But what can I do now ?
Thanks :)
linear-algebra
asked Jul 19 at 10:58
Dima
438214
asked Jul 19 at 10:58
Dima
438214
asked Jul 19 at 10:58
Dima
438214
asked Jul 19 at 10:58
Dima
438214
438214
This works for $ntimes n$ matrices in general. In fact, if $D$ is a division ring, then the center of $M_n(D)$ consists of the scalar matrices $dcdot I_n$, where $I_n$ is the identity matrix and $d$ is an element of the center of $D$.
– A. Pongrácz
Jul 19 at 12:55
add a comment |Â
This works for $ntimes n$ matrices in general. In fact, if $D$ is a division ring, then the center of $M_n(D)$ consists of the scalar matrices $dcdot I_n$, where $I_n$ is the identity matrix and $d$ is an element of the center of $D$.
– A. Pongrácz
Jul 19 at 12:55
This works for $ntimes n$ matrices in general. In fact, if $D$ is a division ring, then the center of $M_n(D)$ consists of the scalar matrices $dcdot I_n$, where $I_n$ is the identity matrix and $d$ is an element of the center of $D$.
– A. Pongrácz
Jul 19 at 12:55
This works for $ntimes n$ matrices in general. In fact, if $D$ is a division ring, then the center of $M_n(D)$ consists of the scalar matrices $dcdot I_n$, where $I_n$ is the identity matrix and $d$ is an element of the center of $D$.
– A. Pongrácz
Jul 19 at 12:55
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
17
down vote
accepted
Let $A=pmatrixa&b\c&d$. For $B=pmatrix1&0\0&0$ we have
$AB=pmatrixa&0\c&0$ and $BA=pmatrixa&b\0&0$. So $AB=BA$
implies $b=c=0$, that is $A=pmatrixa&0\0&d$. Now try, say $B=pmatrix0&1\0&0$.
answered Jul 19 at 11:04
Lord Shark the Unknown
85.5k951112
2
Thank you so much,when I take $B=pmatrix{0&1\0&0 I got $a=d$, then we proved it :)
– Dima
Jul 19 at 11:41
add a comment |Â
up vote
3
down vote
Note:
$$beginpmatrix a_11 & a_12 \ a_21 & a_22 endpmatrix
beginpmatrix b_11 & b_12 \ b_21 & b_22 endpmatrix=
beginpmatrix b_11 & b_12 \ b_21 & b_22 endpmatrix
beginpmatrix a_11 & a_12 \ a_21 & a_22 endpmatrix Rightarrow \
begincasesrequirecancelcancela_11b_11+a_12b_21=cancelb_11a_11+b_12a_21\
a_11b_12+a_12b_22=b_11a_12+b_12a_22\
a_21b_11+a_22b_21=b_21a_11+b_22a_21\
a_21b_12+cancela_22b_22=b_21a_12+cancelb_22a_22endcases$$
From $(1)$, since $b_12$ and $b_21$ can be any number, in particular, $b_12=0$ and $b_21ne 0$, we get: $a_12=0$.
Similarly, for $b_12ne 0$ and $b_21=0$, we get $a_21=0$.
From $(2)$, since $a_12=0$ and $b_12$ is an arbitrary number, we get $a_11b_12=b_12a_22 Rightarrow a_11=a_22$.
answered Jul 19 at 11:50
farruhota
13.7k2632
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
17
down vote
accepted
Let $A=pmatrixa&b\c&d$. For $B=pmatrix1&0\0&0$ we have
$AB=pmatrixa&0\c&0$ and $BA=pmatrixa&b\0&0$. So $AB=BA$
implies $b=c=0$, that is $A=pmatrixa&0\0&d$. Now try, say $B=pmatrix0&1\0&0$.
answered Jul 19 at 11:04
Lord Shark the Unknown
85.5k951112
2
Thank you so much,when I take $B=pmatrix{0&1\0&0 I got $a=d$, then we proved it :)
– Dima
Jul 19 at 11:41
add a comment |Â
up vote
17
down vote
accepted
Let $A=pmatrixa&b\c&d$. For $B=pmatrix1&0\0&0$ we have
$AB=pmatrixa&0\c&0$ and $BA=pmatrixa&b\0&0$. So $AB=BA$
implies $b=c=0$, that is $A=pmatrixa&0\0&d$. Now try, say $B=pmatrix0&1\0&0$.
answered Jul 19 at 11:04
Lord Shark the Unknown
85.5k951112
2
Thank you so much,when I take $B=pmatrix{0&1\0&0 I got $a=d$, then we proved it :)
– Dima
Jul 19 at 11:41
add a comment |Â
up vote
17
down vote
accepted
up vote
17
down vote
accepted
Let $A=pmatrixa&b\c&d$. For $B=pmatrix1&0\0&0$ we have
$AB=pmatrixa&0\c&0$ and $BA=pmatrixa&b\0&0$. So $AB=BA$
implies $b=c=0$, that is $A=pmatrixa&0\0&d$. Now try, say $B=pmatrix0&1\0&0$.
answered Jul 19 at 11:04
Lord Shark the Unknown
85.5k951112
Let $A=pmatrixa&b\c&d$. For $B=pmatrix1&0\0&0$ we have
$AB=pmatrixa&0\c&0$ and $BA=pmatrixa&b\0&0$. So $AB=BA$
implies $b=c=0$, that is $A=pmatrixa&0\0&d$. Now try, say $B=pmatrix0&1\0&0$.
answered Jul 19 at 11:04
Lord Shark the Unknown
85.5k951112
answered Jul 19 at 11:04
Lord Shark the Unknown
85.5k951112
answered Jul 19 at 11:04
Lord Shark the Unknown
85.5k951112
answered Jul 19 at 11:04
Lord Shark the Unknown
85.5k951112
85.5k951112
2
Thank you so much,when I take $B=pmatrix{0&1\0&0 I got $a=d$, then we proved it :)
– Dima
Jul 19 at 11:41
add a comment |Â
2
Thank you so much,when I take $B=pmatrix{0&1\0&0 I got $a=d$, then we proved it :)
– Dima
Jul 19 at 11:41
2
2
Thank you so much,when I take $B=pmatrix{0&1\0&0 I got $a=d$, then we proved it :)
– Dima
Jul 19 at 11:41
Thank you so much,when I take $B=pmatrix{0&1\0&0 I got $a=d$, then we proved it :)
– Dima
Jul 19 at 11:41
add a comment |Â
up vote
3
down vote
Note:
$$beginpmatrix a_11 & a_12 \ a_21 & a_22 endpmatrix
beginpmatrix b_11 & b_12 \ b_21 & b_22 endpmatrix=
beginpmatrix b_11 & b_12 \ b_21 & b_22 endpmatrix
beginpmatrix a_11 & a_12 \ a_21 & a_22 endpmatrix Rightarrow \
begincasesrequirecancelcancela_11b_11+a_12b_21=cancelb_11a_11+b_12a_21\
a_11b_12+a_12b_22=b_11a_12+b_12a_22\
a_21b_11+a_22b_21=b_21a_11+b_22a_21\
a_21b_12+cancela_22b_22=b_21a_12+cancelb_22a_22endcases$$
From $(1)$, since $b_12$ and $b_21$ can be any number, in particular, $b_12=0$ and $b_21ne 0$, we get: $a_12=0$.
Similarly, for $b_12ne 0$ and $b_21=0$, we get $a_21=0$.
From $(2)$, since $a_12=0$ and $b_12$ is an arbitrary number, we get $a_11b_12=b_12a_22 Rightarrow a_11=a_22$.
answered Jul 19 at 11:50
farruhota
13.7k2632
add a comment |Â
up vote
3
down vote
Note:
$$beginpmatrix a_11 & a_12 \ a_21 & a_22 endpmatrix
beginpmatrix b_11 & b_12 \ b_21 & b_22 endpmatrix=
beginpmatrix b_11 & b_12 \ b_21 & b_22 endpmatrix
beginpmatrix a_11 & a_12 \ a_21 & a_22 endpmatrix Rightarrow \
begincasesrequirecancelcancela_11b_11+a_12b_21=cancelb_11a_11+b_12a_21\
a_11b_12+a_12b_22=b_11a_12+b_12a_22\
a_21b_11+a_22b_21=b_21a_11+b_22a_21\
a_21b_12+cancela_22b_22=b_21a_12+cancelb_22a_22endcases$$
From $(1)$, since $b_12$ and $b_21$ can be any number, in particular, $b_12=0$ and $b_21ne 0$, we get: $a_12=0$.
Similarly, for $b_12ne 0$ and $b_21=0$, we get $a_21=0$.
From $(2)$, since $a_12=0$ and $b_12$ is an arbitrary number, we get $a_11b_12=b_12a_22 Rightarrow a_11=a_22$.
answered Jul 19 at 11:50
farruhota
13.7k2632
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Note:
$$beginpmatrix a_11 & a_12 \ a_21 & a_22 endpmatrix
beginpmatrix b_11 & b_12 \ b_21 & b_22 endpmatrix=
beginpmatrix b_11 & b_12 \ b_21 & b_22 endpmatrix
beginpmatrix a_11 & a_12 \ a_21 & a_22 endpmatrix Rightarrow \
begincasesrequirecancelcancela_11b_11+a_12b_21=cancelb_11a_11+b_12a_21\
a_11b_12+a_12b_22=b_11a_12+b_12a_22\
a_21b_11+a_22b_21=b_21a_11+b_22a_21\
a_21b_12+cancela_22b_22=b_21a_12+cancelb_22a_22endcases$$
From $(1)$, since $b_12$ and $b_21$ can be any number, in particular, $b_12=0$ and $b_21ne 0$, we get: $a_12=0$.
Similarly, for $b_12ne 0$ and $b_21=0$, we get $a_21=0$.
From $(2)$, since $a_12=0$ and $b_12$ is an arbitrary number, we get $a_11b_12=b_12a_22 Rightarrow a_11=a_22$.
answered Jul 19 at 11:50
farruhota
13.7k2632
Note:
$$beginpmatrix a_11 & a_12 \ a_21 & a_22 endpmatrix
beginpmatrix b_11 & b_12 \ b_21 & b_22 endpmatrix=
beginpmatrix b_11 & b_12 \ b_21 & b_22 endpmatrix
beginpmatrix a_11 & a_12 \ a_21 & a_22 endpmatrix Rightarrow \
begincasesrequirecancelcancela_11b_11+a_12b_21=cancelb_11a_11+b_12a_21\
a_11b_12+a_12b_22=b_11a_12+b_12a_22\
a_21b_11+a_22b_21=b_21a_11+b_22a_21\
a_21b_12+cancela_22b_22=b_21a_12+cancelb_22a_22endcases$$
From $(1)$, since $b_12$ and $b_21$ can be any number, in particular, $b_12=0$ and $b_21ne 0$, we get: $a_12=0$.
Similarly, for $b_12ne 0$ and $b_21=0$, we get $a_21=0$.
From $(2)$, since $a_12=0$ and $b_12$ is an arbitrary number, we get $a_11b_12=b_12a_22 Rightarrow a_11=a_22$.
answered Jul 19 at 11:50
farruhota
13.7k2632
answered Jul 19 at 11:50
farruhota
13.7k2632
answered Jul 19 at 11:50
farruhota
13.7k2632
answered Jul 19 at 11:50
farruhota
13.7k2632
13.7k2632
add a comment |Â
add a comment |Â
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This works for $ntimes n$ matrices in general. In fact, if $D$ is a division ring, then the center of $M_n(D)$ consists of the scalar matrices $dcdot I_n$, where $I_n$ is the identity matrix and $d$ is an element of the center of $D$.
– A. Pongrácz
Jul 19 at 12:55