Find base of subspace $U=Ain M_3:S^-1AS=D$
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Let $S in M_3(mathbb R)$ is some invertible matrices, find a base and dimension of subspace $U$, where $U$ is set of all matrices $Ain U$ for which $S^-1AS$ is some diagonal matrix.
I only know one base and that is Identical matrix since $S^-1IS=S^-1S=I$ and that is diagonal matrix, is there any other base?
linear-algebra matrices dimension-theory
asked Sep 8 at 10:06
Marko Škorić
4008
add a comment |Â
up vote
2
down vote
favorite
Let $S in M_3(mathbb R)$ is some invertible matrices, find a base and dimension of subspace $U$, where $U$ is set of all matrices $Ain U$ for which $S^-1AS$ is some diagonal matrix.
I only know one base and that is Identical matrix since $S^-1IS=S^-1S=I$ and that is diagonal matrix, is there any other base?
linear-algebra matrices dimension-theory
asked Sep 8 at 10:06
Marko Škorić
4008
Am I understanding this right, that $S$ is fixed for $U$?
– zzuussee
Sep 8 at 10:12
yes you are right
– Marko Å korić
Sep 8 at 10:14
1
According to your definition, $U$ is a singleton, it contains exactly one element, namely $SDS^-1$.
– Ewan Delanoy
Sep 8 at 10:55
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $S in M_3(mathbb R)$ is some invertible matrices, find a base and dimension of subspace $U$, where $U$ is set of all matrices $Ain U$ for which $S^-1AS$ is some diagonal matrix.
I only know one base and that is Identical matrix since $S^-1IS=S^-1S=I$ and that is diagonal matrix, is there any other base?
linear-algebra matrices dimension-theory
asked Sep 8 at 10:06
Marko Škorić
4008
Let $S in M_3(mathbb R)$ is some invertible matrices, find a base and dimension of subspace $U$, where $U$ is set of all matrices $Ain U$ for which $S^-1AS$ is some diagonal matrix.
I only know one base and that is Identical matrix since $S^-1IS=S^-1S=I$ and that is diagonal matrix, is there any other base?
linear-algebra matrices dimension-theory
linear-algebra matrices dimension-theory
asked Sep 8 at 10:06
Marko Škorić
4008
asked Sep 8 at 10:06
Marko Škorić
4008
asked Sep 8 at 10:06
Marko Škorić
4008
asked Sep 8 at 10:06
Marko Škorić
4008
asked Sep 8 at 10:06
Marko Škorić
4008
4008
Am I understanding this right, that $S$ is fixed for $U$?
– zzuussee
Sep 8 at 10:12
yes you are right
– Marko Å korić
Sep 8 at 10:14
1
According to your definition, $U$ is a singleton, it contains exactly one element, namely $SDS^-1$.
– Ewan Delanoy
Sep 8 at 10:55
add a comment |Â
Am I understanding this right, that $S$ is fixed for $U$?
– zzuussee
Sep 8 at 10:12
yes you are right
– Marko Å korić
Sep 8 at 10:14
1
According to your definition, $U$ is a singleton, it contains exactly one element, namely $SDS^-1$.
– Ewan Delanoy
Sep 8 at 10:55
Am I understanding this right, that $S$ is fixed for $U$?
– zzuussee
Sep 8 at 10:12
Am I understanding this right, that $S$ is fixed for $U$?
– zzuussee
Sep 8 at 10:12
yes you are right
– Marko Å korić
Sep 8 at 10:14
yes you are right
– Marko Å korić
Sep 8 at 10:14
1
1
According to your definition, $U$ is a singleton, it contains exactly one element, namely $SDS^-1$.
– Ewan Delanoy
Sep 8 at 10:55
According to your definition, $U$ is a singleton, it contains exactly one element, namely $SDS^-1$.
– Ewan Delanoy
Sep 8 at 10:55
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Let $S in M_n(mathbbR)$ be a fixed invertible matrix. The definition is
$$U = A in M_n(mathbbR) : S^-1AS text is a diagonal matrix$$
Let $A in U$ such that $S^-1AS = D = operatornamediag(lambda_1,ldots, lambda_n)$. We have
$$A = S^-1DS = S^-1(lambda_1E_11 + cdots + lambda_n E_nn)S = lambda_1 S^-1E_11S + cdots+ lambda_n S^-1E_nnS$$
where $(E_rs)_ij = delta_irdelta_js$. Therefore
$$U = operatornamespanS^-1E_11S, ldots, S^-1E_nnS$$
Furthermore, the set $S^-1E_11S, ldots, S^-1E_nnS$ is linearly independent as an image of a linearly independent set $E_11, ldots, E_nn$ by the invertible linear map $X mapsto S^-1XS$.
Hence $S^-1E_11S, ldots, S^-1E_nnS$ is a basis for $U$.
answered Sep 8 at 11:59
mechanodroid
24.6k62245
lets spectrum of $Ain Mn(mathbb R)$ is $lambda1,lambda2,cdots,lambda_n$ and have n element and if I have $W=Min Mn: AM=MD$ can you help me to find a base, I try on the same way but it is not working, i try that $A=SDS^-1$, but even that can not help me, do you have some idea?
– Marko Å korić
Sep 16 at 15:35
@MarkoŠkorić It's not working because this is a different problem. What is $D$, a fixed diagonal matrix?
– mechanodroid
Sep 16 at 15:54
soryy i forgot to write yes it is diagonal matrix which have diagonal entries $lambda1,lambda 2...,lambda n$
– Marko Å korić
Sep 16 at 16:01
1
@MarkoŠkorić Have a look here: math.stackexchange.com/q/2320839/144766
– mechanodroid
Sep 16 at 16:09
thank you so much
– Marko Å korić
Sep 16 at 16:43
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Let $S in M_n(mathbbR)$ be a fixed invertible matrix. The definition is
$$U = A in M_n(mathbbR) : S^-1AS text is a diagonal matrix$$
Let $A in U$ such that $S^-1AS = D = operatornamediag(lambda_1,ldots, lambda_n)$. We have
$$A = S^-1DS = S^-1(lambda_1E_11 + cdots + lambda_n E_nn)S = lambda_1 S^-1E_11S + cdots+ lambda_n S^-1E_nnS$$
where $(E_rs)_ij = delta_irdelta_js$. Therefore
$$U = operatornamespanS^-1E_11S, ldots, S^-1E_nnS$$
Furthermore, the set $S^-1E_11S, ldots, S^-1E_nnS$ is linearly independent as an image of a linearly independent set $E_11, ldots, E_nn$ by the invertible linear map $X mapsto S^-1XS$.
Hence $S^-1E_11S, ldots, S^-1E_nnS$ is a basis for $U$.
answered Sep 8 at 11:59
mechanodroid
24.6k62245
lets spectrum of $Ain Mn(mathbb R)$ is $lambda1,lambda2,cdots,lambda_n$ and have n element and if I have $W=Min Mn: AM=MD$ can you help me to find a base, I try on the same way but it is not working, i try that $A=SDS^-1$, but even that can not help me, do you have some idea?
– Marko Å korić
Sep 16 at 15:35
@MarkoŠkorić It's not working because this is a different problem. What is $D$, a fixed diagonal matrix?
– mechanodroid
Sep 16 at 15:54
soryy i forgot to write yes it is diagonal matrix which have diagonal entries $lambda1,lambda 2...,lambda n$
– Marko Å korić
Sep 16 at 16:01
1
@MarkoŠkorić Have a look here: math.stackexchange.com/q/2320839/144766
– mechanodroid
Sep 16 at 16:09
thank you so much
– Marko Å korić
Sep 16 at 16:43
add a comment |Â
up vote
0
down vote
accepted
Let $S in M_n(mathbbR)$ be a fixed invertible matrix. The definition is
$$U = A in M_n(mathbbR) : S^-1AS text is a diagonal matrix$$
Let $A in U$ such that $S^-1AS = D = operatornamediag(lambda_1,ldots, lambda_n)$. We have
$$A = S^-1DS = S^-1(lambda_1E_11 + cdots + lambda_n E_nn)S = lambda_1 S^-1E_11S + cdots+ lambda_n S^-1E_nnS$$
where $(E_rs)_ij = delta_irdelta_js$. Therefore
$$U = operatornamespanS^-1E_11S, ldots, S^-1E_nnS$$
Furthermore, the set $S^-1E_11S, ldots, S^-1E_nnS$ is linearly independent as an image of a linearly independent set $E_11, ldots, E_nn$ by the invertible linear map $X mapsto S^-1XS$.
Hence $S^-1E_11S, ldots, S^-1E_nnS$ is a basis for $U$.
answered Sep 8 at 11:59
mechanodroid
24.6k62245
lets spectrum of $Ain Mn(mathbb R)$ is $lambda1,lambda2,cdots,lambda_n$ and have n element and if I have $W=Min Mn: AM=MD$ can you help me to find a base, I try on the same way but it is not working, i try that $A=SDS^-1$, but even that can not help me, do you have some idea?
– Marko Å korić
Sep 16 at 15:35
@MarkoŠkorić It's not working because this is a different problem. What is $D$, a fixed diagonal matrix?
– mechanodroid
Sep 16 at 15:54
soryy i forgot to write yes it is diagonal matrix which have diagonal entries $lambda1,lambda 2...,lambda n$
– Marko Å korić
Sep 16 at 16:01
1
@MarkoŠkorić Have a look here: math.stackexchange.com/q/2320839/144766
– mechanodroid
Sep 16 at 16:09
thank you so much
– Marko Å korić
Sep 16 at 16:43
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Let $S in M_n(mathbbR)$ be a fixed invertible matrix. The definition is
$$U = A in M_n(mathbbR) : S^-1AS text is a diagonal matrix$$
Let $A in U$ such that $S^-1AS = D = operatornamediag(lambda_1,ldots, lambda_n)$. We have
$$A = S^-1DS = S^-1(lambda_1E_11 + cdots + lambda_n E_nn)S = lambda_1 S^-1E_11S + cdots+ lambda_n S^-1E_nnS$$
where $(E_rs)_ij = delta_irdelta_js$. Therefore
$$U = operatornamespanS^-1E_11S, ldots, S^-1E_nnS$$
Furthermore, the set $S^-1E_11S, ldots, S^-1E_nnS$ is linearly independent as an image of a linearly independent set $E_11, ldots, E_nn$ by the invertible linear map $X mapsto S^-1XS$.
Hence $S^-1E_11S, ldots, S^-1E_nnS$ is a basis for $U$.
answered Sep 8 at 11:59
mechanodroid
24.6k62245
Let $S in M_n(mathbbR)$ be a fixed invertible matrix. The definition is
$$U = A in M_n(mathbbR) : S^-1AS text is a diagonal matrix$$
Let $A in U$ such that $S^-1AS = D = operatornamediag(lambda_1,ldots, lambda_n)$. We have
$$A = S^-1DS = S^-1(lambda_1E_11 + cdots + lambda_n E_nn)S = lambda_1 S^-1E_11S + cdots+ lambda_n S^-1E_nnS$$
where $(E_rs)_ij = delta_irdelta_js$. Therefore
$$U = operatornamespanS^-1E_11S, ldots, S^-1E_nnS$$
Furthermore, the set $S^-1E_11S, ldots, S^-1E_nnS$ is linearly independent as an image of a linearly independent set $E_11, ldots, E_nn$ by the invertible linear map $X mapsto S^-1XS$.
Hence $S^-1E_11S, ldots, S^-1E_nnS$ is a basis for $U$.
answered Sep 8 at 11:59
mechanodroid
24.6k62245
answered Sep 8 at 11:59
mechanodroid
24.6k62245
answered Sep 8 at 11:59
mechanodroid
24.6k62245
answered Sep 8 at 11:59
mechanodroid
24.6k62245
24.6k62245
lets spectrum of $Ain Mn(mathbb R)$ is $lambda1,lambda2,cdots,lambda_n$ and have n element and if I have $W=Min Mn: AM=MD$ can you help me to find a base, I try on the same way but it is not working, i try that $A=SDS^-1$, but even that can not help me, do you have some idea?
– Marko Å korić
Sep 16 at 15:35
@MarkoŠkorić It's not working because this is a different problem. What is $D$, a fixed diagonal matrix?
– mechanodroid
Sep 16 at 15:54
soryy i forgot to write yes it is diagonal matrix which have diagonal entries $lambda1,lambda 2...,lambda n$
– Marko Å korić
Sep 16 at 16:01
1
@MarkoŠkorić Have a look here: math.stackexchange.com/q/2320839/144766
– mechanodroid
Sep 16 at 16:09
thank you so much
– Marko Å korić
Sep 16 at 16:43
add a comment |Â
lets spectrum of $Ain Mn(mathbb R)$ is $lambda1,lambda2,cdots,lambda_n$ and have n element and if I have $W=Min Mn: AM=MD$ can you help me to find a base, I try on the same way but it is not working, i try that $A=SDS^-1$, but even that can not help me, do you have some idea?
– Marko Å korić
Sep 16 at 15:35
@MarkoŠkorić It's not working because this is a different problem. What is $D$, a fixed diagonal matrix?
– mechanodroid
Sep 16 at 15:54
soryy i forgot to write yes it is diagonal matrix which have diagonal entries $lambda1,lambda 2...,lambda n$
– Marko Å korić
Sep 16 at 16:01
1
@MarkoŠkorić Have a look here: math.stackexchange.com/q/2320839/144766
– mechanodroid
Sep 16 at 16:09
thank you so much
– Marko Å korić
Sep 16 at 16:43
lets spectrum of $Ain Mn(mathbb R)$ is $lambda1,lambda2,cdots,lambda_n$ and have n element and if I have $W=Min Mn: AM=MD$ can you help me to find a base, I try on the same way but it is not working, i try that $A=SDS^-1$, but even that can not help me, do you have some idea?
– Marko Å korić
Sep 16 at 15:35
lets spectrum of $Ain Mn(mathbb R)$ is $lambda1,lambda2,cdots,lambda_n$ and have n element and if I have $W=Min Mn: AM=MD$ can you help me to find a base, I try on the same way but it is not working, i try that $A=SDS^-1$, but even that can not help me, do you have some idea?
– Marko Å korić
Sep 16 at 15:35
@MarkoŠkorić It's not working because this is a different problem. What is $D$, a fixed diagonal matrix?
– mechanodroid
Sep 16 at 15:54
@MarkoŠkorić It's not working because this is a different problem. What is $D$, a fixed diagonal matrix?
– mechanodroid
Sep 16 at 15:54
soryy i forgot to write yes it is diagonal matrix which have diagonal entries $lambda1,lambda 2...,lambda n$
– Marko Å korić
Sep 16 at 16:01
soryy i forgot to write yes it is diagonal matrix which have diagonal entries $lambda1,lambda 2...,lambda n$
– Marko Å korić
Sep 16 at 16:01
1
1
@MarkoŠkorić Have a look here: math.stackexchange.com/q/2320839/144766
– mechanodroid
Sep 16 at 16:09
@MarkoŠkorić Have a look here: math.stackexchange.com/q/2320839/144766
– mechanodroid
Sep 16 at 16:09
thank you so much
– Marko Å korić
Sep 16 at 16:43
thank you so much
– Marko Å korić
Sep 16 at 16:43
add a comment |Â
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Am I understanding this right, that $S$ is fixed for $U$?
– zzuussee
Sep 8 at 10:12
yes you are right
– Marko Å korić
Sep 8 at 10:14
1
According to your definition, $U$ is a singleton, it contains exactly one element, namely $SDS^-1$.
– Ewan Delanoy
Sep 8 at 10:55