Proving $A$ is similar to $B$ when $operatornameadj A$ is similar to $operatornameadj B$
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$A,B$ are two $6 times 6$ real matrices and $A$ is invertible. I need to prove that if $operatornameadj A$ is similar to $operatornameadj B$ then $A$ is similar to $B$.
$operatornameadjA = M^-1(operatornameadjB) M$ but can't think of a direction to the solution?
linear-algebra
edited Aug 27 at 13:39
Jendrik Stelzner
7,63121037
asked Aug 27 at 12:08
bm1125
38016
add a comment |Â
up vote
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down vote
favorite
1
$A,B$ are two $6 times 6$ real matrices and $A$ is invertible. I need to prove that if $operatornameadj A$ is similar to $operatornameadj B$ then $A$ is similar to $B$.
$operatornameadjA = M^-1(operatornameadjB) M$ but can't think of a direction to the solution?
linear-algebra
edited Aug 27 at 13:39
Jendrik Stelzner
7,63121037
asked Aug 27 at 12:08
bm1125
38016
add a comment |Â
up vote
0
down vote
favorite
1
up vote
0
down vote
favorite
1
1
$A,B$ are two $6 times 6$ real matrices and $A$ is invertible. I need to prove that if $operatornameadj A$ is similar to $operatornameadj B$ then $A$ is similar to $B$.
$operatornameadjA = M^-1(operatornameadjB) M$ but can't think of a direction to the solution?
linear-algebra
edited Aug 27 at 13:39
Jendrik Stelzner
7,63121037
asked Aug 27 at 12:08
bm1125
38016
$A,B$ are two $6 times 6$ real matrices and $A$ is invertible. I need to prove that if $operatornameadj A$ is similar to $operatornameadj B$ then $A$ is similar to $B$.
$operatornameadjA = M^-1(operatornameadjB) M$ but can't think of a direction to the solution?
linear-algebra
edited Aug 27 at 13:39
Jendrik Stelzner
7,63121037
asked Aug 27 at 12:08
bm1125
38016
edited Aug 27 at 13:39
Jendrik Stelzner
7,63121037
edited Aug 27 at 13:39
Jendrik Stelzner
7,63121037
edited Aug 27 at 13:39
Jendrik Stelzner
7,63121037
7,63121037
asked Aug 27 at 12:08
bm1125
38016
asked Aug 27 at 12:08
bm1125
38016
asked Aug 27 at 12:08
bm1125
38016
38016
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Hint: use the relations
beginalign
AoperatornameadjA&=det(A)I,tag1\
BoperatornameadjB&=det(B)Itag2
endalign
to prove first that $det A=det B$.
One way to do that: prove that
$$
det Ane 0implies detoperatornameadjAne 0implies
detoperatornameadjBne 0implies det Bne 0.
$$
Then take the determinant of boths side in (1) and (2).
answered Aug 27 at 12:30
A.Γ.
20.6k22353
thanks!! yet I still couldn't understand what am I missing. I defined $ operatornameadjA = A^-1 |A| I $ and the same with B and from the details of the question $ operatornameadjA = P^-1 cdot operatornameadjB cdot P $ so its just $ A^-1 cdot |A| cdot I = P^-1 cdot B^-1 cdot |B| cdot I cdot P $ and can't get any further from here.. Oh and btw, why would proving detA = det B would help me? I mean two matrices could have the same determinant & trace & rank and yet not be similar.. correct?
– bm1125
Aug 27 at 12:56
@bm1125 You cannot use $B^-1$ until you prove it exists, e.g. prove that $det Bne 0$. Once you prove the determinants are equal, the rest is quite simple algebra calculation from the same relations and similarity of $operatornameadjA$ and $operatornameadjB$.
– A.Γ.
Aug 27 at 12:58
right! thanks again. So $ det(operatornameadjA) = det(A)^n-1 = det(A)^5 $ and so $ det(operatornameadjB) = det(operatornameadjA $ because they are similar and thus $ det(B) not = 0 $ and $ B$ is invertible. I'm not sure how to find invertible matrix such that $ A = P^-1 B P $ ??
– bm1125
Aug 27 at 13:26
@bm1125 You can e.g. continue from what you wrote above $$ A^-1 cdot |A| cdot I = P^-1 cdot B^-1 cdot |B| cdot I cdot P$$ Cancel the deteminants (equal) and take the inverse of each side.
– A.Γ.
Aug 27 at 14:06
1
@bm1125 Yes, because then $$A=(A^-1)^-1=(P^-1B^-1P)^-1=P^-1BP.$$
– A.Γ.
Aug 27 at 14:23
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint: use the relations
beginalign
AoperatornameadjA&=det(A)I,tag1\
BoperatornameadjB&=det(B)Itag2
endalign
to prove first that $det A=det B$.
One way to do that: prove that
$$
det Ane 0implies detoperatornameadjAne 0implies
detoperatornameadjBne 0implies det Bne 0.
$$
Then take the determinant of boths side in (1) and (2).
answered Aug 27 at 12:30
A.Γ.
20.6k22353
thanks!! yet I still couldn't understand what am I missing. I defined $ operatornameadjA = A^-1 |A| I $ and the same with B and from the details of the question $ operatornameadjA = P^-1 cdot operatornameadjB cdot P $ so its just $ A^-1 cdot |A| cdot I = P^-1 cdot B^-1 cdot |B| cdot I cdot P $ and can't get any further from here.. Oh and btw, why would proving detA = det B would help me? I mean two matrices could have the same determinant & trace & rank and yet not be similar.. correct?
– bm1125
Aug 27 at 12:56
@bm1125 You cannot use $B^-1$ until you prove it exists, e.g. prove that $det Bne 0$. Once you prove the determinants are equal, the rest is quite simple algebra calculation from the same relations and similarity of $operatornameadjA$ and $operatornameadjB$.
– A.Γ.
Aug 27 at 12:58
right! thanks again. So $ det(operatornameadjA) = det(A)^n-1 = det(A)^5 $ and so $ det(operatornameadjB) = det(operatornameadjA $ because they are similar and thus $ det(B) not = 0 $ and $ B$ is invertible. I'm not sure how to find invertible matrix such that $ A = P^-1 B P $ ??
– bm1125
Aug 27 at 13:26
@bm1125 You can e.g. continue from what you wrote above $$ A^-1 cdot |A| cdot I = P^-1 cdot B^-1 cdot |B| cdot I cdot P$$ Cancel the deteminants (equal) and take the inverse of each side.
– A.Γ.
Aug 27 at 14:06
1
@bm1125 Yes, because then $$A=(A^-1)^-1=(P^-1B^-1P)^-1=P^-1BP.$$
– A.Γ.
Aug 27 at 14:23
 |Â
show 1 more comment
up vote
2
down vote
accepted
Hint: use the relations
beginalign
AoperatornameadjA&=det(A)I,tag1\
BoperatornameadjB&=det(B)Itag2
endalign
to prove first that $det A=det B$.
One way to do that: prove that
$$
det Ane 0implies detoperatornameadjAne 0implies
detoperatornameadjBne 0implies det Bne 0.
$$
Then take the determinant of boths side in (1) and (2).
answered Aug 27 at 12:30
A.Γ.
20.6k22353
thanks!! yet I still couldn't understand what am I missing. I defined $ operatornameadjA = A^-1 |A| I $ and the same with B and from the details of the question $ operatornameadjA = P^-1 cdot operatornameadjB cdot P $ so its just $ A^-1 cdot |A| cdot I = P^-1 cdot B^-1 cdot |B| cdot I cdot P $ and can't get any further from here.. Oh and btw, why would proving detA = det B would help me? I mean two matrices could have the same determinant & trace & rank and yet not be similar.. correct?
– bm1125
Aug 27 at 12:56
@bm1125 You cannot use $B^-1$ until you prove it exists, e.g. prove that $det Bne 0$. Once you prove the determinants are equal, the rest is quite simple algebra calculation from the same relations and similarity of $operatornameadjA$ and $operatornameadjB$.
– A.Γ.
Aug 27 at 12:58
right! thanks again. So $ det(operatornameadjA) = det(A)^n-1 = det(A)^5 $ and so $ det(operatornameadjB) = det(operatornameadjA $ because they are similar and thus $ det(B) not = 0 $ and $ B$ is invertible. I'm not sure how to find invertible matrix such that $ A = P^-1 B P $ ??
– bm1125
Aug 27 at 13:26
@bm1125 You can e.g. continue from what you wrote above $$ A^-1 cdot |A| cdot I = P^-1 cdot B^-1 cdot |B| cdot I cdot P$$ Cancel the deteminants (equal) and take the inverse of each side.
– A.Γ.
Aug 27 at 14:06
1
@bm1125 Yes, because then $$A=(A^-1)^-1=(P^-1B^-1P)^-1=P^-1BP.$$
– A.Γ.
Aug 27 at 14:23
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint: use the relations
beginalign
AoperatornameadjA&=det(A)I,tag1\
BoperatornameadjB&=det(B)Itag2
endalign
to prove first that $det A=det B$.
One way to do that: prove that
$$
det Ane 0implies detoperatornameadjAne 0implies
detoperatornameadjBne 0implies det Bne 0.
$$
Then take the determinant of boths side in (1) and (2).
answered Aug 27 at 12:30
A.Γ.
20.6k22353
Hint: use the relations
beginalign
AoperatornameadjA&=det(A)I,tag1\
BoperatornameadjB&=det(B)Itag2
endalign
to prove first that $det A=det B$.
One way to do that: prove that
$$
det Ane 0implies detoperatornameadjAne 0implies
detoperatornameadjBne 0implies det Bne 0.
$$
Then take the determinant of boths side in (1) and (2).
answered Aug 27 at 12:30
A.Γ.
20.6k22353
edited Aug 27 at 13:06
answered Aug 27 at 12:30
A.Γ.
20.6k22353
answered Aug 27 at 12:30
A.Γ.
20.6k22353
answered Aug 27 at 12:30
A.Γ.
20.6k22353
20.6k22353
thanks!! yet I still couldn't understand what am I missing. I defined $ operatornameadjA = A^-1 |A| I $ and the same with B and from the details of the question $ operatornameadjA = P^-1 cdot operatornameadjB cdot P $ so its just $ A^-1 cdot |A| cdot I = P^-1 cdot B^-1 cdot |B| cdot I cdot P $ and can't get any further from here.. Oh and btw, why would proving detA = det B would help me? I mean two matrices could have the same determinant & trace & rank and yet not be similar.. correct?
– bm1125
Aug 27 at 12:56
@bm1125 You cannot use $B^-1$ until you prove it exists, e.g. prove that $det Bne 0$. Once you prove the determinants are equal, the rest is quite simple algebra calculation from the same relations and similarity of $operatornameadjA$ and $operatornameadjB$.
– A.Γ.
Aug 27 at 12:58
right! thanks again. So $ det(operatornameadjA) = det(A)^n-1 = det(A)^5 $ and so $ det(operatornameadjB) = det(operatornameadjA $ because they are similar and thus $ det(B) not = 0 $ and $ B$ is invertible. I'm not sure how to find invertible matrix such that $ A = P^-1 B P $ ??
– bm1125
Aug 27 at 13:26
@bm1125 You can e.g. continue from what you wrote above $$ A^-1 cdot |A| cdot I = P^-1 cdot B^-1 cdot |B| cdot I cdot P$$ Cancel the deteminants (equal) and take the inverse of each side.
– A.Γ.
Aug 27 at 14:06
1
@bm1125 Yes, because then $$A=(A^-1)^-1=(P^-1B^-1P)^-1=P^-1BP.$$
– A.Γ.
Aug 27 at 14:23
 |Â
show 1 more comment
thanks!! yet I still couldn't understand what am I missing. I defined $ operatornameadjA = A^-1 |A| I $ and the same with B and from the details of the question $ operatornameadjA = P^-1 cdot operatornameadjB cdot P $ so its just $ A^-1 cdot |A| cdot I = P^-1 cdot B^-1 cdot |B| cdot I cdot P $ and can't get any further from here.. Oh and btw, why would proving detA = det B would help me? I mean two matrices could have the same determinant & trace & rank and yet not be similar.. correct?
– bm1125
Aug 27 at 12:56
@bm1125 You cannot use $B^-1$ until you prove it exists, e.g. prove that $det Bne 0$. Once you prove the determinants are equal, the rest is quite simple algebra calculation from the same relations and similarity of $operatornameadjA$ and $operatornameadjB$.
– A.Γ.
Aug 27 at 12:58
right! thanks again. So $ det(operatornameadjA) = det(A)^n-1 = det(A)^5 $ and so $ det(operatornameadjB) = det(operatornameadjA $ because they are similar and thus $ det(B) not = 0 $ and $ B$ is invertible. I'm not sure how to find invertible matrix such that $ A = P^-1 B P $ ??
– bm1125
Aug 27 at 13:26
@bm1125 You can e.g. continue from what you wrote above $$ A^-1 cdot |A| cdot I = P^-1 cdot B^-1 cdot |B| cdot I cdot P$$ Cancel the deteminants (equal) and take the inverse of each side.
– A.Γ.
Aug 27 at 14:06
1
@bm1125 Yes, because then $$A=(A^-1)^-1=(P^-1B^-1P)^-1=P^-1BP.$$
– A.Γ.
Aug 27 at 14:23
thanks!! yet I still couldn't understand what am I missing. I defined $ operatornameadjA = A^-1 |A| I $ and the same with B and from the details of the question $ operatornameadjA = P^-1 cdot operatornameadjB cdot P $ so its just $ A^-1 cdot |A| cdot I = P^-1 cdot B^-1 cdot |B| cdot I cdot P $ and can't get any further from here.. Oh and btw, why would proving detA = det B would help me? I mean two matrices could have the same determinant & trace & rank and yet not be similar.. correct?
– bm1125
Aug 27 at 12:56
thanks!! yet I still couldn't understand what am I missing. I defined $ operatornameadjA = A^-1 |A| I $ and the same with B and from the details of the question $ operatornameadjA = P^-1 cdot operatornameadjB cdot P $ so its just $ A^-1 cdot |A| cdot I = P^-1 cdot B^-1 cdot |B| cdot I cdot P $ and can't get any further from here.. Oh and btw, why would proving detA = det B would help me? I mean two matrices could have the same determinant & trace & rank and yet not be similar.. correct?
– bm1125
Aug 27 at 12:56
@bm1125 You cannot use $B^-1$ until you prove it exists, e.g. prove that $det Bne 0$. Once you prove the determinants are equal, the rest is quite simple algebra calculation from the same relations and similarity of $operatornameadjA$ and $operatornameadjB$.
– A.Γ.
Aug 27 at 12:58
@bm1125 You cannot use $B^-1$ until you prove it exists, e.g. prove that $det Bne 0$. Once you prove the determinants are equal, the rest is quite simple algebra calculation from the same relations and similarity of $operatornameadjA$ and $operatornameadjB$.
– A.Γ.
Aug 27 at 12:58
right! thanks again. So $ det(operatornameadjA) = det(A)^n-1 = det(A)^5 $ and so $ det(operatornameadjB) = det(operatornameadjA $ because they are similar and thus $ det(B) not = 0 $ and $ B$ is invertible. I'm not sure how to find invertible matrix such that $ A = P^-1 B P $ ??
– bm1125
Aug 27 at 13:26
right! thanks again. So $ det(operatornameadjA) = det(A)^n-1 = det(A)^5 $ and so $ det(operatornameadjB) = det(operatornameadjA $ because they are similar and thus $ det(B) not = 0 $ and $ B$ is invertible. I'm not sure how to find invertible matrix such that $ A = P^-1 B P $ ??
– bm1125
Aug 27 at 13:26
@bm1125 You can e.g. continue from what you wrote above $$ A^-1 cdot |A| cdot I = P^-1 cdot B^-1 cdot |B| cdot I cdot P$$ Cancel the deteminants (equal) and take the inverse of each side.
– A.Γ.
Aug 27 at 14:06
@bm1125 You can e.g. continue from what you wrote above $$ A^-1 cdot |A| cdot I = P^-1 cdot B^-1 cdot |B| cdot I cdot P$$ Cancel the deteminants (equal) and take the inverse of each side.
– A.Γ.
Aug 27 at 14:06
1
1
@bm1125 Yes, because then $$A=(A^-1)^-1=(P^-1B^-1P)^-1=P^-1BP.$$
– A.Γ.
Aug 27 at 14:23
@bm1125 Yes, because then $$A=(A^-1)^-1=(P^-1B^-1P)^-1=P^-1BP.$$
– A.Γ.
Aug 27 at 14:23
 |Â
show 1 more comment
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