Mixing
Similar matrices with same invertible matrix.
$begingroup$
If $A$ and $B$ are similar matrices, then
$B = P^{-1}AP$ and $A = P^{-1}BP$ for some invertible matrix $P$. True or false.
All I know is if $A, B$ and $C$ are similar matrices then there is equivalence relation within them. Can someone please look into this and let me know whether is it true or false?. If is it true, then why and if false, provide me the justification.
linear-algebra
edited Jan 1 at 21:21
mechanodroid
27.1k62446
asked Jan 1 at 19:05
Shashank DwivediShashank Dwivedi
735
$endgroup$
|
show 6 more comments
$begingroup$
If $A$ and $B$ are similar matrices, then
$B = P^{-1}AP$ and $A = P^{-1}BP$ for some invertible matrix $P$. True or false.
All I know is if $A, B$ and $C$ are similar matrices then there is equivalence relation within them. Can someone please look into this and let me know whether is it true or false?. If is it true, then why and if false, provide me the justification.
linear-algebra
edited Jan 1 at 21:21
mechanodroid
27.1k62446
asked Jan 1 at 19:05
Shashank DwivediShashank Dwivedi
735
$endgroup$
$begingroup$
What does “there is equivalence relation within them” mean?
$endgroup$
– José Carlos Santos
Jan 1 at 19:09
$begingroup$
I meant Suppose A, B and C are square matrices of size n. Then, A is similar to A (Reflexive), If A is similar to B, then B is similar to A(Symmetric), and It is transitive too. However, I am unable to understand whether this is even relevant with what has been asked in the question.
$endgroup$
– Shashank Dwivedi
Jan 1 at 19:12
$begingroup$
It is not relevant.
$endgroup$
– José Carlos Santos
Jan 1 at 19:12
2
$begingroup$
The formula you have given is usually how similar matrices are defined, so it is true. There is other criteria for similar matrices, such as equal determinants and equal traces. So if you use the definition given in your question to check if the criteria hold that will work
$endgroup$
– Scosh_lr
Jan 1 at 19:17
2
$begingroup$
If $B=P^{-1}AP$ then $A=PBP^{-1}$. You claim $A=P^{-1}BP$. This is not obvious and I strongly suspect it is false, although finding a specific counterexample may be tricky.
$endgroup$
– Ben W
Jan 1 at 19:34
|
show 6 more comments
$begingroup$
If $A$ and $B$ are similar matrices, then
$B = P^{-1}AP$ and $A = P^{-1}BP$ for some invertible matrix $P$. True or false.
All I know is if $A, B$ and $C$ are similar matrices then there is equivalence relation within them. Can someone please look into this and let me know whether is it true or false?. If is it true, then why and if false, provide me the justification.
linear-algebra
edited Jan 1 at 21:21
mechanodroid
27.1k62446
asked Jan 1 at 19:05
Shashank DwivediShashank Dwivedi
735
$endgroup$
If $A$ and $B$ are similar matrices, then
$B = P^{-1}AP$ and $A = P^{-1}BP$ for some invertible matrix $P$. True or false.
All I know is if $A, B$ and $C$ are similar matrices then there is equivalence relation within them. Can someone please look into this and let me know whether is it true or false?. If is it true, then why and if false, provide me the justification.
linear-algebra
linear-algebra
edited Jan 1 at 21:21
mechanodroid
27.1k62446
asked Jan 1 at 19:05
Shashank DwivediShashank Dwivedi
735
edited Jan 1 at 21:21
mechanodroid
27.1k62446
asked Jan 1 at 19:05
Shashank DwivediShashank Dwivedi
735
edited Jan 1 at 21:21
mechanodroid
27.1k62446
edited Jan 1 at 21:21
mechanodroid
27.1k62446
edited Jan 1 at 21:21
mechanodroid
27.1k62446
27.1k62446
asked Jan 1 at 19:05
Shashank DwivediShashank Dwivedi
735
asked Jan 1 at 19:05
Shashank DwivediShashank Dwivedi
735
asked Jan 1 at 19:05
Shashank DwivediShashank Dwivedi
735
735
$begingroup$
What does “there is equivalence relation within them” mean?
$endgroup$
– José Carlos Santos
Jan 1 at 19:09
$begingroup$
I meant Suppose A, B and C are square matrices of size n. Then, A is similar to A (Reflexive), If A is similar to B, then B is similar to A(Symmetric), and It is transitive too. However, I am unable to understand whether this is even relevant with what has been asked in the question.
$endgroup$
– Shashank Dwivedi
Jan 1 at 19:12
$begingroup$
It is not relevant.
$endgroup$
– José Carlos Santos
Jan 1 at 19:12
2
$begingroup$
The formula you have given is usually how similar matrices are defined, so it is true. There is other criteria for similar matrices, such as equal determinants and equal traces. So if you use the definition given in your question to check if the criteria hold that will work
$endgroup$
– Scosh_lr
Jan 1 at 19:17
2
$begingroup$
If $B=P^{-1}AP$ then $A=PBP^{-1}$. You claim $A=P^{-1}BP$. This is not obvious and I strongly suspect it is false, although finding a specific counterexample may be tricky.
$endgroup$
– Ben W
Jan 1 at 19:34
|
show 6 more comments
$begingroup$
What does “there is equivalence relation within them” mean?
$endgroup$
– José Carlos Santos
Jan 1 at 19:09
$begingroup$
I meant Suppose A, B and C are square matrices of size n. Then, A is similar to A (Reflexive), If A is similar to B, then B is similar to A(Symmetric), and It is transitive too. However, I am unable to understand whether this is even relevant with what has been asked in the question.
$endgroup$
– Shashank Dwivedi
Jan 1 at 19:12
$begingroup$
It is not relevant.
$endgroup$
– José Carlos Santos
Jan 1 at 19:12
2
$begingroup$
The formula you have given is usually how similar matrices are defined, so it is true. There is other criteria for similar matrices, such as equal determinants and equal traces. So if you use the definition given in your question to check if the criteria hold that will work
$endgroup$
– Scosh_lr
Jan 1 at 19:17
2
$begingroup$
If $B=P^{-1}AP$ then $A=PBP^{-1}$. You claim $A=P^{-1}BP$. This is not obvious and I strongly suspect it is false, although finding a specific counterexample may be tricky.
$endgroup$
– Ben W
Jan 1 at 19:34
$begingroup$
What does “there is equivalence relation within them” mean?
$endgroup$
– José Carlos Santos
Jan 1 at 19:09
$begingroup$
What does “there is equivalence relation within them” mean?
$endgroup$
– José Carlos Santos
Jan 1 at 19:09
$begingroup$
I meant Suppose A, B and C are square matrices of size n. Then, A is similar to A (Reflexive), If A is similar to B, then B is similar to A(Symmetric), and It is transitive too. However, I am unable to understand whether this is even relevant with what has been asked in the question.
$endgroup$
– Shashank Dwivedi
Jan 1 at 19:12
$begingroup$
I meant Suppose A, B and C are square matrices of size n. Then, A is similar to A (Reflexive), If A is similar to B, then B is similar to A(Symmetric), and It is transitive too. However, I am unable to understand whether this is even relevant with what has been asked in the question.
$endgroup$
– Shashank Dwivedi
Jan 1 at 19:12
$begingroup$
It is not relevant.
$endgroup$
– José Carlos Santos
Jan 1 at 19:12
$begingroup$
It is not relevant.
$endgroup$
– José Carlos Santos
Jan 1 at 19:12
2
2
$begingroup$
The formula you have given is usually how similar matrices are defined, so it is true. There is other criteria for similar matrices, such as equal determinants and equal traces. So if you use the definition given in your question to check if the criteria hold that will work
$endgroup$
– Scosh_lr
Jan 1 at 19:17
$begingroup$
The formula you have given is usually how similar matrices are defined, so it is true. There is other criteria for similar matrices, such as equal determinants and equal traces. So if you use the definition given in your question to check if the criteria hold that will work
$endgroup$
– Scosh_lr
Jan 1 at 19:17
2
2
$begingroup$
If $B=P^{-1}AP$ then $A=PBP^{-1}$. You claim $A=P^{-1}BP$. This is not obvious and I strongly suspect it is false, although finding a specific counterexample may be tricky.
$endgroup$
– Ben W
Jan 1 at 19:34
$begingroup$
If $B=P^{-1}AP$ then $A=PBP^{-1}$. You claim $A=P^{-1}BP$. This is not obvious and I strongly suspect it is false, although finding a specific counterexample may be tricky.
$endgroup$
– Ben W
Jan 1 at 19:34
|
show 6 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Note that the problem statement says that $B=P^{-1}AP$ and $A=P^{-1}BP$ --- rather than $A=PBP^{-1}$. As it stands, the statement is false.
Consider the case where $A=operatorname{diag}(1,2,3)$ and $B=S^{-1}AS$, where $S=pmatrix{0&1&0\ 0&0&1\ 1&0&0}$. By construction, the two matrices are similar to each other. Recall that diagonal matrices with distinct diagonal entries only commute with diagonal matrices. If $B=P^{-1}AP$ for some matrix invertible $P$, then $(PS^{-1})A=A(PS^{-1})$. Therefore $PS^{-1}$ is a diagonal matrix, meaning that
$$
P=pmatrix{0&a&0\ 0&0&b\ c&0&0} text{ for some } a,b,cne0.tag{1}
$$
Now, if we also have $A=P^{-1}BP$, then $A=P^{-1}BP=P^{-1}(P^{-1}AP)P=P^{-2}AP^2$. Therefore $P^2A=AP^2$, i.e. $P^2$ is a diagonal matrix. Yet this is not possible for any $P$ in the form of $(1)$.
answered Jan 1 at 20:25
user1551user1551
72k566126
$endgroup$
$begingroup$
Well, this is pretty strange because, $B = S^{-1}AS = pmatrix{1&-1\ 0&2}$. And we can easily see that $S^{-1}BS = A$. Therefore this is not a counterexample.
$endgroup$
– Euler Pythagoras
Jan 1 at 20:41
$begingroup$
@EulerPythagoras Oops, I've picked a wrong $S$ (with $S^2=I$!!!). It should be easy to fix that.
$endgroup$
– user1551
Jan 1 at 20:44
add a comment |
$begingroup$
The answer is no.
If $A = P^{-1}BP$ and $B = P^{-1}BP$ then also $B = PBP^{-1}$ and hence $P^2B = BP^2$.
Take $P = begin{bmatrix} 1 & 1 \ 0 & 1end{bmatrix}$ and $B = begin{bmatrix} 1 & 2 \ 0 & 3end{bmatrix}$.
We have
$$P^2B = begin{bmatrix} 1 & 2 \ 0 & 1end{bmatrix}begin{bmatrix} 1 & 2 \ 0 & 3end{bmatrix} = begin{bmatrix} 1 & 8 \ 0 & 1end{bmatrix}$$
$$BP^2 = begin{bmatrix} 1 & 2 \ 0 & 3end{bmatrix}begin{bmatrix} 1 & 2 \ 0 & 1end{bmatrix} = begin{bmatrix} 1 & 4 \ 0 & 3end{bmatrix}$$
so they are not equal.
answered Jan 1 at 21:20
mechanodroidmechanodroid
27.1k62446
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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oldest
votes
$begingroup$
Note that the problem statement says that $B=P^{-1}AP$ and $A=P^{-1}BP$ --- rather than $A=PBP^{-1}$. As it stands, the statement is false.
Consider the case where $A=operatorname{diag}(1,2,3)$ and $B=S^{-1}AS$, where $S=pmatrix{0&1&0\ 0&0&1\ 1&0&0}$. By construction, the two matrices are similar to each other. Recall that diagonal matrices with distinct diagonal entries only commute with diagonal matrices. If $B=P^{-1}AP$ for some matrix invertible $P$, then $(PS^{-1})A=A(PS^{-1})$. Therefore $PS^{-1}$ is a diagonal matrix, meaning that
$$
P=pmatrix{0&a&0\ 0&0&b\ c&0&0} text{ for some } a,b,cne0.tag{1}
$$
Now, if we also have $A=P^{-1}BP$, then $A=P^{-1}BP=P^{-1}(P^{-1}AP)P=P^{-2}AP^2$. Therefore $P^2A=AP^2$, i.e. $P^2$ is a diagonal matrix. Yet this is not possible for any $P$ in the form of $(1)$.
answered Jan 1 at 20:25
user1551user1551
72k566126
$endgroup$
$begingroup$
Well, this is pretty strange because, $B = S^{-1}AS = pmatrix{1&-1\ 0&2}$. And we can easily see that $S^{-1}BS = A$. Therefore this is not a counterexample.
$endgroup$
– Euler Pythagoras
Jan 1 at 20:41
$begingroup$
@EulerPythagoras Oops, I've picked a wrong $S$ (with $S^2=I$!!!). It should be easy to fix that.
$endgroup$
– user1551
Jan 1 at 20:44
add a comment |
$begingroup$
Note that the problem statement says that $B=P^{-1}AP$ and $A=P^{-1}BP$ --- rather than $A=PBP^{-1}$. As it stands, the statement is false.
Consider the case where $A=operatorname{diag}(1,2,3)$ and $B=S^{-1}AS$, where $S=pmatrix{0&1&0\ 0&0&1\ 1&0&0}$. By construction, the two matrices are similar to each other. Recall that diagonal matrices with distinct diagonal entries only commute with diagonal matrices. If $B=P^{-1}AP$ for some matrix invertible $P$, then $(PS^{-1})A=A(PS^{-1})$. Therefore $PS^{-1}$ is a diagonal matrix, meaning that
$$
P=pmatrix{0&a&0\ 0&0&b\ c&0&0} text{ for some } a,b,cne0.tag{1}
$$
Now, if we also have $A=P^{-1}BP$, then $A=P^{-1}BP=P^{-1}(P^{-1}AP)P=P^{-2}AP^2$. Therefore $P^2A=AP^2$, i.e. $P^2$ is a diagonal matrix. Yet this is not possible for any $P$ in the form of $(1)$.
answered Jan 1 at 20:25
user1551user1551
72k566126
$endgroup$
$begingroup$
Well, this is pretty strange because, $B = S^{-1}AS = pmatrix{1&-1\ 0&2}$. And we can easily see that $S^{-1}BS = A$. Therefore this is not a counterexample.
$endgroup$
– Euler Pythagoras
Jan 1 at 20:41
$begingroup$
@EulerPythagoras Oops, I've picked a wrong $S$ (with $S^2=I$!!!). It should be easy to fix that.
$endgroup$
– user1551
Jan 1 at 20:44
add a comment |
$begingroup$
Note that the problem statement says that $B=P^{-1}AP$ and $A=P^{-1}BP$ --- rather than $A=PBP^{-1}$. As it stands, the statement is false.
Consider the case where $A=operatorname{diag}(1,2,3)$ and $B=S^{-1}AS$, where $S=pmatrix{0&1&0\ 0&0&1\ 1&0&0}$. By construction, the two matrices are similar to each other. Recall that diagonal matrices with distinct diagonal entries only commute with diagonal matrices. If $B=P^{-1}AP$ for some matrix invertible $P$, then $(PS^{-1})A=A(PS^{-1})$. Therefore $PS^{-1}$ is a diagonal matrix, meaning that
$$
P=pmatrix{0&a&0\ 0&0&b\ c&0&0} text{ for some } a,b,cne0.tag{1}
$$
Now, if we also have $A=P^{-1}BP$, then $A=P^{-1}BP=P^{-1}(P^{-1}AP)P=P^{-2}AP^2$. Therefore $P^2A=AP^2$, i.e. $P^2$ is a diagonal matrix. Yet this is not possible for any $P$ in the form of $(1)$.
answered Jan 1 at 20:25
user1551user1551
72k566126
$endgroup$
Note that the problem statement says that $B=P^{-1}AP$ and $A=P^{-1}BP$ --- rather than $A=PBP^{-1}$. As it stands, the statement is false.
Consider the case where $A=operatorname{diag}(1,2,3)$ and $B=S^{-1}AS$, where $S=pmatrix{0&1&0\ 0&0&1\ 1&0&0}$. By construction, the two matrices are similar to each other. Recall that diagonal matrices with distinct diagonal entries only commute with diagonal matrices. If $B=P^{-1}AP$ for some matrix invertible $P$, then $(PS^{-1})A=A(PS^{-1})$. Therefore $PS^{-1}$ is a diagonal matrix, meaning that
$$
P=pmatrix{0&a&0\ 0&0&b\ c&0&0} text{ for some } a,b,cne0.tag{1}
$$
Now, if we also have $A=P^{-1}BP$, then $A=P^{-1}BP=P^{-1}(P^{-1}AP)P=P^{-2}AP^2$. Therefore $P^2A=AP^2$, i.e. $P^2$ is a diagonal matrix. Yet this is not possible for any $P$ in the form of $(1)$.
answered Jan 1 at 20:25
user1551user1551
72k566126
edited Jan 1 at 21:11
answered Jan 1 at 20:25
user1551user1551
72k566126
answered Jan 1 at 20:25
user1551user1551
72k566126
answered Jan 1 at 20:25
user1551user1551
72k566126
72k566126
$begingroup$
Well, this is pretty strange because, $B = S^{-1}AS = pmatrix{1&-1\ 0&2}$. And we can easily see that $S^{-1}BS = A$. Therefore this is not a counterexample.
$endgroup$
– Euler Pythagoras
Jan 1 at 20:41
$begingroup$
@EulerPythagoras Oops, I've picked a wrong $S$ (with $S^2=I$!!!). It should be easy to fix that.
$endgroup$
– user1551
Jan 1 at 20:44
add a comment |
$begingroup$
Well, this is pretty strange because, $B = S^{-1}AS = pmatrix{1&-1\ 0&2}$. And we can easily see that $S^{-1}BS = A$. Therefore this is not a counterexample.
$endgroup$
– Euler Pythagoras
Jan 1 at 20:41
$begingroup$
@EulerPythagoras Oops, I've picked a wrong $S$ (with $S^2=I$!!!). It should be easy to fix that.
$endgroup$
– user1551
Jan 1 at 20:44
$begingroup$
Well, this is pretty strange because, $B = S^{-1}AS = pmatrix{1&-1\ 0&2}$. And we can easily see that $S^{-1}BS = A$. Therefore this is not a counterexample.
$endgroup$
– Euler Pythagoras
Jan 1 at 20:41
$begingroup$
Well, this is pretty strange because, $B = S^{-1}AS = pmatrix{1&-1\ 0&2}$. And we can easily see that $S^{-1}BS = A$. Therefore this is not a counterexample.
$endgroup$
– Euler Pythagoras
Jan 1 at 20:41
$begingroup$
@EulerPythagoras Oops, I've picked a wrong $S$ (with $S^2=I$!!!). It should be easy to fix that.
$endgroup$
– user1551
Jan 1 at 20:44
$begingroup$
@EulerPythagoras Oops, I've picked a wrong $S$ (with $S^2=I$!!!). It should be easy to fix that.
$endgroup$
– user1551
Jan 1 at 20:44
add a comment |
$begingroup$
The answer is no.
If $A = P^{-1}BP$ and $B = P^{-1}BP$ then also $B = PBP^{-1}$ and hence $P^2B = BP^2$.
Take $P = begin{bmatrix} 1 & 1 \ 0 & 1end{bmatrix}$ and $B = begin{bmatrix} 1 & 2 \ 0 & 3end{bmatrix}$.
We have
$$P^2B = begin{bmatrix} 1 & 2 \ 0 & 1end{bmatrix}begin{bmatrix} 1 & 2 \ 0 & 3end{bmatrix} = begin{bmatrix} 1 & 8 \ 0 & 1end{bmatrix}$$
$$BP^2 = begin{bmatrix} 1 & 2 \ 0 & 3end{bmatrix}begin{bmatrix} 1 & 2 \ 0 & 1end{bmatrix} = begin{bmatrix} 1 & 4 \ 0 & 3end{bmatrix}$$
so they are not equal.
answered Jan 1 at 21:20
mechanodroidmechanodroid
27.1k62446
$endgroup$
add a comment |
$begingroup$
The answer is no.
If $A = P^{-1}BP$ and $B = P^{-1}BP$ then also $B = PBP^{-1}$ and hence $P^2B = BP^2$.
Take $P = begin{bmatrix} 1 & 1 \ 0 & 1end{bmatrix}$ and $B = begin{bmatrix} 1 & 2 \ 0 & 3end{bmatrix}$.
We have
$$P^2B = begin{bmatrix} 1 & 2 \ 0 & 1end{bmatrix}begin{bmatrix} 1 & 2 \ 0 & 3end{bmatrix} = begin{bmatrix} 1 & 8 \ 0 & 1end{bmatrix}$$
$$BP^2 = begin{bmatrix} 1 & 2 \ 0 & 3end{bmatrix}begin{bmatrix} 1 & 2 \ 0 & 1end{bmatrix} = begin{bmatrix} 1 & 4 \ 0 & 3end{bmatrix}$$
so they are not equal.
answered Jan 1 at 21:20
mechanodroidmechanodroid
27.1k62446
$endgroup$
add a comment |
$begingroup$
The answer is no.
If $A = P^{-1}BP$ and $B = P^{-1}BP$ then also $B = PBP^{-1}$ and hence $P^2B = BP^2$.
Take $P = begin{bmatrix} 1 & 1 \ 0 & 1end{bmatrix}$ and $B = begin{bmatrix} 1 & 2 \ 0 & 3end{bmatrix}$.
We have
$$P^2B = begin{bmatrix} 1 & 2 \ 0 & 1end{bmatrix}begin{bmatrix} 1 & 2 \ 0 & 3end{bmatrix} = begin{bmatrix} 1 & 8 \ 0 & 1end{bmatrix}$$
$$BP^2 = begin{bmatrix} 1 & 2 \ 0 & 3end{bmatrix}begin{bmatrix} 1 & 2 \ 0 & 1end{bmatrix} = begin{bmatrix} 1 & 4 \ 0 & 3end{bmatrix}$$
so they are not equal.
answered Jan 1 at 21:20
mechanodroidmechanodroid
27.1k62446
$endgroup$
The answer is no.
If $A = P^{-1}BP$ and $B = P^{-1}BP$ then also $B = PBP^{-1}$ and hence $P^2B = BP^2$.
Take $P = begin{bmatrix} 1 & 1 \ 0 & 1end{bmatrix}$ and $B = begin{bmatrix} 1 & 2 \ 0 & 3end{bmatrix}$.
We have
$$P^2B = begin{bmatrix} 1 & 2 \ 0 & 1end{bmatrix}begin{bmatrix} 1 & 2 \ 0 & 3end{bmatrix} = begin{bmatrix} 1 & 8 \ 0 & 1end{bmatrix}$$
$$BP^2 = begin{bmatrix} 1 & 2 \ 0 & 3end{bmatrix}begin{bmatrix} 1 & 2 \ 0 & 1end{bmatrix} = begin{bmatrix} 1 & 4 \ 0 & 3end{bmatrix}$$
so they are not equal.
answered Jan 1 at 21:20
mechanodroidmechanodroid
27.1k62446
answered Jan 1 at 21:20
mechanodroidmechanodroid
27.1k62446
answered Jan 1 at 21:20
mechanodroidmechanodroid
27.1k62446
answered Jan 1 at 21:20
mechanodroidmechanodroid
27.1k62446
27.1k62446
add a comment |
add a comment |
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What does “there is equivalence relation within them” mean?
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– José Carlos Santos
Jan 1 at 19:09
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I meant Suppose A, B and C are square matrices of size n. Then, A is similar to A (Reflexive), If A is similar to B, then B is similar to A(Symmetric), and It is transitive too. However, I am unable to understand whether this is even relevant with what has been asked in the question.
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– Shashank Dwivedi
Jan 1 at 19:12
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It is not relevant.
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– José Carlos Santos
Jan 1 at 19:12
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The formula you have given is usually how similar matrices are defined, so it is true. There is other criteria for similar matrices, such as equal determinants and equal traces. So if you use the definition given in your question to check if the criteria hold that will work
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– Scosh_lr
Jan 1 at 19:17
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If $B=P^{-1}AP$ then $A=PBP^{-1}$. You claim $A=P^{-1}BP$. This is not obvious and I strongly suspect it is false, although finding a specific counterexample may be tricky.
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– Ben W
Jan 1 at 19:34