Similar Invertible Matrices
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If $B$ is similar to $A$, show that there exists an invertible matrix $S$ and another marix $T$ such that $A=ST$ and $B=TS$.
So I know that $B$ being similar to $A$ implies $M^{-1}AM=B$. Can I just say let $A$ be a product of 2 invertible matrices $S$ and $T$ then manipulate $M$ to be either $S$ or $T$?
linear-algebra matrices proof-verification
asked Dec 13 '18 at 21:48
FundementalJTheoremFundementalJTheorem
424
$endgroup$
add a comment |
$begingroup$
If $B$ is similar to $A$, show that there exists an invertible matrix $S$ and another marix $T$ such that $A=ST$ and $B=TS$.
So I know that $B$ being similar to $A$ implies $M^{-1}AM=B$. Can I just say let $A$ be a product of 2 invertible matrices $S$ and $T$ then manipulate $M$ to be either $S$ or $T$?
linear-algebra matrices proof-verification
asked Dec 13 '18 at 21:48
FundementalJTheoremFundementalJTheorem
424
$endgroup$
$begingroup$
You can do this to get an idea of what you must get, but it's not a valid proof. You must show a construction without this supposition.
$endgroup$
– Lucas Henrique
Dec 13 '18 at 21:55
add a comment |
$begingroup$
If $B$ is similar to $A$, show that there exists an invertible matrix $S$ and another marix $T$ such that $A=ST$ and $B=TS$.
So I know that $B$ being similar to $A$ implies $M^{-1}AM=B$. Can I just say let $A$ be a product of 2 invertible matrices $S$ and $T$ then manipulate $M$ to be either $S$ or $T$?
linear-algebra matrices proof-verification
asked Dec 13 '18 at 21:48
FundementalJTheoremFundementalJTheorem
424
$endgroup$
If $B$ is similar to $A$, show that there exists an invertible matrix $S$ and another marix $T$ such that $A=ST$ and $B=TS$.
So I know that $B$ being similar to $A$ implies $M^{-1}AM=B$. Can I just say let $A$ be a product of 2 invertible matrices $S$ and $T$ then manipulate $M$ to be either $S$ or $T$?
linear-algebra matrices proof-verification
linear-algebra matrices proof-verification
asked Dec 13 '18 at 21:48
FundementalJTheoremFundementalJTheorem
424
asked Dec 13 '18 at 21:48
FundementalJTheoremFundementalJTheorem
424
asked Dec 13 '18 at 21:48
FundementalJTheoremFundementalJTheorem
424
asked Dec 13 '18 at 21:48
FundementalJTheoremFundementalJTheorem
424
asked Dec 13 '18 at 21:48
FundementalJTheoremFundementalJTheorem
424
424
$begingroup$
You can do this to get an idea of what you must get, but it's not a valid proof. You must show a construction without this supposition.
$endgroup$
– Lucas Henrique
Dec 13 '18 at 21:55
add a comment |
$begingroup$
You can do this to get an idea of what you must get, but it's not a valid proof. You must show a construction without this supposition.
$endgroup$
– Lucas Henrique
Dec 13 '18 at 21:55
$begingroup$
You can do this to get an idea of what you must get, but it's not a valid proof. You must show a construction without this supposition.
$endgroup$
– Lucas Henrique
Dec 13 '18 at 21:55
$begingroup$
You can do this to get an idea of what you must get, but it's not a valid proof. You must show a construction without this supposition.
$endgroup$
– Lucas Henrique
Dec 13 '18 at 21:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $A$ and $B$ be similar. Then we have $A=S(BS^{-1})=ST$ with $T=BS^{-1}$. It follows that
$$
TS=BS^{-1}S=B.
$$
answered Dec 13 '18 at 21:56
Dietrich BurdeDietrich Burde
79.2k647103
$endgroup$
add a comment |
$begingroup$
Write the similarity condition as $AM=MB$.
Can you see, that you are quite close to the wanted product presentation?
answered Dec 13 '18 at 21:56
HannoHanno
2,223428
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let $A$ and $B$ be similar. Then we have $A=S(BS^{-1})=ST$ with $T=BS^{-1}$. It follows that
$$
TS=BS^{-1}S=B.
$$
answered Dec 13 '18 at 21:56
Dietrich BurdeDietrich Burde
79.2k647103
$endgroup$
add a comment |
$begingroup$
Let $A$ and $B$ be similar. Then we have $A=S(BS^{-1})=ST$ with $T=BS^{-1}$. It follows that
$$
TS=BS^{-1}S=B.
$$
answered Dec 13 '18 at 21:56
Dietrich BurdeDietrich Burde
79.2k647103
$endgroup$
add a comment |
$begingroup$
Let $A$ and $B$ be similar. Then we have $A=S(BS^{-1})=ST$ with $T=BS^{-1}$. It follows that
$$
TS=BS^{-1}S=B.
$$
answered Dec 13 '18 at 21:56
Dietrich BurdeDietrich Burde
79.2k647103
$endgroup$
Let $A$ and $B$ be similar. Then we have $A=S(BS^{-1})=ST$ with $T=BS^{-1}$. It follows that
$$
TS=BS^{-1}S=B.
$$
answered Dec 13 '18 at 21:56
Dietrich BurdeDietrich Burde
79.2k647103
answered Dec 13 '18 at 21:56
Dietrich BurdeDietrich Burde
79.2k647103
answered Dec 13 '18 at 21:56
Dietrich BurdeDietrich Burde
79.2k647103
answered Dec 13 '18 at 21:56
Dietrich BurdeDietrich Burde
79.2k647103
79.2k647103
add a comment |
add a comment |
$begingroup$
Write the similarity condition as $AM=MB$.
Can you see, that you are quite close to the wanted product presentation?
answered Dec 13 '18 at 21:56
HannoHanno
2,223428
$endgroup$
add a comment |
$begingroup$
Write the similarity condition as $AM=MB$.
Can you see, that you are quite close to the wanted product presentation?
answered Dec 13 '18 at 21:56
HannoHanno
2,223428
$endgroup$
add a comment |
$begingroup$
Write the similarity condition as $AM=MB$.
Can you see, that you are quite close to the wanted product presentation?
answered Dec 13 '18 at 21:56
HannoHanno
2,223428
$endgroup$
Write the similarity condition as $AM=MB$.
Can you see, that you are quite close to the wanted product presentation?
answered Dec 13 '18 at 21:56
HannoHanno
2,223428
edited Dec 14 '18 at 11:49
answered Dec 13 '18 at 21:56
HannoHanno
2,223428
answered Dec 13 '18 at 21:56
HannoHanno
2,223428
answered Dec 13 '18 at 21:56
HannoHanno
2,223428
2,223428
add a comment |
add a comment |
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$begingroup$
You can do this to get an idea of what you must get, but it's not a valid proof. You must show a construction without this supposition.
$endgroup$
– Lucas Henrique
Dec 13 '18 at 21:55