Does this result for inverses of linear transformation holds for infinite dimensional case.
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Q.If V is an arbitrary vector space over $F$ and if $T$ belongs $Hom(V,V)$ is right-invertible with a unique right inverse, prove that its invertible. This was question asked in Topics in Algebra by I.N. Herstein.
I was able to prove this result by using the argument by contradiction, let us assume $T$ is not invertible and So there exist a $Snot=0$ such that $ST=TS=0$ and let $S_1$ is unique right inverse, now $ S_1TS=IS=S=0$, which is a contradiction. I had question if the vector space is not finite dimensional does the statement holds true? If yes, does my argument above holds.
linear-algebra linear-transformations
asked Nov 14 at 15:37
Ash Pd
809
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Q.If V is an arbitrary vector space over $F$ and if $T$ belongs $Hom(V,V)$ is right-invertible with a unique right inverse, prove that its invertible. This was question asked in Topics in Algebra by I.N. Herstein.
I was able to prove this result by using the argument by contradiction, let us assume $T$ is not invertible and So there exist a $Snot=0$ such that $ST=TS=0$ and let $S_1$ is unique right inverse, now $ S_1TS=IS=S=0$, which is a contradiction. I had question if the vector space is not finite dimensional does the statement holds true? If yes, does my argument above holds.
linear-algebra linear-transformations
asked Nov 14 at 15:37
Ash Pd
809
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Q.If V is an arbitrary vector space over $F$ and if $T$ belongs $Hom(V,V)$ is right-invertible with a unique right inverse, prove that its invertible. This was question asked in Topics in Algebra by I.N. Herstein.
I was able to prove this result by using the argument by contradiction, let us assume $T$ is not invertible and So there exist a $Snot=0$ such that $ST=TS=0$ and let $S_1$ is unique right inverse, now $ S_1TS=IS=S=0$, which is a contradiction. I had question if the vector space is not finite dimensional does the statement holds true? If yes, does my argument above holds.
linear-algebra linear-transformations
asked Nov 14 at 15:37
Ash Pd
809
Q.If V is an arbitrary vector space over $F$ and if $T$ belongs $Hom(V,V)$ is right-invertible with a unique right inverse, prove that its invertible. This was question asked in Topics in Algebra by I.N. Herstein.
I was able to prove this result by using the argument by contradiction, let us assume $T$ is not invertible and So there exist a $Snot=0$ such that $ST=TS=0$ and let $S_1$ is unique right inverse, now $ S_1TS=IS=S=0$, which is a contradiction. I had question if the vector space is not finite dimensional does the statement holds true? If yes, does my argument above holds.
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Nov 14 at 15:37
Ash Pd
809
asked Nov 14 at 15:37
Ash Pd
809
asked Nov 14 at 15:37
Ash Pd
809
asked Nov 14 at 15:37
Ash Pd
809
asked Nov 14 at 15:37
Ash Pd
809
809
add a comment |
add a comment |
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In general the statement is not true. The problem is that $S_1T$ is the identity on the image of $T$ which could be smaller than $V$. Think for example in the space of polynomial ring the multiplication by $x$ which is linear and has an inverse in the image (i.e. divide by $x$ by shifting all the coefficient) but clearly is not invertible on the all space (a polynomial of degree zero has not pre image)
answered Nov 14 at 15:53
ALG
55413
This provide some light, but could uniqueness of right inverse have some effect on the statement in infinite dimensional case?
– Ash Pd
Nov 15 at 11:42
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
In general the statement is not true. The problem is that $S_1T$ is the identity on the image of $T$ which could be smaller than $V$. Think for example in the space of polynomial ring the multiplication by $x$ which is linear and has an inverse in the image (i.e. divide by $x$ by shifting all the coefficient) but clearly is not invertible on the all space (a polynomial of degree zero has not pre image)
answered Nov 14 at 15:53
ALG
55413
This provide some light, but could uniqueness of right inverse have some effect on the statement in infinite dimensional case?
– Ash Pd
Nov 15 at 11:42
add a comment |
up vote
1
down vote
In general the statement is not true. The problem is that $S_1T$ is the identity on the image of $T$ which could be smaller than $V$. Think for example in the space of polynomial ring the multiplication by $x$ which is linear and has an inverse in the image (i.e. divide by $x$ by shifting all the coefficient) but clearly is not invertible on the all space (a polynomial of degree zero has not pre image)
answered Nov 14 at 15:53
ALG
55413
This provide some light, but could uniqueness of right inverse have some effect on the statement in infinite dimensional case?
– Ash Pd
Nov 15 at 11:42
add a comment |
up vote
1
down vote
up vote
1
down vote
In general the statement is not true. The problem is that $S_1T$ is the identity on the image of $T$ which could be smaller than $V$. Think for example in the space of polynomial ring the multiplication by $x$ which is linear and has an inverse in the image (i.e. divide by $x$ by shifting all the coefficient) but clearly is not invertible on the all space (a polynomial of degree zero has not pre image)
answered Nov 14 at 15:53
ALG
55413
In general the statement is not true. The problem is that $S_1T$ is the identity on the image of $T$ which could be smaller than $V$. Think for example in the space of polynomial ring the multiplication by $x$ which is linear and has an inverse in the image (i.e. divide by $x$ by shifting all the coefficient) but clearly is not invertible on the all space (a polynomial of degree zero has not pre image)
answered Nov 14 at 15:53
ALG
55413
answered Nov 14 at 15:53
ALG
55413
answered Nov 14 at 15:53
ALG
55413
answered Nov 14 at 15:53
ALG
55413
55413
This provide some light, but could uniqueness of right inverse have some effect on the statement in infinite dimensional case?
– Ash Pd
Nov 15 at 11:42
add a comment |
This provide some light, but could uniqueness of right inverse have some effect on the statement in infinite dimensional case?
– Ash Pd
Nov 15 at 11:42
This provide some light, but could uniqueness of right inverse have some effect on the statement in infinite dimensional case?
– Ash Pd
Nov 15 at 11:42
This provide some light, but could uniqueness of right inverse have some effect on the statement in infinite dimensional case?
– Ash Pd
Nov 15 at 11:42
add a comment |
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