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Let T be a linear operator on a vector space V over F. Prove that $Ker~ Tsubset Ker~ T^2$ and $Im~ T^2subset...
$begingroup$
Let T be a linear operator on a vector space V over F. Prove that $Ker~ Tsubset Ker~ T^2$ and $Im~ T^2subset Im~ T$
I know the definition of $Ker~ T$ & $Im~ T$. Using those, I am unable to solve the problem. Please help with elementary properties of Linear Transformation.
linear-algebra vector-spaces linear-transformations
asked Nov 30 '16 at 10:19
user1942348user1942348
1,3901934
$endgroup$
add a comment |
$begingroup$
Let T be a linear operator on a vector space V over F. Prove that $Ker~ Tsubset Ker~ T^2$ and $Im~ T^2subset Im~ T$
I know the definition of $Ker~ T$ & $Im~ T$. Using those, I am unable to solve the problem. Please help with elementary properties of Linear Transformation.
linear-algebra vector-spaces linear-transformations
asked Nov 30 '16 at 10:19
user1942348user1942348
1,3901934
$endgroup$
add a comment |
$begingroup$
Let T be a linear operator on a vector space V over F. Prove that $Ker~ Tsubset Ker~ T^2$ and $Im~ T^2subset Im~ T$
I know the definition of $Ker~ T$ & $Im~ T$. Using those, I am unable to solve the problem. Please help with elementary properties of Linear Transformation.
linear-algebra vector-spaces linear-transformations
asked Nov 30 '16 at 10:19
user1942348user1942348
1,3901934
$endgroup$
Let T be a linear operator on a vector space V over F. Prove that $Ker~ Tsubset Ker~ T^2$ and $Im~ T^2subset Im~ T$
I know the definition of $Ker~ T$ & $Im~ T$. Using those, I am unable to solve the problem. Please help with elementary properties of Linear Transformation.
linear-algebra vector-spaces linear-transformations
linear-algebra vector-spaces linear-transformations
asked Nov 30 '16 at 10:19
user1942348user1942348
1,3901934
asked Nov 30 '16 at 10:19
user1942348user1942348
1,3901934
asked Nov 30 '16 at 10:19
user1942348user1942348
1,3901934
asked Nov 30 '16 at 10:19
user1942348user1942348
1,3901934
asked Nov 30 '16 at 10:19
user1942348user1942348
1,3901934
1,3901934
add a comment |
add a comment |
1 Answer
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$begingroup$
For $vin ker(T)$ we have $T(v)=0$ and so espacially $T^2(v)=T(T(v))=T(0)=0$ and therefore $vin ker(T^2)$.
For $vin im(T^2)$ we have a $win V$ with $T^2(w)=v$ and so especially $T(T(w))=v$ which means $v$ is the image of $T(w)$ and so $vin im(T)$.
$endgroup$
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
For $vin ker(T)$ we have $T(v)=0$ and so espacially $T^2(v)=T(T(v))=T(0)=0$ and therefore $vin ker(T^2)$.
For $vin im(T^2)$ we have a $win V$ with $T^2(w)=v$ and so especially $T(T(w))=v$ which means $v$ is the image of $T(w)$ and so $vin im(T)$.
$endgroup$
add a comment |
$begingroup$
For $vin ker(T)$ we have $T(v)=0$ and so espacially $T^2(v)=T(T(v))=T(0)=0$ and therefore $vin ker(T^2)$.
For $vin im(T^2)$ we have a $win V$ with $T^2(w)=v$ and so especially $T(T(w))=v$ which means $v$ is the image of $T(w)$ and so $vin im(T)$.
$endgroup$
add a comment |
$begingroup$
For $vin ker(T)$ we have $T(v)=0$ and so espacially $T^2(v)=T(T(v))=T(0)=0$ and therefore $vin ker(T^2)$.
For $vin im(T^2)$ we have a $win V$ with $T^2(w)=v$ and so especially $T(T(w))=v$ which means $v$ is the image of $T(w)$ and so $vin im(T)$.
$endgroup$
For $vin ker(T)$ we have $T(v)=0$ and so espacially $T^2(v)=T(T(v))=T(0)=0$ and therefore $vin ker(T^2)$.
For $vin im(T^2)$ we have a $win V$ with $T^2(w)=v$ and so especially $T(T(w))=v$ which means $v$ is the image of $T(w)$ and so $vin im(T)$.
answered Nov 30 '16 at 10:24
user302982
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