Show that the Image of $0$ under a Linear Mapping is also $0$
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In my book of Linear Algebra, I have the following exercises:
Let $T: V to W$ be a linear map from one vector space to another. Show that $T(0) = 0$.
I'm somehow having a block.
For me, it is natural to express it like: If $T(0)$ were not in $Ker(T)$ and thus contradiction, because $Ker(T)$ is a subspace by default (the way it is constructed).
How can we show that with more elementarily? (with less abstraction.)
linear-algebra linear-transformations alternative-proof
edited Jan 15 at 23:22
José Carlos Santos
177k24139251
asked Jan 15 at 23:19
freehumoristfreehumorist
351214
$endgroup$
add a comment |
$begingroup$
In my book of Linear Algebra, I have the following exercises:
Let $T: V to W$ be a linear map from one vector space to another. Show that $T(0) = 0$.
I'm somehow having a block.
For me, it is natural to express it like: If $T(0)$ were not in $Ker(T)$ and thus contradiction, because $Ker(T)$ is a subspace by default (the way it is constructed).
How can we show that with more elementarily? (with less abstraction.)
linear-algebra linear-transformations alternative-proof
edited Jan 15 at 23:22
José Carlos Santos
177k24139251
asked Jan 15 at 23:19
freehumoristfreehumorist
351214
$endgroup$
2
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Hint: Try to deduce $T(0)=T(0)+T(0)$ somehow.
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– BigbearZzz
Jan 15 at 23:24
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@BigbearZzz i will
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– freehumorist
Jan 15 at 23:31
1
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By definition of linear mapping we know that for any vector $ v in V $ $ T(0 v) = 0 t(V) $
$endgroup$
– CrabMan
Jan 27 at 13:45
add a comment |
$begingroup$
In my book of Linear Algebra, I have the following exercises:
Let $T: V to W$ be a linear map from one vector space to another. Show that $T(0) = 0$.
I'm somehow having a block.
For me, it is natural to express it like: If $T(0)$ were not in $Ker(T)$ and thus contradiction, because $Ker(T)$ is a subspace by default (the way it is constructed).
How can we show that with more elementarily? (with less abstraction.)
linear-algebra linear-transformations alternative-proof
edited Jan 15 at 23:22
José Carlos Santos
177k24139251
asked Jan 15 at 23:19
freehumoristfreehumorist
351214
$endgroup$
In my book of Linear Algebra, I have the following exercises:
Let $T: V to W$ be a linear map from one vector space to another. Show that $T(0) = 0$.
I'm somehow having a block.
For me, it is natural to express it like: If $T(0)$ were not in $Ker(T)$ and thus contradiction, because $Ker(T)$ is a subspace by default (the way it is constructed).
How can we show that with more elementarily? (with less abstraction.)
linear-algebra linear-transformations alternative-proof
linear-algebra linear-transformations alternative-proof
edited Jan 15 at 23:22
José Carlos Santos
177k24139251
asked Jan 15 at 23:19
freehumoristfreehumorist
351214
edited Jan 15 at 23:22
José Carlos Santos
177k24139251
asked Jan 15 at 23:19
freehumoristfreehumorist
351214
edited Jan 15 at 23:22
José Carlos Santos
177k24139251
edited Jan 15 at 23:22
José Carlos Santos
177k24139251
edited Jan 15 at 23:22
José Carlos Santos
177k24139251
177k24139251
asked Jan 15 at 23:19
freehumoristfreehumorist
351214
asked Jan 15 at 23:19
freehumoristfreehumorist
351214
asked Jan 15 at 23:19
freehumoristfreehumorist
351214
351214
2
$begingroup$
Hint: Try to deduce $T(0)=T(0)+T(0)$ somehow.
$endgroup$
– BigbearZzz
Jan 15 at 23:24
$begingroup$
@BigbearZzz i will
$endgroup$
– freehumorist
Jan 15 at 23:31
1
$begingroup$
By definition of linear mapping we know that for any vector $ v in V $ $ T(0 v) = 0 t(V) $
$endgroup$
– CrabMan
Jan 27 at 13:45
add a comment |
2
$begingroup$
Hint: Try to deduce $T(0)=T(0)+T(0)$ somehow.
$endgroup$
– BigbearZzz
Jan 15 at 23:24
$begingroup$
@BigbearZzz i will
$endgroup$
– freehumorist
Jan 15 at 23:31
1
$begingroup$
By definition of linear mapping we know that for any vector $ v in V $ $ T(0 v) = 0 t(V) $
$endgroup$
– CrabMan
Jan 27 at 13:45
2
2
$begingroup$
Hint: Try to deduce $T(0)=T(0)+T(0)$ somehow.
$endgroup$
– BigbearZzz
Jan 15 at 23:24
$begingroup$
Hint: Try to deduce $T(0)=T(0)+T(0)$ somehow.
$endgroup$
– BigbearZzz
Jan 15 at 23:24
$begingroup$
@BigbearZzz i will
$endgroup$
– freehumorist
Jan 15 at 23:31
$begingroup$
@BigbearZzz i will
$endgroup$
– freehumorist
Jan 15 at 23:31
1
1
$begingroup$
By definition of linear mapping we know that for any vector $ v in V $ $ T(0 v) = 0 t(V) $
$endgroup$
– CrabMan
Jan 27 at 13:45
$begingroup$
By definition of linear mapping we know that for any vector $ v in V $ $ T(0 v) = 0 t(V) $
$endgroup$
– CrabMan
Jan 27 at 13:45
add a comment |
1 Answer
1
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$begingroup$
Hint: Begin with $0=0+0implies T(0)=T(0+0)$.
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
177k24139251
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add a comment |
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$begingroup$
Hint: Begin with $0=0+0implies T(0)=T(0+0)$.
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
177k24139251
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add a comment |
$begingroup$
Hint: Begin with $0=0+0implies T(0)=T(0+0)$.
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
177k24139251
$endgroup$
add a comment |
$begingroup$
Hint: Begin with $0=0+0implies T(0)=T(0+0)$.
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
177k24139251
$endgroup$
Hint: Begin with $0=0+0implies T(0)=T(0+0)$.
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
177k24139251
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
177k24139251
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
177k24139251
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
177k24139251
177k24139251
add a comment |
add a comment |
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2
$begingroup$
Hint: Try to deduce $T(0)=T(0)+T(0)$ somehow.
$endgroup$
– BigbearZzz
Jan 15 at 23:24
$begingroup$
@BigbearZzz i will
$endgroup$
– freehumorist
Jan 15 at 23:31
1
$begingroup$
By definition of linear mapping we know that for any vector $ v in V $ $ T(0 v) = 0 t(V) $
$endgroup$
– CrabMan
Jan 27 at 13:45