Mixing
Need help to understand the uniqueness of linear maps on basis of domain.
$begingroup$
I am reading Linear Algebra Done Right. Now, I understand that the meaning of this statement is that if we know the basis of $V$ and how $T$ acts on the basis ($Tv_j = w_j$), we will know how it acts on all the vectors in $V$ ($T: V mapsto W$).
I follow the proof of existence part where it proves the property of additivity and homogeneity of linear map of the following $$T(c_1v_1 +...+ c_nv_n) = c_1w_1 +...+ c_nw_n$$
However, I don't understand the uniqueness part.
To prove uniqueness, now suppose that $T in mathcal{L}(V,W)$ and that $Tv_j = w_j$, for $j = 1,...,n$. Let $c_1,...,c_n in mathbb{F}$. The homogeneity of $T$ implies that $T(c_jv_j) = c_jT(v_j) = c_jw_j$ for $j = 1,...,n$. The additivity of $T$ now implies that
$$T(c_1v_1 +...+ c_nv_n) = c_1w_1 +...+ c_nw_n$$.
Thus $T$ is uniquely determined on $span(v_1,...,v_n)$ by the equation above. Because $v_1,...,v_n$ is a basis of $V$, this implies that $T$ is uniquely determined on $V$.
- What is meant by $T$ is uniquely determined on $span(v_1,...,v_n)$?
- What is meant by Because $v_1,...,v_n$ is a basis of $V$, this implies that $T$ is uniquely determined on $V$?
- What is meant by unique? Unique means no other form of $T$?
- Why uniqueness is important in the proof?
linear-algebra linear-transformations proof-explanation
edited Jan 13 at 9:15
José Carlos Santos
161k22127232
asked Jan 13 at 8:53
JOHN JOHN
3449
$endgroup$
add a comment |
$begingroup$
I am reading Linear Algebra Done Right. Now, I understand that the meaning of this statement is that if we know the basis of $V$ and how $T$ acts on the basis ($Tv_j = w_j$), we will know how it acts on all the vectors in $V$ ($T: V mapsto W$).
I follow the proof of existence part where it proves the property of additivity and homogeneity of linear map of the following $$T(c_1v_1 +...+ c_nv_n) = c_1w_1 +...+ c_nw_n$$
However, I don't understand the uniqueness part.
To prove uniqueness, now suppose that $T in mathcal{L}(V,W)$ and that $Tv_j = w_j$, for $j = 1,...,n$. Let $c_1,...,c_n in mathbb{F}$. The homogeneity of $T$ implies that $T(c_jv_j) = c_jT(v_j) = c_jw_j$ for $j = 1,...,n$. The additivity of $T$ now implies that
$$T(c_1v_1 +...+ c_nv_n) = c_1w_1 +...+ c_nw_n$$.
Thus $T$ is uniquely determined on $span(v_1,...,v_n)$ by the equation above. Because $v_1,...,v_n$ is a basis of $V$, this implies that $T$ is uniquely determined on $V$.
- What is meant by $T$ is uniquely determined on $span(v_1,...,v_n)$?
- What is meant by Because $v_1,...,v_n$ is a basis of $V$, this implies that $T$ is uniquely determined on $V$?
- What is meant by unique? Unique means no other form of $T$?
- Why uniqueness is important in the proof?
linear-algebra linear-transformations proof-explanation
edited Jan 13 at 9:15
José Carlos Santos
161k22127232
asked Jan 13 at 8:53
JOHN JOHN
3449
$endgroup$
add a comment |
$begingroup$
I am reading Linear Algebra Done Right. Now, I understand that the meaning of this statement is that if we know the basis of $V$ and how $T$ acts on the basis ($Tv_j = w_j$), we will know how it acts on all the vectors in $V$ ($T: V mapsto W$).
I follow the proof of existence part where it proves the property of additivity and homogeneity of linear map of the following $$T(c_1v_1 +...+ c_nv_n) = c_1w_1 +...+ c_nw_n$$
However, I don't understand the uniqueness part.
To prove uniqueness, now suppose that $T in mathcal{L}(V,W)$ and that $Tv_j = w_j$, for $j = 1,...,n$. Let $c_1,...,c_n in mathbb{F}$. The homogeneity of $T$ implies that $T(c_jv_j) = c_jT(v_j) = c_jw_j$ for $j = 1,...,n$. The additivity of $T$ now implies that
$$T(c_1v_1 +...+ c_nv_n) = c_1w_1 +...+ c_nw_n$$.
Thus $T$ is uniquely determined on $span(v_1,...,v_n)$ by the equation above. Because $v_1,...,v_n$ is a basis of $V$, this implies that $T$ is uniquely determined on $V$.
- What is meant by $T$ is uniquely determined on $span(v_1,...,v_n)$?
- What is meant by Because $v_1,...,v_n$ is a basis of $V$, this implies that $T$ is uniquely determined on $V$?
- What is meant by unique? Unique means no other form of $T$?
- Why uniqueness is important in the proof?
linear-algebra linear-transformations proof-explanation
edited Jan 13 at 9:15
José Carlos Santos
161k22127232
asked Jan 13 at 8:53
JOHN JOHN
3449
$endgroup$
I am reading Linear Algebra Done Right. Now, I understand that the meaning of this statement is that if we know the basis of $V$ and how $T$ acts on the basis ($Tv_j = w_j$), we will know how it acts on all the vectors in $V$ ($T: V mapsto W$).
I follow the proof of existence part where it proves the property of additivity and homogeneity of linear map of the following $$T(c_1v_1 +...+ c_nv_n) = c_1w_1 +...+ c_nw_n$$
However, I don't understand the uniqueness part.
To prove uniqueness, now suppose that $T in mathcal{L}(V,W)$ and that $Tv_j = w_j$, for $j = 1,...,n$. Let $c_1,...,c_n in mathbb{F}$. The homogeneity of $T$ implies that $T(c_jv_j) = c_jT(v_j) = c_jw_j$ for $j = 1,...,n$. The additivity of $T$ now implies that
$$T(c_1v_1 +...+ c_nv_n) = c_1w_1 +...+ c_nw_n$$.
Thus $T$ is uniquely determined on $span(v_1,...,v_n)$ by the equation above. Because $v_1,...,v_n$ is a basis of $V$, this implies that $T$ is uniquely determined on $V$.
- What is meant by $T$ is uniquely determined on $span(v_1,...,v_n)$?
- What is meant by Because $v_1,...,v_n$ is a basis of $V$, this implies that $T$ is uniquely determined on $V$?
- What is meant by unique? Unique means no other form of $T$?
- Why uniqueness is important in the proof?
linear-algebra linear-transformations proof-explanation
linear-algebra linear-transformations proof-explanation
edited Jan 13 at 9:15
José Carlos Santos
161k22127232
asked Jan 13 at 8:53
JOHN JOHN
3449
edited Jan 13 at 9:15
José Carlos Santos
161k22127232
asked Jan 13 at 8:53
JOHN JOHN
3449
edited Jan 13 at 9:15
José Carlos Santos
161k22127232
edited Jan 13 at 9:15
José Carlos Santos
161k22127232
edited Jan 13 at 9:15
José Carlos Santos
161k22127232
161k22127232
asked Jan 13 at 8:53
JOHN JOHN
3449
asked Jan 13 at 8:53
JOHN JOHN
3449
asked Jan 13 at 8:53
JOHN JOHN
3449
3449
add a comment |
add a comment |
1 Answer
1
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$begingroup$
- It means that that map $T$ is the only linear map defined on $operatorname{span}(v_1,ldots,v_n)$ for which the given condition ($forall jin{1,2,ldots,n}):Tv_j=w_j$) holds.
- Since ${v_1,ldots,v_n}$ is a basis, $operatorname{span}(v_1,ldots,v_n)=V$. Therefore, asserting that $T$ is uniquely determind in $operatorname{span}(v_1,ldots,v_n)$ is the same thing as asserting that it is uniquely determind on $V$.
- Yes. The map $T$ is the only linear map from $V$ into $W$ for which the condition $forall jin{1,2,ldots,n}):Tv_j=w_j$ holds.
- Because the statment that you are proving states that $T$ is unique. Therefore, you have to prove it.
answered Jan 13 at 9:10
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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votes
$begingroup$
- It means that that map $T$ is the only linear map defined on $operatorname{span}(v_1,ldots,v_n)$ for which the given condition ($forall jin{1,2,ldots,n}):Tv_j=w_j$) holds.
- Since ${v_1,ldots,v_n}$ is a basis, $operatorname{span}(v_1,ldots,v_n)=V$. Therefore, asserting that $T$ is uniquely determind in $operatorname{span}(v_1,ldots,v_n)$ is the same thing as asserting that it is uniquely determind on $V$.
- Yes. The map $T$ is the only linear map from $V$ into $W$ for which the condition $forall jin{1,2,ldots,n}):Tv_j=w_j$ holds.
- Because the statment that you are proving states that $T$ is unique. Therefore, you have to prove it.
answered Jan 13 at 9:10
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
add a comment |
$begingroup$
- It means that that map $T$ is the only linear map defined on $operatorname{span}(v_1,ldots,v_n)$ for which the given condition ($forall jin{1,2,ldots,n}):Tv_j=w_j$) holds.
- Since ${v_1,ldots,v_n}$ is a basis, $operatorname{span}(v_1,ldots,v_n)=V$. Therefore, asserting that $T$ is uniquely determind in $operatorname{span}(v_1,ldots,v_n)$ is the same thing as asserting that it is uniquely determind on $V$.
- Yes. The map $T$ is the only linear map from $V$ into $W$ for which the condition $forall jin{1,2,ldots,n}):Tv_j=w_j$ holds.
- Because the statment that you are proving states that $T$ is unique. Therefore, you have to prove it.
answered Jan 13 at 9:10
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
add a comment |
$begingroup$
- It means that that map $T$ is the only linear map defined on $operatorname{span}(v_1,ldots,v_n)$ for which the given condition ($forall jin{1,2,ldots,n}):Tv_j=w_j$) holds.
- Since ${v_1,ldots,v_n}$ is a basis, $operatorname{span}(v_1,ldots,v_n)=V$. Therefore, asserting that $T$ is uniquely determind in $operatorname{span}(v_1,ldots,v_n)$ is the same thing as asserting that it is uniquely determind on $V$.
- Yes. The map $T$ is the only linear map from $V$ into $W$ for which the condition $forall jin{1,2,ldots,n}):Tv_j=w_j$ holds.
- Because the statment that you are proving states that $T$ is unique. Therefore, you have to prove it.
answered Jan 13 at 9:10
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
- It means that that map $T$ is the only linear map defined on $operatorname{span}(v_1,ldots,v_n)$ for which the given condition ($forall jin{1,2,ldots,n}):Tv_j=w_j$) holds.
- Since ${v_1,ldots,v_n}$ is a basis, $operatorname{span}(v_1,ldots,v_n)=V$. Therefore, asserting that $T$ is uniquely determind in $operatorname{span}(v_1,ldots,v_n)$ is the same thing as asserting that it is uniquely determind on $V$.
- Yes. The map $T$ is the only linear map from $V$ into $W$ for which the condition $forall jin{1,2,ldots,n}):Tv_j=w_j$ holds.
- Because the statment that you are proving states that $T$ is unique. Therefore, you have to prove it.
answered Jan 13 at 9:10
José Carlos SantosJosé Carlos Santos
161k22127232
answered Jan 13 at 9:10
José Carlos SantosJosé Carlos Santos
161k22127232
answered Jan 13 at 9:10
José Carlos SantosJosé Carlos Santos
161k22127232
answered Jan 13 at 9:10
José Carlos SantosJosé Carlos Santos
161k22127232
161k22127232
add a comment |
add a comment |
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