Compute the dim$Z(v)$ in case $v neq 0$, $Z(v) = T in L(V, W) $$T: Vrightarrow W$ is an injective linear transformation when restricted to subspace $A$ of $V$. Then can we conclude that $dim(A) = dim T(A)$some corollaries of the rank - nullity theoremIs there a linear transformation $psi: Mat_3x2(ℝ) mapsto ℝ_8[x]$ that is surjective?$dim(V) = dim T(V) + dim T^-1(0)$What does it mean when dim(V)=rankTmotivation and theoritical clarification of some linear transformation related conceptsSimplest Proof of the Dimensions theoremWhy is a map to a smaller dimensional space not injective?Is $textRank (T) = textDim(V)$ all the time?$renewcommandImtextIm$Prove that $dim(ker(T))=dim(ker(T^2))$ if $dim(Im(T))=dim(Im(T^2))$
Fixing conmutation for high voltage switching with power mosfet
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Compute the dim$Z(v)$ in case $v neq 0$, $Z(v) = T in L(V, W) $
$T: Vrightarrow W$ is an injective linear transformation when restricted to subspace $A$ of $V$. Then can we conclude that $dim(A) = dim T(A)$some corollaries of the rank - nullity theoremIs there a linear transformation $psi: Mat_3x2(ℝ) mapsto ℝ_8[x]$ that is surjective?$dim(V) = dim T(V) + dim T^-1(0)$What does it mean when dim(V)=rankTmotivation and theoritical clarification of some linear transformation related conceptsSimplest Proof of the Dimensions theoremWhy is a map to a smaller dimensional space not injective?Is $textRank (T) = textDim(V)$ all the time?$renewcommandImtextIm$Prove that $dim(ker(T))=dim(ker(T^2))$ if $dim(Im(T))=dim(Im(T^2))$
$begingroup$
I want to find the dimZ(v).
V and W are finite dimensional vector spaces over $F$. and v is a vector of V and T is a linear Transformation.
Since $v neq 0$ I know $N(T) neq 0$, $T(v) = 0$ $forall T(v) in Z(v).$ This implies $1leq dim N(T) leq dimV$. I don't know where to go from here. I think you can't get anywhere without knowing what v is or F is.
N(T) is the nullity of T
linear-algebra
asked 2 days ago
ConorConor
205
$endgroup$
add a comment |
$begingroup$
I want to find the dimZ(v).
V and W are finite dimensional vector spaces over $F$. and v is a vector of V and T is a linear Transformation.
Since $v neq 0$ I know $N(T) neq 0$, $T(v) = 0$ $forall T(v) in Z(v).$ This implies $1leq dim N(T) leq dimV$. I don't know where to go from here. I think you can't get anywhere without knowing what v is or F is.
N(T) is the nullity of T
linear-algebra
asked 2 days ago
ConorConor
205
$endgroup$
1
$begingroup$
I'm sorry. I'm new to mathjax/Latex and this site so im having a little difficulty.
$endgroup$
– Conor
2 days ago
add a comment |
$begingroup$
I want to find the dimZ(v).
V and W are finite dimensional vector spaces over $F$. and v is a vector of V and T is a linear Transformation.
Since $v neq 0$ I know $N(T) neq 0$, $T(v) = 0$ $forall T(v) in Z(v).$ This implies $1leq dim N(T) leq dimV$. I don't know where to go from here. I think you can't get anywhere without knowing what v is or F is.
N(T) is the nullity of T
linear-algebra
asked 2 days ago
ConorConor
205
$endgroup$
I want to find the dimZ(v).
V and W are finite dimensional vector spaces over $F$. and v is a vector of V and T is a linear Transformation.
Since $v neq 0$ I know $N(T) neq 0$, $T(v) = 0$ $forall T(v) in Z(v).$ This implies $1leq dim N(T) leq dimV$. I don't know where to go from here. I think you can't get anywhere without knowing what v is or F is.
N(T) is the nullity of T
linear-algebra
linear-algebra
asked 2 days ago
ConorConor
205
asked 2 days ago
ConorConor
205
edited 2 days ago
Conor
asked 2 days ago
ConorConor
205
asked 2 days ago
ConorConor
205
asked 2 days ago
ConorConor
205
205
1
$begingroup$
I'm sorry. I'm new to mathjax/Latex and this site so im having a little difficulty.
$endgroup$
– Conor
2 days ago
add a comment |
1
$begingroup$
I'm sorry. I'm new to mathjax/Latex and this site so im having a little difficulty.
$endgroup$
– Conor
2 days ago
1
1
$begingroup$
I'm sorry. I'm new to mathjax/Latex and this site so im having a little difficulty.
$endgroup$
– Conor
2 days ago
$begingroup$
I'm sorry. I'm new to mathjax/Latex and this site so im having a little difficulty.
$endgroup$
– Conor
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Observe that the map
$$phi:L(V,W)to W,,,,textdefined by;;phi T:=Tv$$
is linear (proof?), and precisely $;kerphi=Z(v);$ . On the other hand, $;phi;$ is "clearly" surjective (proof? This is not trivial and perhaps is the hardest part of all this). Now use the dimensions theorem to deduce $;dim Z(v);$ ...
answered 2 days ago
DonAntonioDonAntonio
179k1494233
$endgroup$
$begingroup$
$L(V, W)$ are all transformations from V to W. How can you map a transformation to a vector space?
$endgroup$
– Conor
2 days ago
$begingroup$
$;L(V,W);$ is a vector space, and I defined above a map between vector spaces which is linear...Is anything unclear?
$endgroup$
– DonAntonio
2 days ago
$begingroup$
By Tv you mean T(v)?
$endgroup$
– Conor
2 days ago
$begingroup$
@Conor I mean "the map $T$ applied on the vector $v$" . Yes, I guess another way to write that is as $;T(v);$ ...It is the same, of course.
$endgroup$
– DonAntonio
2 days ago
$begingroup$
By the way, I think $$dim Z(V)=left(dim V-1right)dim W$$ according to the answer above.
$endgroup$
– DonAntonio
2 days ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Observe that the map
$$phi:L(V,W)to W,,,,textdefined by;;phi T:=Tv$$
is linear (proof?), and precisely $;kerphi=Z(v);$ . On the other hand, $;phi;$ is "clearly" surjective (proof? This is not trivial and perhaps is the hardest part of all this). Now use the dimensions theorem to deduce $;dim Z(v);$ ...
answered 2 days ago
DonAntonioDonAntonio
179k1494233
$endgroup$
$begingroup$
$L(V, W)$ are all transformations from V to W. How can you map a transformation to a vector space?
$endgroup$
– Conor
2 days ago
$begingroup$
$;L(V,W);$ is a vector space, and I defined above a map between vector spaces which is linear...Is anything unclear?
$endgroup$
– DonAntonio
2 days ago
$begingroup$
By Tv you mean T(v)?
$endgroup$
– Conor
2 days ago
$begingroup$
@Conor I mean "the map $T$ applied on the vector $v$" . Yes, I guess another way to write that is as $;T(v);$ ...It is the same, of course.
$endgroup$
– DonAntonio
2 days ago
$begingroup$
By the way, I think $$dim Z(V)=left(dim V-1right)dim W$$ according to the answer above.
$endgroup$
– DonAntonio
2 days ago
add a comment |
$begingroup$
Observe that the map
$$phi:L(V,W)to W,,,,textdefined by;;phi T:=Tv$$
is linear (proof?), and precisely $;kerphi=Z(v);$ . On the other hand, $;phi;$ is "clearly" surjective (proof? This is not trivial and perhaps is the hardest part of all this). Now use the dimensions theorem to deduce $;dim Z(v);$ ...
answered 2 days ago
DonAntonioDonAntonio
179k1494233
$endgroup$
$begingroup$
$L(V, W)$ are all transformations from V to W. How can you map a transformation to a vector space?
$endgroup$
– Conor
2 days ago
$begingroup$
$;L(V,W);$ is a vector space, and I defined above a map between vector spaces which is linear...Is anything unclear?
$endgroup$
– DonAntonio
2 days ago
$begingroup$
By Tv you mean T(v)?
$endgroup$
– Conor
2 days ago
$begingroup$
@Conor I mean "the map $T$ applied on the vector $v$" . Yes, I guess another way to write that is as $;T(v);$ ...It is the same, of course.
$endgroup$
– DonAntonio
2 days ago
$begingroup$
By the way, I think $$dim Z(V)=left(dim V-1right)dim W$$ according to the answer above.
$endgroup$
– DonAntonio
2 days ago
add a comment |
$begingroup$
Observe that the map
$$phi:L(V,W)to W,,,,textdefined by;;phi T:=Tv$$
is linear (proof?), and precisely $;kerphi=Z(v);$ . On the other hand, $;phi;$ is "clearly" surjective (proof? This is not trivial and perhaps is the hardest part of all this). Now use the dimensions theorem to deduce $;dim Z(v);$ ...
answered 2 days ago
DonAntonioDonAntonio
179k1494233
$endgroup$
Observe that the map
$$phi:L(V,W)to W,,,,textdefined by;;phi T:=Tv$$
is linear (proof?), and precisely $;kerphi=Z(v);$ . On the other hand, $;phi;$ is "clearly" surjective (proof? This is not trivial and perhaps is the hardest part of all this). Now use the dimensions theorem to deduce $;dim Z(v);$ ...
answered 2 days ago
DonAntonioDonAntonio
179k1494233
answered 2 days ago
DonAntonioDonAntonio
179k1494233
answered 2 days ago
DonAntonioDonAntonio
179k1494233
answered 2 days ago
DonAntonioDonAntonio
179k1494233
179k1494233
$begingroup$
$L(V, W)$ are all transformations from V to W. How can you map a transformation to a vector space?
$endgroup$
– Conor
2 days ago
$begingroup$
$;L(V,W);$ is a vector space, and I defined above a map between vector spaces which is linear...Is anything unclear?
$endgroup$
– DonAntonio
2 days ago
$begingroup$
By Tv you mean T(v)?
$endgroup$
– Conor
2 days ago
$begingroup$
@Conor I mean "the map $T$ applied on the vector $v$" . Yes, I guess another way to write that is as $;T(v);$ ...It is the same, of course.
$endgroup$
– DonAntonio
2 days ago
$begingroup$
By the way, I think $$dim Z(V)=left(dim V-1right)dim W$$ according to the answer above.
$endgroup$
– DonAntonio
2 days ago
add a comment |
$begingroup$
$L(V, W)$ are all transformations from V to W. How can you map a transformation to a vector space?
$endgroup$
– Conor
2 days ago
$begingroup$
$;L(V,W);$ is a vector space, and I defined above a map between vector spaces which is linear...Is anything unclear?
$endgroup$
– DonAntonio
2 days ago
$begingroup$
By Tv you mean T(v)?
$endgroup$
– Conor
2 days ago
$begingroup$
@Conor I mean "the map $T$ applied on the vector $v$" . Yes, I guess another way to write that is as $;T(v);$ ...It is the same, of course.
$endgroup$
– DonAntonio
2 days ago
$begingroup$
By the way, I think $$dim Z(V)=left(dim V-1right)dim W$$ according to the answer above.
$endgroup$
– DonAntonio
2 days ago
$begingroup$
$L(V, W)$ are all transformations from V to W. How can you map a transformation to a vector space?
$endgroup$
– Conor
2 days ago
$begingroup$
$L(V, W)$ are all transformations from V to W. How can you map a transformation to a vector space?
$endgroup$
– Conor
2 days ago
$begingroup$
$;L(V,W);$ is a vector space, and I defined above a map between vector spaces which is linear...Is anything unclear?
$endgroup$
– DonAntonio
2 days ago
$begingroup$
$;L(V,W);$ is a vector space, and I defined above a map between vector spaces which is linear...Is anything unclear?
$endgroup$
– DonAntonio
2 days ago
$begingroup$
By Tv you mean T(v)?
$endgroup$
– Conor
2 days ago
$begingroup$
By Tv you mean T(v)?
$endgroup$
– Conor
2 days ago
$begingroup$
@Conor I mean "the map $T$ applied on the vector $v$" . Yes, I guess another way to write that is as $;T(v);$ ...It is the same, of course.
$endgroup$
– DonAntonio
2 days ago
$begingroup$
@Conor I mean "the map $T$ applied on the vector $v$" . Yes, I guess another way to write that is as $;T(v);$ ...It is the same, of course.
$endgroup$
– DonAntonio
2 days ago
$begingroup$
By the way, I think $$dim Z(V)=left(dim V-1right)dim W$$ according to the answer above.
$endgroup$
– DonAntonio
2 days ago
$begingroup$
By the way, I think $$dim Z(V)=left(dim V-1right)dim W$$ according to the answer above.
$endgroup$
– DonAntonio
2 days ago
add a comment |
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$begingroup$
I'm sorry. I'm new to mathjax/Latex and this site so im having a little difficulty.
$endgroup$
– Conor
2 days ago