Can I say dim range $T = $ dim null $S$ + range $S$ if $S in mathcalL(V,W)$, and $T in mathcalL(U,V)$?If two vector spaces $V$ and $W$ are isomorphic and $V$ is F-D then $W$ is F-D. Furthermore, $textdim V = textdim W$.Dimension of Range and Null Space of Composition of Two Linear Maps$textnull,T^ksubsetneqtextnull,T^k+1$ and $textrange,T^ksupsetneqtextrange,T^k+1$ for all $kinmathbbN$If $T in mathcalL(V)$ is diagonalizable and V is infinite dimensional, then $V = null (T) oplus range (T)$.If $operatornamerangeT' = operatornamespan(varphi)$, then $operatornamenull T = operatornamenull varphi$Does $textnull T^m = textnull T^m+1$ if and only if $textrange T^m = textrange T^m+1$ hold in infinite dimensional vector spaces?An intuitive argument demonstrating $dimoperatornamerangeSTleqmindimoperatornamerangeS,dimoperatornamerangeT$Proving that $dimoperatornamenullSTleqdimoperatornamenullT+dimoperatornamenullS$.How to prove that $textnull ;T_1 subset textnull ;T_2$ implies $textdim(range; T_1) geq textdim(range T_2)$?A map to a larger dimensional space is not surjective
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Can I say dim range $T = $ dim null $S$ + range $S$ if $S in mathcalL(V,W)$, and $T in mathcalL(U,V)$?
If two vector spaces $V$ and $W$ are isomorphic and $V$ is F-D then $W$ is F-D. Furthermore, $textdim V = textdim W$.Dimension of Range and Null Space of Composition of Two Linear Maps$textnull,T^ksubsetneqtextnull,T^k+1$ and $textrange,T^ksupsetneqtextrange,T^k+1$ for all $kinmathbbN$If $T in mathcalL(V)$ is diagonalizable and V is infinite dimensional, then $V = null (T) oplus range (T)$.If $operatornamerangeT' = operatornamespan(varphi)$, then $operatornamenull T = operatornamenull varphi$Does $textnull T^m = textnull T^m+1$ if and only if $textrange T^m = textrange T^m+1$ hold in infinite dimensional vector spaces?An intuitive argument demonstrating $dimoperatornamerangeSTleqmindimoperatornamerangeS,dimoperatornamerangeT$Proving that $dimoperatornamenullSTleqdimoperatornamenullT+dimoperatornamenullS$.How to prove that $textnull ;T_1 subset textnull ;T_2$ implies $textdim(range; T_1) geq textdim(range T_2)$?A map to a larger dimensional space is not surjective
$begingroup$
Can I say dim range $T = $ dim null $S$ + range $S$ if $S in mathcalL(V,W)$, and $T in mathcalL(U,V)$?
The reason I asked this is because I am doing Q22 of section 3.B on Linear Algebra Done Right.
Suppose $U$ and $V$ are finite-dimensional vector spaces and $S in mathcalL(V,W)$, and $T in mathcalL(U,V)$. Prove that $$textdim null ST leq textdim null S + textdim null T$$
My failed trail is as follow:
dim range $T = $ dim null $S$ + dim range $S$ (from S)
dim $U$ = dim null $T$ + dim range $T$ (from T)
dim $U$ = dim null $ST$ + dim range $ST$ (from $ST in mathcalL(U,W)$)
then
dim null $T$ + dim range $T$ = dim null $ST$ + dim range $ST$
dim null $T$ + dim null $S$ + dim range $S$ = dim null $ST$ + dim range $ST$
dim null $ST$ = dim null $T$ + dim null $S$ + dim range $S$ - dim range $ST$
The trial seems not working. Maybe range $T = $ dim null $S$ + dim range $S$ is wrong in the first place.
linear-algebra linear-transformations
asked yesterday
JOHN JOHN
4209
$endgroup$
add a comment |
$begingroup$
Can I say dim range $T = $ dim null $S$ + range $S$ if $S in mathcalL(V,W)$, and $T in mathcalL(U,V)$?
The reason I asked this is because I am doing Q22 of section 3.B on Linear Algebra Done Right.
Suppose $U$ and $V$ are finite-dimensional vector spaces and $S in mathcalL(V,W)$, and $T in mathcalL(U,V)$. Prove that $$textdim null ST leq textdim null S + textdim null T$$
My failed trail is as follow:
dim range $T = $ dim null $S$ + dim range $S$ (from S)
dim $U$ = dim null $T$ + dim range $T$ (from T)
dim $U$ = dim null $ST$ + dim range $ST$ (from $ST in mathcalL(U,W)$)
then
dim null $T$ + dim range $T$ = dim null $ST$ + dim range $ST$
dim null $T$ + dim null $S$ + dim range $S$ = dim null $ST$ + dim range $ST$
dim null $ST$ = dim null $T$ + dim null $S$ + dim range $S$ - dim range $ST$
The trial seems not working. Maybe range $T = $ dim null $S$ + dim range $S$ is wrong in the first place.
linear-algebra linear-transformations
asked yesterday
JOHN JOHN
4209
$endgroup$
$begingroup$
Your first step is wrong indeed. There is nothing which says that $T$ should depend on $S$. Moreover, your notation is a bit weird in that statement. You say the range (a set) of $T$ is a number plus another set. You can define adding numbers to sets but I am not sure what your thought was?
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
If ai were you I would more argue in the trend: given a 'vector' in the nullspace of $ST$, show that you can relate this to a vector in the nullspace of $S$ or $T$
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
@StanTendijck The notation was a typo. My thought was since $ST(u)$ can be written as $S(T(u))$, I thought the range of $T$ is the domain of $S$.
$endgroup$
– JOHN
yesterday
$begingroup$
If $T=0$, then $dim ,textRan, T=0$.
$endgroup$
– Yiorgos S. Smyrlis
yesterday
$begingroup$
@John ok so what you need for it to work is that the range of $T$ is a subset of the domain of $S$. Per definition this is true since $T$s domain is part of $V$ which is exactly the domain of $S$ per definition.
$endgroup$
– Stan Tendijck
yesterday
add a comment |
$begingroup$
Can I say dim range $T = $ dim null $S$ + range $S$ if $S in mathcalL(V,W)$, and $T in mathcalL(U,V)$?
The reason I asked this is because I am doing Q22 of section 3.B on Linear Algebra Done Right.
Suppose $U$ and $V$ are finite-dimensional vector spaces and $S in mathcalL(V,W)$, and $T in mathcalL(U,V)$. Prove that $$textdim null ST leq textdim null S + textdim null T$$
My failed trail is as follow:
dim range $T = $ dim null $S$ + dim range $S$ (from S)
dim $U$ = dim null $T$ + dim range $T$ (from T)
dim $U$ = dim null $ST$ + dim range $ST$ (from $ST in mathcalL(U,W)$)
then
dim null $T$ + dim range $T$ = dim null $ST$ + dim range $ST$
dim null $T$ + dim null $S$ + dim range $S$ = dim null $ST$ + dim range $ST$
dim null $ST$ = dim null $T$ + dim null $S$ + dim range $S$ - dim range $ST$
The trial seems not working. Maybe range $T = $ dim null $S$ + dim range $S$ is wrong in the first place.
linear-algebra linear-transformations
asked yesterday
JOHN JOHN
4209
$endgroup$
Can I say dim range $T = $ dim null $S$ + range $S$ if $S in mathcalL(V,W)$, and $T in mathcalL(U,V)$?
The reason I asked this is because I am doing Q22 of section 3.B on Linear Algebra Done Right.
Suppose $U$ and $V$ are finite-dimensional vector spaces and $S in mathcalL(V,W)$, and $T in mathcalL(U,V)$. Prove that $$textdim null ST leq textdim null S + textdim null T$$
My failed trail is as follow:
dim range $T = $ dim null $S$ + dim range $S$ (from S)
dim $U$ = dim null $T$ + dim range $T$ (from T)
dim $U$ = dim null $ST$ + dim range $ST$ (from $ST in mathcalL(U,W)$)
then
dim null $T$ + dim range $T$ = dim null $ST$ + dim range $ST$
dim null $T$ + dim null $S$ + dim range $S$ = dim null $ST$ + dim range $ST$
dim null $ST$ = dim null $T$ + dim null $S$ + dim range $S$ - dim range $ST$
The trial seems not working. Maybe range $T = $ dim null $S$ + dim range $S$ is wrong in the first place.
linear-algebra linear-transformations
linear-algebra linear-transformations
asked yesterday
JOHN JOHN
4209
asked yesterday
JOHN JOHN
4209
edited yesterday
JOHN
asked yesterday
JOHN JOHN
4209
asked yesterday
JOHN JOHN
4209
asked yesterday
JOHN JOHN
4209
4209
$begingroup$
Your first step is wrong indeed. There is nothing which says that $T$ should depend on $S$. Moreover, your notation is a bit weird in that statement. You say the range (a set) of $T$ is a number plus another set. You can define adding numbers to sets but I am not sure what your thought was?
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
If ai were you I would more argue in the trend: given a 'vector' in the nullspace of $ST$, show that you can relate this to a vector in the nullspace of $S$ or $T$
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
@StanTendijck The notation was a typo. My thought was since $ST(u)$ can be written as $S(T(u))$, I thought the range of $T$ is the domain of $S$.
$endgroup$
– JOHN
yesterday
$begingroup$
If $T=0$, then $dim ,textRan, T=0$.
$endgroup$
– Yiorgos S. Smyrlis
yesterday
$begingroup$
@John ok so what you need for it to work is that the range of $T$ is a subset of the domain of $S$. Per definition this is true since $T$s domain is part of $V$ which is exactly the domain of $S$ per definition.
$endgroup$
– Stan Tendijck
yesterday
add a comment |
$begingroup$
Your first step is wrong indeed. There is nothing which says that $T$ should depend on $S$. Moreover, your notation is a bit weird in that statement. You say the range (a set) of $T$ is a number plus another set. You can define adding numbers to sets but I am not sure what your thought was?
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
If ai were you I would more argue in the trend: given a 'vector' in the nullspace of $ST$, show that you can relate this to a vector in the nullspace of $S$ or $T$
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
@StanTendijck The notation was a typo. My thought was since $ST(u)$ can be written as $S(T(u))$, I thought the range of $T$ is the domain of $S$.
$endgroup$
– JOHN
yesterday
$begingroup$
If $T=0$, then $dim ,textRan, T=0$.
$endgroup$
– Yiorgos S. Smyrlis
yesterday
$begingroup$
@John ok so what you need for it to work is that the range of $T$ is a subset of the domain of $S$. Per definition this is true since $T$s domain is part of $V$ which is exactly the domain of $S$ per definition.
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
Your first step is wrong indeed. There is nothing which says that $T$ should depend on $S$. Moreover, your notation is a bit weird in that statement. You say the range (a set) of $T$ is a number plus another set. You can define adding numbers to sets but I am not sure what your thought was?
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
Your first step is wrong indeed. There is nothing which says that $T$ should depend on $S$. Moreover, your notation is a bit weird in that statement. You say the range (a set) of $T$ is a number plus another set. You can define adding numbers to sets but I am not sure what your thought was?
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
If ai were you I would more argue in the trend: given a 'vector' in the nullspace of $ST$, show that you can relate this to a vector in the nullspace of $S$ or $T$
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
If ai were you I would more argue in the trend: given a 'vector' in the nullspace of $ST$, show that you can relate this to a vector in the nullspace of $S$ or $T$
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
@StanTendijck The notation was a typo. My thought was since $ST(u)$ can be written as $S(T(u))$, I thought the range of $T$ is the domain of $S$.
$endgroup$
– JOHN
yesterday
$begingroup$
@StanTendijck The notation was a typo. My thought was since $ST(u)$ can be written as $S(T(u))$, I thought the range of $T$ is the domain of $S$.
$endgroup$
– JOHN
yesterday
$begingroup$
If $T=0$, then $dim ,textRan, T=0$.
$endgroup$
– Yiorgos S. Smyrlis
yesterday
$begingroup$
If $T=0$, then $dim ,textRan, T=0$.
$endgroup$
– Yiorgos S. Smyrlis
yesterday
$begingroup$
@John ok so what you need for it to work is that the range of $T$ is a subset of the domain of $S$. Per definition this is true since $T$s domain is part of $V$ which is exactly the domain of $S$ per definition.
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
@John ok so what you need for it to work is that the range of $T$ is a subset of the domain of $S$. Per definition this is true since $T$s domain is part of $V$ which is exactly the domain of $S$ per definition.
$endgroup$
– Stan Tendijck
yesterday
add a comment |
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$begingroup$
Your first step is wrong indeed. There is nothing which says that $T$ should depend on $S$. Moreover, your notation is a bit weird in that statement. You say the range (a set) of $T$ is a number plus another set. You can define adding numbers to sets but I am not sure what your thought was?
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
If ai were you I would more argue in the trend: given a 'vector' in the nullspace of $ST$, show that you can relate this to a vector in the nullspace of $S$ or $T$
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
@StanTendijck The notation was a typo. My thought was since $ST(u)$ can be written as $S(T(u))$, I thought the range of $T$ is the domain of $S$.
$endgroup$
– JOHN
yesterday
$begingroup$
If $T=0$, then $dim ,textRan, T=0$.
$endgroup$
– Yiorgos S. Smyrlis
yesterday
$begingroup$
@John ok so what you need for it to work is that the range of $T$ is a subset of the domain of $S$. Per definition this is true since $T$s domain is part of $V$ which is exactly the domain of $S$ per definition.
$endgroup$
– Stan Tendijck
yesterday