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let $V_1$ and $V_2$ be subspaces of $V$. then which of the following statements is Corrects?
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Let $T$ be a linear transformation on a finite dimensional vector space
$V$ and let $V_1$ and $V_2$ be subspaces of $V$. then which of the following statements is Corrects ?
$(a)$ $T(V_1cap V_2) = T(V_1) cap T(V_2).$
$(b)$ $T(V_1 cup V_2) = T(V_1) cup T(V_2).$
My attempts : i know that intersection two subspaces is subspaces and union of two subspace need not to be subspaces
so option $a)$ is correct
option $b)$ is not corrects
Is it true ??
linear-algebra linear-transformations
edited Aug 6 at 21:40
José Carlos Santos
115k1698177
asked Aug 6 at 21:31
Messi fifa
1718
add a comment |Â
up vote
0
down vote
favorite
Let $T$ be a linear transformation on a finite dimensional vector space
$V$ and let $V_1$ and $V_2$ be subspaces of $V$. then which of the following statements is Corrects ?
$(a)$ $T(V_1cap V_2) = T(V_1) cap T(V_2).$
$(b)$ $T(V_1 cup V_2) = T(V_1) cup T(V_2).$
My attempts : i know that intersection two subspaces is subspaces and union of two subspace need not to be subspaces
so option $a)$ is correct
option $b)$ is not corrects
Is it true ??
linear-algebra linear-transformations
edited Aug 6 at 21:40
José Carlos Santos
115k1698177
asked Aug 6 at 21:31
Messi fifa
1718
Have you tried to write down a standard double-inclusion argument, or perhaps a counterexample, in either case? That's where I'd start.
– The Count
Aug 6 at 21:37
1
Try approaching the question via definition and element chasing. Suppose that $xin T(V_1cap V_2)$. Then there must be some $yin V_1cap V_2$ such that $T(y)=x$. Does it follow then that $xin T(V_1)$? Does it also follow that $xin T(V_2)$? How about whether or not $xin T(V_1)cap T(V_2)$? (The element $y$ that I point out the existence of plays a critical role in these last few steps). Now... there is one more direction left for part (a) and two directions to do for part (b). Can you continue?
– JMoravitz
Aug 6 at 21:37
Your remarks about intersection and union are correct, nevertheless, that does not completely verify nor refute the equalities. Even if you have established that both sides are subspaces, you still have to show them to be equal.
– zzuussee
Aug 6 at 21:39
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $T$ be a linear transformation on a finite dimensional vector space
$V$ and let $V_1$ and $V_2$ be subspaces of $V$. then which of the following statements is Corrects ?
$(a)$ $T(V_1cap V_2) = T(V_1) cap T(V_2).$
$(b)$ $T(V_1 cup V_2) = T(V_1) cup T(V_2).$
My attempts : i know that intersection two subspaces is subspaces and union of two subspace need not to be subspaces
so option $a)$ is correct
option $b)$ is not corrects
Is it true ??
linear-algebra linear-transformations
edited Aug 6 at 21:40
José Carlos Santos
115k1698177
asked Aug 6 at 21:31
Messi fifa
1718
Let $T$ be a linear transformation on a finite dimensional vector space
$V$ and let $V_1$ and $V_2$ be subspaces of $V$. then which of the following statements is Corrects ?
$(a)$ $T(V_1cap V_2) = T(V_1) cap T(V_2).$
$(b)$ $T(V_1 cup V_2) = T(V_1) cup T(V_2).$
My attempts : i know that intersection two subspaces is subspaces and union of two subspace need not to be subspaces
so option $a)$ is correct
option $b)$ is not corrects
Is it true ??
linear-algebra linear-transformations
edited Aug 6 at 21:40
José Carlos Santos
115k1698177
asked Aug 6 at 21:31
Messi fifa
1718
edited Aug 6 at 21:40
José Carlos Santos
115k1698177
edited Aug 6 at 21:40
José Carlos Santos
115k1698177
edited Aug 6 at 21:40
José Carlos Santos
115k1698177
115k1698177
asked Aug 6 at 21:31
Messi fifa
1718
asked Aug 6 at 21:31
Messi fifa
1718
asked Aug 6 at 21:31
Messi fifa
1718
1718
Have you tried to write down a standard double-inclusion argument, or perhaps a counterexample, in either case? That's where I'd start.
– The Count
Aug 6 at 21:37
1
Try approaching the question via definition and element chasing. Suppose that $xin T(V_1cap V_2)$. Then there must be some $yin V_1cap V_2$ such that $T(y)=x$. Does it follow then that $xin T(V_1)$? Does it also follow that $xin T(V_2)$? How about whether or not $xin T(V_1)cap T(V_2)$? (The element $y$ that I point out the existence of plays a critical role in these last few steps). Now... there is one more direction left for part (a) and two directions to do for part (b). Can you continue?
– JMoravitz
Aug 6 at 21:37
Your remarks about intersection and union are correct, nevertheless, that does not completely verify nor refute the equalities. Even if you have established that both sides are subspaces, you still have to show them to be equal.
– zzuussee
Aug 6 at 21:39
add a comment |Â
Have you tried to write down a standard double-inclusion argument, or perhaps a counterexample, in either case? That's where I'd start.
– The Count
Aug 6 at 21:37
1
Try approaching the question via definition and element chasing. Suppose that $xin T(V_1cap V_2)$. Then there must be some $yin V_1cap V_2$ such that $T(y)=x$. Does it follow then that $xin T(V_1)$? Does it also follow that $xin T(V_2)$? How about whether or not $xin T(V_1)cap T(V_2)$? (The element $y$ that I point out the existence of plays a critical role in these last few steps). Now... there is one more direction left for part (a) and two directions to do for part (b). Can you continue?
– JMoravitz
Aug 6 at 21:37
Your remarks about intersection and union are correct, nevertheless, that does not completely verify nor refute the equalities. Even if you have established that both sides are subspaces, you still have to show them to be equal.
– zzuussee
Aug 6 at 21:39
Have you tried to write down a standard double-inclusion argument, or perhaps a counterexample, in either case? That's where I'd start.
– The Count
Aug 6 at 21:37
Have you tried to write down a standard double-inclusion argument, or perhaps a counterexample, in either case? That's where I'd start.
– The Count
Aug 6 at 21:37
1
1
Try approaching the question via definition and element chasing. Suppose that $xin T(V_1cap V_2)$. Then there must be some $yin V_1cap V_2$ such that $T(y)=x$. Does it follow then that $xin T(V_1)$? Does it also follow that $xin T(V_2)$? How about whether or not $xin T(V_1)cap T(V_2)$? (The element $y$ that I point out the existence of plays a critical role in these last few steps). Now... there is one more direction left for part (a) and two directions to do for part (b). Can you continue?
– JMoravitz
Aug 6 at 21:37
Try approaching the question via definition and element chasing. Suppose that $xin T(V_1cap V_2)$. Then there must be some $yin V_1cap V_2$ such that $T(y)=x$. Does it follow then that $xin T(V_1)$? Does it also follow that $xin T(V_2)$? How about whether or not $xin T(V_1)cap T(V_2)$? (The element $y$ that I point out the existence of plays a critical role in these last few steps). Now... there is one more direction left for part (a) and two directions to do for part (b). Can you continue?
– JMoravitz
Aug 6 at 21:37
Your remarks about intersection and union are correct, nevertheless, that does not completely verify nor refute the equalities. Even if you have established that both sides are subspaces, you still have to show them to be equal.
– zzuussee
Aug 6 at 21:39
Your remarks about intersection and union are correct, nevertheless, that does not completely verify nor refute the equalities. Even if you have established that both sides are subspaces, you still have to show them to be equal.
– zzuussee
Aug 6 at 21:39
add a comment |Â
3 Answers
3
active
oldest
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up vote
2
down vote
accepted
(a) is false. Take $TcolonmathbbR^2longrightarrowmathbbR^2$ defined by $f(x,y)=(x+y,0)$. If$$V_1=,xinmathbbRtext and V_2=(0,x),,$$then $T(V_1cap V_2)=T(0)=0$, whereas $T(V_1)cap T(V_2)=,xinmathbbR$.
(b) this holds for every funtion (linear or otherwise).
answered Aug 6 at 21:37
José Carlos Santos
115k1698177
for b) if i will take ur counter example as u have taken for option a) then it will not True ..can elaborate option b) more ...@Jose sir
– Messi fifa
Aug 6 at 21:45
1
@Messififa $T(V_1cup V_2)=T(V_1)cup T(V_2)=,xinmathbbR$.
– José Carlos Santos
Aug 6 at 21:48
add a comment |Â
up vote
1
down vote
No, in contrary $$T(V_1 cup V_2) = T(V_1) cup T(V_2)$$is the true choice. Note that $$ V_1 cup V_2$$ does not have to be a subspace for the above statement to be true.
answered Aug 6 at 21:41
Mohammad Riazi-Kermani
27.8k41852
add a comment |Â
up vote
1
down vote
Your reasoning is wrong - the question didn't ask whether any of those sets were subspaces.
And your conclusions are both wrong as well. Another answer already shows that (a) is false. In fact (b) is true.
If $yin T(V_1cup V_2)$ then there exists $xin V_1cup V_2$ with $y=T(x)$. But $xin V_1cup V_2$ says $xin V_1$ or $xin V_2$; hence $y=T(x)$ is either in $T(V_1)$ or $T(V_2)$, so $yin T(V_1)cup T(V_2)$.
So $T(V_1cup V_2)subset T(V_1)cup T(V_2)$. Conversely, $V_1subset V_1cup V_2$ shows that $T(V_1)subset T(V_1cup V_2)$. Similarly $T(V_2)subset T(V_1cup V_2)$, hence $T(V_1)cup T(V_2)subset T(V_1cup V_2)$.
answered Aug 6 at 21:47
David C. Ullrich
54.4k33684
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
(a) is false. Take $TcolonmathbbR^2longrightarrowmathbbR^2$ defined by $f(x,y)=(x+y,0)$. If$$V_1=,xinmathbbRtext and V_2=(0,x),,$$then $T(V_1cap V_2)=T(0)=0$, whereas $T(V_1)cap T(V_2)=,xinmathbbR$.
(b) this holds for every funtion (linear or otherwise).
answered Aug 6 at 21:37
José Carlos Santos
115k1698177
for b) if i will take ur counter example as u have taken for option a) then it will not True ..can elaborate option b) more ...@Jose sir
– Messi fifa
Aug 6 at 21:45
1
@Messififa $T(V_1cup V_2)=T(V_1)cup T(V_2)=,xinmathbbR$.
– José Carlos Santos
Aug 6 at 21:48
add a comment |Â
up vote
2
down vote
accepted
(a) is false. Take $TcolonmathbbR^2longrightarrowmathbbR^2$ defined by $f(x,y)=(x+y,0)$. If$$V_1=,xinmathbbRtext and V_2=(0,x),,$$then $T(V_1cap V_2)=T(0)=0$, whereas $T(V_1)cap T(V_2)=,xinmathbbR$.
(b) this holds for every funtion (linear or otherwise).
answered Aug 6 at 21:37
José Carlos Santos
115k1698177
for b) if i will take ur counter example as u have taken for option a) then it will not True ..can elaborate option b) more ...@Jose sir
– Messi fifa
Aug 6 at 21:45
1
@Messififa $T(V_1cup V_2)=T(V_1)cup T(V_2)=,xinmathbbR$.
– José Carlos Santos
Aug 6 at 21:48
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
(a) is false. Take $TcolonmathbbR^2longrightarrowmathbbR^2$ defined by $f(x,y)=(x+y,0)$. If$$V_1=,xinmathbbRtext and V_2=(0,x),,$$then $T(V_1cap V_2)=T(0)=0$, whereas $T(V_1)cap T(V_2)=,xinmathbbR$.
(b) this holds for every funtion (linear or otherwise).
answered Aug 6 at 21:37
José Carlos Santos
115k1698177
(a) is false. Take $TcolonmathbbR^2longrightarrowmathbbR^2$ defined by $f(x,y)=(x+y,0)$. If$$V_1=,xinmathbbRtext and V_2=(0,x),,$$then $T(V_1cap V_2)=T(0)=0$, whereas $T(V_1)cap T(V_2)=,xinmathbbR$.
(b) this holds for every funtion (linear or otherwise).
answered Aug 6 at 21:37
José Carlos Santos
115k1698177
answered Aug 6 at 21:37
José Carlos Santos
115k1698177
answered Aug 6 at 21:37
José Carlos Santos
115k1698177
answered Aug 6 at 21:37
José Carlos Santos
115k1698177
115k1698177
for b) if i will take ur counter example as u have taken for option a) then it will not True ..can elaborate option b) more ...@Jose sir
– Messi fifa
Aug 6 at 21:45
1
@Messififa $T(V_1cup V_2)=T(V_1)cup T(V_2)=,xinmathbbR$.
– José Carlos Santos
Aug 6 at 21:48
add a comment |Â
for b) if i will take ur counter example as u have taken for option a) then it will not True ..can elaborate option b) more ...@Jose sir
– Messi fifa
Aug 6 at 21:45
1
@Messififa $T(V_1cup V_2)=T(V_1)cup T(V_2)=,xinmathbbR$.
– José Carlos Santos
Aug 6 at 21:48
for b) if i will take ur counter example as u have taken for option a) then it will not True ..can elaborate option b) more ...@Jose sir
– Messi fifa
Aug 6 at 21:45
for b) if i will take ur counter example as u have taken for option a) then it will not True ..can elaborate option b) more ...@Jose sir
– Messi fifa
Aug 6 at 21:45
1
1
@Messififa $T(V_1cup V_2)=T(V_1)cup T(V_2)=,xinmathbbR$.
– José Carlos Santos
Aug 6 at 21:48
@Messififa $T(V_1cup V_2)=T(V_1)cup T(V_2)=,xinmathbbR$.
– José Carlos Santos
Aug 6 at 21:48
add a comment |Â
up vote
1
down vote
No, in contrary $$T(V_1 cup V_2) = T(V_1) cup T(V_2)$$is the true choice. Note that $$ V_1 cup V_2$$ does not have to be a subspace for the above statement to be true.
answered Aug 6 at 21:41
Mohammad Riazi-Kermani
27.8k41852
add a comment |Â
up vote
1
down vote
No, in contrary $$T(V_1 cup V_2) = T(V_1) cup T(V_2)$$is the true choice. Note that $$ V_1 cup V_2$$ does not have to be a subspace for the above statement to be true.
answered Aug 6 at 21:41
Mohammad Riazi-Kermani
27.8k41852
add a comment |Â
up vote
1
down vote
up vote
1
down vote
No, in contrary $$T(V_1 cup V_2) = T(V_1) cup T(V_2)$$is the true choice. Note that $$ V_1 cup V_2$$ does not have to be a subspace for the above statement to be true.
answered Aug 6 at 21:41
Mohammad Riazi-Kermani
27.8k41852
No, in contrary $$T(V_1 cup V_2) = T(V_1) cup T(V_2)$$is the true choice. Note that $$ V_1 cup V_2$$ does not have to be a subspace for the above statement to be true.
answered Aug 6 at 21:41
Mohammad Riazi-Kermani
27.8k41852
answered Aug 6 at 21:41
Mohammad Riazi-Kermani
27.8k41852
answered Aug 6 at 21:41
Mohammad Riazi-Kermani
27.8k41852
answered Aug 6 at 21:41
Mohammad Riazi-Kermani
27.8k41852
27.8k41852
add a comment |Â
add a comment |Â
up vote
1
down vote
Your reasoning is wrong - the question didn't ask whether any of those sets were subspaces.
And your conclusions are both wrong as well. Another answer already shows that (a) is false. In fact (b) is true.
If $yin T(V_1cup V_2)$ then there exists $xin V_1cup V_2$ with $y=T(x)$. But $xin V_1cup V_2$ says $xin V_1$ or $xin V_2$; hence $y=T(x)$ is either in $T(V_1)$ or $T(V_2)$, so $yin T(V_1)cup T(V_2)$.
So $T(V_1cup V_2)subset T(V_1)cup T(V_2)$. Conversely, $V_1subset V_1cup V_2$ shows that $T(V_1)subset T(V_1cup V_2)$. Similarly $T(V_2)subset T(V_1cup V_2)$, hence $T(V_1)cup T(V_2)subset T(V_1cup V_2)$.
answered Aug 6 at 21:47
David C. Ullrich
54.4k33684
add a comment |Â
up vote
1
down vote
Your reasoning is wrong - the question didn't ask whether any of those sets were subspaces.
And your conclusions are both wrong as well. Another answer already shows that (a) is false. In fact (b) is true.
If $yin T(V_1cup V_2)$ then there exists $xin V_1cup V_2$ with $y=T(x)$. But $xin V_1cup V_2$ says $xin V_1$ or $xin V_2$; hence $y=T(x)$ is either in $T(V_1)$ or $T(V_2)$, so $yin T(V_1)cup T(V_2)$.
So $T(V_1cup V_2)subset T(V_1)cup T(V_2)$. Conversely, $V_1subset V_1cup V_2$ shows that $T(V_1)subset T(V_1cup V_2)$. Similarly $T(V_2)subset T(V_1cup V_2)$, hence $T(V_1)cup T(V_2)subset T(V_1cup V_2)$.
answered Aug 6 at 21:47
David C. Ullrich
54.4k33684
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your reasoning is wrong - the question didn't ask whether any of those sets were subspaces.
And your conclusions are both wrong as well. Another answer already shows that (a) is false. In fact (b) is true.
If $yin T(V_1cup V_2)$ then there exists $xin V_1cup V_2$ with $y=T(x)$. But $xin V_1cup V_2$ says $xin V_1$ or $xin V_2$; hence $y=T(x)$ is either in $T(V_1)$ or $T(V_2)$, so $yin T(V_1)cup T(V_2)$.
So $T(V_1cup V_2)subset T(V_1)cup T(V_2)$. Conversely, $V_1subset V_1cup V_2$ shows that $T(V_1)subset T(V_1cup V_2)$. Similarly $T(V_2)subset T(V_1cup V_2)$, hence $T(V_1)cup T(V_2)subset T(V_1cup V_2)$.
answered Aug 6 at 21:47
David C. Ullrich
54.4k33684
Your reasoning is wrong - the question didn't ask whether any of those sets were subspaces.
And your conclusions are both wrong as well. Another answer already shows that (a) is false. In fact (b) is true.
If $yin T(V_1cup V_2)$ then there exists $xin V_1cup V_2$ with $y=T(x)$. But $xin V_1cup V_2$ says $xin V_1$ or $xin V_2$; hence $y=T(x)$ is either in $T(V_1)$ or $T(V_2)$, so $yin T(V_1)cup T(V_2)$.
So $T(V_1cup V_2)subset T(V_1)cup T(V_2)$. Conversely, $V_1subset V_1cup V_2$ shows that $T(V_1)subset T(V_1cup V_2)$. Similarly $T(V_2)subset T(V_1cup V_2)$, hence $T(V_1)cup T(V_2)subset T(V_1cup V_2)$.
answered Aug 6 at 21:47
David C. Ullrich
54.4k33684
answered Aug 6 at 21:47
David C. Ullrich
54.4k33684
answered Aug 6 at 21:47
David C. Ullrich
54.4k33684
answered Aug 6 at 21:47
David C. Ullrich
54.4k33684
54.4k33684
add a comment |Â
add a comment |Â
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Have you tried to write down a standard double-inclusion argument, or perhaps a counterexample, in either case? That's where I'd start.
– The Count
Aug 6 at 21:37
1
Try approaching the question via definition and element chasing. Suppose that $xin T(V_1cap V_2)$. Then there must be some $yin V_1cap V_2$ such that $T(y)=x$. Does it follow then that $xin T(V_1)$? Does it also follow that $xin T(V_2)$? How about whether or not $xin T(V_1)cap T(V_2)$? (The element $y$ that I point out the existence of plays a critical role in these last few steps). Now... there is one more direction left for part (a) and two directions to do for part (b). Can you continue?
– JMoravitz
Aug 6 at 21:37
Your remarks about intersection and union are correct, nevertheless, that does not completely verify nor refute the equalities. Even if you have established that both sides are subspaces, you still have to show them to be equal.
– zzuussee
Aug 6 at 21:39