How can I prove that?
$begingroup$
If
T
:
V
→
W
is a linear transformation.
If $ker T$ has ${u_1,dots,u_n}$ as basis.
If image of $T$ has ${T(v_1),dots,T(v_m)}$ as basis.
How can I prove that
$$
{v_1,dots,v_m,u_1,dots,u_n}
$$
is a basis for $V$?
I'm a inicial student so I don't have a lot of knowlodge.
linear-algebra vector-spaces linear-transformations
edited Jan 16 at 10:16
José Carlos Santos
159k22126231
asked Jan 16 at 9:31
HarryHarry
83
$endgroup$
add a comment |
$begingroup$
If
T
:
V
→
W
is a linear transformation.
If $ker T$ has ${u_1,dots,u_n}$ as basis.
If image of $T$ has ${T(v_1),dots,T(v_m)}$ as basis.
How can I prove that
$$
{v_1,dots,v_m,u_1,dots,u_n}
$$
is a basis for $V$?
I'm a inicial student so I don't have a lot of knowlodge.
linear-algebra vector-spaces linear-transformations
edited Jan 16 at 10:16
José Carlos Santos
159k22126231
asked Jan 16 at 9:31
HarryHarry
83
$endgroup$
add a comment |
$begingroup$
If
T
:
V
→
W
is a linear transformation.
If $ker T$ has ${u_1,dots,u_n}$ as basis.
If image of $T$ has ${T(v_1),dots,T(v_m)}$ as basis.
How can I prove that
$$
{v_1,dots,v_m,u_1,dots,u_n}
$$
is a basis for $V$?
I'm a inicial student so I don't have a lot of knowlodge.
linear-algebra vector-spaces linear-transformations
edited Jan 16 at 10:16
José Carlos Santos
159k22126231
asked Jan 16 at 9:31
HarryHarry
83
$endgroup$
If
T
:
V
→
W
is a linear transformation.
If $ker T$ has ${u_1,dots,u_n}$ as basis.
If image of $T$ has ${T(v_1),dots,T(v_m)}$ as basis.
How can I prove that
$$
{v_1,dots,v_m,u_1,dots,u_n}
$$
is a basis for $V$?
I'm a inicial student so I don't have a lot of knowlodge.
linear-algebra vector-spaces linear-transformations
linear-algebra vector-spaces linear-transformations
edited Jan 16 at 10:16
José Carlos Santos
159k22126231
asked Jan 16 at 9:31
HarryHarry
83
edited Jan 16 at 10:16
José Carlos Santos
159k22126231
asked Jan 16 at 9:31
HarryHarry
83
edited Jan 16 at 10:16
José Carlos Santos
159k22126231
edited Jan 16 at 10:16
José Carlos Santos
159k22126231
edited Jan 16 at 10:16
José Carlos Santos
159k22126231
159k22126231
asked Jan 16 at 9:31
HarryHarry
83
asked Jan 16 at 9:31
HarryHarry
83
asked Jan 16 at 9:31
HarryHarry
83
83
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $B={v_1,ldots,v_m,u_1,ldots,u_n}$. There are two things that must be checked: that $B$ spans $V$ and that it is linearly independent.
In order to prove that $operatorname{span}(B)=V$. Take $vin V$. Then $f(v)$ can be written as $alpha_1 f(v_1)+cdots+alpha_m f(v_m)$. In other words, $f(v)=f(alpha_1v_1+cdots+alpha_mv_m)$. But then $v-(alpha_1v_1+cdots+alpha_mv_m)inker f$ and therefore it can be written as $beta_1u_1+cdots+beta_nw_n$. So$$v=alpha_1v_1+cdots+alpha_mv_m+beta_1u_1+cdots+beta_nw_n.$$
On the other hand, if$$alpha_1v_1+cdots+alpha_mv_m+beta_1u_1+cdots+beta_nw_n=0,tag1$$then$$0=f(0)=alpha_1 f(v_1)+cdots+alpha_mf(v_m)$$and therefore every $alpha_j$ is equal to $0$. But then $(1)$ means that $beta_1u_1+cdots+beta_nw_n=0$ and so every $beta_k$ is equal to $0$ too.
answered Jan 16 at 9:38
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
add a comment |
$begingroup$
Highlights of proof:
Take $;vin V;$ , then
$$Tvintext{Im},Timplies Tv=sum_{k=1}^m a_kTv_kimplies Tleft(v-sum_{k=1}^m a_kv_kright)=0implies v-sum_{k=1}^m a_kv_kinker T$$
and now you try to complete the last few steps to finish the proof, after you're sure you can justify all the steps above.
This is practically the proof of the well-known and very important Dimensions Theorem for finite dimensional spaces (in fact, only $;V;$ , the definition domain, has to be finite dimensional)
answered Jan 16 at 9:37
DonAntonioDonAntonio
178k1493229
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let $B={v_1,ldots,v_m,u_1,ldots,u_n}$. There are two things that must be checked: that $B$ spans $V$ and that it is linearly independent.
In order to prove that $operatorname{span}(B)=V$. Take $vin V$. Then $f(v)$ can be written as $alpha_1 f(v_1)+cdots+alpha_m f(v_m)$. In other words, $f(v)=f(alpha_1v_1+cdots+alpha_mv_m)$. But then $v-(alpha_1v_1+cdots+alpha_mv_m)inker f$ and therefore it can be written as $beta_1u_1+cdots+beta_nw_n$. So$$v=alpha_1v_1+cdots+alpha_mv_m+beta_1u_1+cdots+beta_nw_n.$$
On the other hand, if$$alpha_1v_1+cdots+alpha_mv_m+beta_1u_1+cdots+beta_nw_n=0,tag1$$then$$0=f(0)=alpha_1 f(v_1)+cdots+alpha_mf(v_m)$$and therefore every $alpha_j$ is equal to $0$. But then $(1)$ means that $beta_1u_1+cdots+beta_nw_n=0$ and so every $beta_k$ is equal to $0$ too.
answered Jan 16 at 9:38
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
add a comment |
$begingroup$
Let $B={v_1,ldots,v_m,u_1,ldots,u_n}$. There are two things that must be checked: that $B$ spans $V$ and that it is linearly independent.
In order to prove that $operatorname{span}(B)=V$. Take $vin V$. Then $f(v)$ can be written as $alpha_1 f(v_1)+cdots+alpha_m f(v_m)$. In other words, $f(v)=f(alpha_1v_1+cdots+alpha_mv_m)$. But then $v-(alpha_1v_1+cdots+alpha_mv_m)inker f$ and therefore it can be written as $beta_1u_1+cdots+beta_nw_n$. So$$v=alpha_1v_1+cdots+alpha_mv_m+beta_1u_1+cdots+beta_nw_n.$$
On the other hand, if$$alpha_1v_1+cdots+alpha_mv_m+beta_1u_1+cdots+beta_nw_n=0,tag1$$then$$0=f(0)=alpha_1 f(v_1)+cdots+alpha_mf(v_m)$$and therefore every $alpha_j$ is equal to $0$. But then $(1)$ means that $beta_1u_1+cdots+beta_nw_n=0$ and so every $beta_k$ is equal to $0$ too.
answered Jan 16 at 9:38
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
add a comment |
$begingroup$
Let $B={v_1,ldots,v_m,u_1,ldots,u_n}$. There are two things that must be checked: that $B$ spans $V$ and that it is linearly independent.
In order to prove that $operatorname{span}(B)=V$. Take $vin V$. Then $f(v)$ can be written as $alpha_1 f(v_1)+cdots+alpha_m f(v_m)$. In other words, $f(v)=f(alpha_1v_1+cdots+alpha_mv_m)$. But then $v-(alpha_1v_1+cdots+alpha_mv_m)inker f$ and therefore it can be written as $beta_1u_1+cdots+beta_nw_n$. So$$v=alpha_1v_1+cdots+alpha_mv_m+beta_1u_1+cdots+beta_nw_n.$$
On the other hand, if$$alpha_1v_1+cdots+alpha_mv_m+beta_1u_1+cdots+beta_nw_n=0,tag1$$then$$0=f(0)=alpha_1 f(v_1)+cdots+alpha_mf(v_m)$$and therefore every $alpha_j$ is equal to $0$. But then $(1)$ means that $beta_1u_1+cdots+beta_nw_n=0$ and so every $beta_k$ is equal to $0$ too.
answered Jan 16 at 9:38
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
Let $B={v_1,ldots,v_m,u_1,ldots,u_n}$. There are two things that must be checked: that $B$ spans $V$ and that it is linearly independent.
In order to prove that $operatorname{span}(B)=V$. Take $vin V$. Then $f(v)$ can be written as $alpha_1 f(v_1)+cdots+alpha_m f(v_m)$. In other words, $f(v)=f(alpha_1v_1+cdots+alpha_mv_m)$. But then $v-(alpha_1v_1+cdots+alpha_mv_m)inker f$ and therefore it can be written as $beta_1u_1+cdots+beta_nw_n$. So$$v=alpha_1v_1+cdots+alpha_mv_m+beta_1u_1+cdots+beta_nw_n.$$
On the other hand, if$$alpha_1v_1+cdots+alpha_mv_m+beta_1u_1+cdots+beta_nw_n=0,tag1$$then$$0=f(0)=alpha_1 f(v_1)+cdots+alpha_mf(v_m)$$and therefore every $alpha_j$ is equal to $0$. But then $(1)$ means that $beta_1u_1+cdots+beta_nw_n=0$ and so every $beta_k$ is equal to $0$ too.
answered Jan 16 at 9:38
José Carlos SantosJosé Carlos Santos
159k22126231
answered Jan 16 at 9:38
José Carlos SantosJosé Carlos Santos
159k22126231
answered Jan 16 at 9:38
José Carlos SantosJosé Carlos Santos
159k22126231
answered Jan 16 at 9:38
José Carlos SantosJosé Carlos Santos
159k22126231
159k22126231
add a comment |
add a comment |
$begingroup$
Highlights of proof:
Take $;vin V;$ , then
$$Tvintext{Im},Timplies Tv=sum_{k=1}^m a_kTv_kimplies Tleft(v-sum_{k=1}^m a_kv_kright)=0implies v-sum_{k=1}^m a_kv_kinker T$$
and now you try to complete the last few steps to finish the proof, after you're sure you can justify all the steps above.
This is practically the proof of the well-known and very important Dimensions Theorem for finite dimensional spaces (in fact, only $;V;$ , the definition domain, has to be finite dimensional)
answered Jan 16 at 9:37
DonAntonioDonAntonio
178k1493229
$endgroup$
add a comment |
$begingroup$
Highlights of proof:
Take $;vin V;$ , then
$$Tvintext{Im},Timplies Tv=sum_{k=1}^m a_kTv_kimplies Tleft(v-sum_{k=1}^m a_kv_kright)=0implies v-sum_{k=1}^m a_kv_kinker T$$
and now you try to complete the last few steps to finish the proof, after you're sure you can justify all the steps above.
This is practically the proof of the well-known and very important Dimensions Theorem for finite dimensional spaces (in fact, only $;V;$ , the definition domain, has to be finite dimensional)
answered Jan 16 at 9:37
DonAntonioDonAntonio
178k1493229
$endgroup$
add a comment |
$begingroup$
Highlights of proof:
Take $;vin V;$ , then
$$Tvintext{Im},Timplies Tv=sum_{k=1}^m a_kTv_kimplies Tleft(v-sum_{k=1}^m a_kv_kright)=0implies v-sum_{k=1}^m a_kv_kinker T$$
and now you try to complete the last few steps to finish the proof, after you're sure you can justify all the steps above.
This is practically the proof of the well-known and very important Dimensions Theorem for finite dimensional spaces (in fact, only $;V;$ , the definition domain, has to be finite dimensional)
answered Jan 16 at 9:37
DonAntonioDonAntonio
178k1493229
$endgroup$
Highlights of proof:
Take $;vin V;$ , then
$$Tvintext{Im},Timplies Tv=sum_{k=1}^m a_kTv_kimplies Tleft(v-sum_{k=1}^m a_kv_kright)=0implies v-sum_{k=1}^m a_kv_kinker T$$
and now you try to complete the last few steps to finish the proof, after you're sure you can justify all the steps above.
This is practically the proof of the well-known and very important Dimensions Theorem for finite dimensional spaces (in fact, only $;V;$ , the definition domain, has to be finite dimensional)
answered Jan 16 at 9:37
DonAntonioDonAntonio
178k1493229
answered Jan 16 at 9:37
DonAntonioDonAntonio
178k1493229
answered Jan 16 at 9:37
DonAntonioDonAntonio
178k1493229
answered Jan 16 at 9:37
DonAntonioDonAntonio
178k1493229
178k1493229
add a comment |
add a comment |
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