linear algebra and linear functionals
$begingroup$
Let V a vector space and $dim_mathbb{K} V = n$. Proof that if m < n and $f_1,..,f_m$ be linear functionals in $mathscr{L}(V,mathbb{K}$). Then, there is a vector nonzero $v in$ V such that $f_i(v)=0$ for $i=1,..,m.$
I am trying to prove the following. Let B = {$v_1,..,v_m,v_{m+1},...,v_n$} a basis of V, I suppose that $hspace{0.4cm}$ {$f_1,..,f_m$} $subset$ B* basis of V*. Then let $v_{m+1}$ $in$ B, we have that
$f_i(v_{m+1})=0, hspace{0.5cm}forall i=1,...n. $
But, if {$f_1,..,f_m$} $notsubset$ B*? what do i do?
linear-algebra
asked Jan 25 at 15:52
user638057user638057
776
$endgroup$
add a comment |
$begingroup$
Let V a vector space and $dim_mathbb{K} V = n$. Proof that if m < n and $f_1,..,f_m$ be linear functionals in $mathscr{L}(V,mathbb{K}$). Then, there is a vector nonzero $v in$ V such that $f_i(v)=0$ for $i=1,..,m.$
I am trying to prove the following. Let B = {$v_1,..,v_m,v_{m+1},...,v_n$} a basis of V, I suppose that $hspace{0.4cm}$ {$f_1,..,f_m$} $subset$ B* basis of V*. Then let $v_{m+1}$ $in$ B, we have that
$f_i(v_{m+1})=0, hspace{0.5cm}forall i=1,...n. $
But, if {$f_1,..,f_m$} $notsubset$ B*? what do i do?
linear-algebra
asked Jan 25 at 15:52
user638057user638057
776
$endgroup$
add a comment |
$begingroup$
Let V a vector space and $dim_mathbb{K} V = n$. Proof that if m < n and $f_1,..,f_m$ be linear functionals in $mathscr{L}(V,mathbb{K}$). Then, there is a vector nonzero $v in$ V such that $f_i(v)=0$ for $i=1,..,m.$
I am trying to prove the following. Let B = {$v_1,..,v_m,v_{m+1},...,v_n$} a basis of V, I suppose that $hspace{0.4cm}$ {$f_1,..,f_m$} $subset$ B* basis of V*. Then let $v_{m+1}$ $in$ B, we have that
$f_i(v_{m+1})=0, hspace{0.5cm}forall i=1,...n. $
But, if {$f_1,..,f_m$} $notsubset$ B*? what do i do?
linear-algebra
asked Jan 25 at 15:52
user638057user638057
776
$endgroup$
Let V a vector space and $dim_mathbb{K} V = n$. Proof that if m < n and $f_1,..,f_m$ be linear functionals in $mathscr{L}(V,mathbb{K}$). Then, there is a vector nonzero $v in$ V such that $f_i(v)=0$ for $i=1,..,m.$
I am trying to prove the following. Let B = {$v_1,..,v_m,v_{m+1},...,v_n$} a basis of V, I suppose that $hspace{0.4cm}$ {$f_1,..,f_m$} $subset$ B* basis of V*. Then let $v_{m+1}$ $in$ B, we have that
$f_i(v_{m+1})=0, hspace{0.5cm}forall i=1,...n. $
But, if {$f_1,..,f_m$} $notsubset$ B*? what do i do?
linear-algebra
linear-algebra
asked Jan 25 at 15:52
user638057user638057
776
asked Jan 25 at 15:52
user638057user638057
776
asked Jan 25 at 15:52
user638057user638057
776
asked Jan 25 at 15:52
user638057user638057
776
asked Jan 25 at 15:52
user638057user638057
776
776
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: The map $T:V to Bbb F^m$ defined by
$$
T(v) = (f_1(v),dots,f_m(v))
$$
is a linear map from the $n$-dimensional space $V$ to the smaller space $Bbb F^m$. By the rank-nullity theorem, $T$ cannot have a trivial kernel.
answered Jan 25 at 17:25
OmnomnomnomOmnomnomnom
128k791186
$endgroup$
$begingroup$
Thanks a lot. Since $Dim_mathbb{F} KerT >0 $ then there is a nonzero vector $v in V$ such that $v in KerT$, this implies that $f_i(v)=0 forall i=1,...,m.$
$endgroup$
– user638057
Jan 26 at 17:48
add a comment |
$begingroup$
Hint: $ker f_i$ has dimension at least $n-1$.
Use the dimension formula to prove that the intersection of an $n-a$ dimensional subspace with an $n-b$ dimensional subspace has dimension at least $n-(a+b)$. Then use induction.
answered Jan 25 at 16:00
BerciBerci
61.3k23674
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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active
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$begingroup$
Hint: The map $T:V to Bbb F^m$ defined by
$$
T(v) = (f_1(v),dots,f_m(v))
$$
is a linear map from the $n$-dimensional space $V$ to the smaller space $Bbb F^m$. By the rank-nullity theorem, $T$ cannot have a trivial kernel.
answered Jan 25 at 17:25
OmnomnomnomOmnomnomnom
128k791186
$endgroup$
$begingroup$
Thanks a lot. Since $Dim_mathbb{F} KerT >0 $ then there is a nonzero vector $v in V$ such that $v in KerT$, this implies that $f_i(v)=0 forall i=1,...,m.$
$endgroup$
– user638057
Jan 26 at 17:48
add a comment |
$begingroup$
Hint: The map $T:V to Bbb F^m$ defined by
$$
T(v) = (f_1(v),dots,f_m(v))
$$
is a linear map from the $n$-dimensional space $V$ to the smaller space $Bbb F^m$. By the rank-nullity theorem, $T$ cannot have a trivial kernel.
answered Jan 25 at 17:25
OmnomnomnomOmnomnomnom
128k791186
$endgroup$
$begingroup$
Thanks a lot. Since $Dim_mathbb{F} KerT >0 $ then there is a nonzero vector $v in V$ such that $v in KerT$, this implies that $f_i(v)=0 forall i=1,...,m.$
$endgroup$
– user638057
Jan 26 at 17:48
add a comment |
$begingroup$
Hint: The map $T:V to Bbb F^m$ defined by
$$
T(v) = (f_1(v),dots,f_m(v))
$$
is a linear map from the $n$-dimensional space $V$ to the smaller space $Bbb F^m$. By the rank-nullity theorem, $T$ cannot have a trivial kernel.
answered Jan 25 at 17:25
OmnomnomnomOmnomnomnom
128k791186
$endgroup$
Hint: The map $T:V to Bbb F^m$ defined by
$$
T(v) = (f_1(v),dots,f_m(v))
$$
is a linear map from the $n$-dimensional space $V$ to the smaller space $Bbb F^m$. By the rank-nullity theorem, $T$ cannot have a trivial kernel.
answered Jan 25 at 17:25
OmnomnomnomOmnomnomnom
128k791186
answered Jan 25 at 17:25
OmnomnomnomOmnomnomnom
128k791186
answered Jan 25 at 17:25
OmnomnomnomOmnomnomnom
128k791186
answered Jan 25 at 17:25
OmnomnomnomOmnomnomnom
128k791186
128k791186
$begingroup$
Thanks a lot. Since $Dim_mathbb{F} KerT >0 $ then there is a nonzero vector $v in V$ such that $v in KerT$, this implies that $f_i(v)=0 forall i=1,...,m.$
$endgroup$
– user638057
Jan 26 at 17:48
add a comment |
$begingroup$
Thanks a lot. Since $Dim_mathbb{F} KerT >0 $ then there is a nonzero vector $v in V$ such that $v in KerT$, this implies that $f_i(v)=0 forall i=1,...,m.$
$endgroup$
– user638057
Jan 26 at 17:48
$begingroup$
Thanks a lot. Since $Dim_mathbb{F} KerT >0 $ then there is a nonzero vector $v in V$ such that $v in KerT$, this implies that $f_i(v)=0 forall i=1,...,m.$
$endgroup$
– user638057
Jan 26 at 17:48
$begingroup$
Thanks a lot. Since $Dim_mathbb{F} KerT >0 $ then there is a nonzero vector $v in V$ such that $v in KerT$, this implies that $f_i(v)=0 forall i=1,...,m.$
$endgroup$
– user638057
Jan 26 at 17:48
add a comment |
$begingroup$
Hint: $ker f_i$ has dimension at least $n-1$.
Use the dimension formula to prove that the intersection of an $n-a$ dimensional subspace with an $n-b$ dimensional subspace has dimension at least $n-(a+b)$. Then use induction.
answered Jan 25 at 16:00
BerciBerci
61.3k23674
$endgroup$
add a comment |
$begingroup$
Hint: $ker f_i$ has dimension at least $n-1$.
Use the dimension formula to prove that the intersection of an $n-a$ dimensional subspace with an $n-b$ dimensional subspace has dimension at least $n-(a+b)$. Then use induction.
answered Jan 25 at 16:00
BerciBerci
61.3k23674
$endgroup$
add a comment |
$begingroup$
Hint: $ker f_i$ has dimension at least $n-1$.
Use the dimension formula to prove that the intersection of an $n-a$ dimensional subspace with an $n-b$ dimensional subspace has dimension at least $n-(a+b)$. Then use induction.
answered Jan 25 at 16:00
BerciBerci
61.3k23674
$endgroup$
Hint: $ker f_i$ has dimension at least $n-1$.
Use the dimension formula to prove that the intersection of an $n-a$ dimensional subspace with an $n-b$ dimensional subspace has dimension at least $n-(a+b)$. Then use induction.
answered Jan 25 at 16:00
BerciBerci
61.3k23674
answered Jan 25 at 16:00
BerciBerci
61.3k23674
answered Jan 25 at 16:00
BerciBerci
61.3k23674
answered Jan 25 at 16:00
BerciBerci
61.3k23674
61.3k23674
add a comment |
add a comment |
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