Let L and M be finite dimensional linear spaces over the field F. Prove that $ dim L/L_{0}= dim M/M_{0}$.
$begingroup$
Let $L$ and $M$ be finite dimensional linear spaces over the field $F$ and let $g : L times M to F$ be a bilinear mapping. We shall call the set
$L_{0} = {l ∈ L : g(l, m) = 0text{ for all }m ∈ M}$
the left kernel of $g$ and the set
$M_{0} = {m ∈ M : g(l, m) = 0text{ for all }l ∈ L}$
the right kernel of $g$. Prove that $ dim L/L_{0} = dim M/M_{0}$.
linear-algebra
edited Nov 25 '18 at 12:42
Bernard
119k639113
asked Nov 25 '18 at 12:37
LenaLena
163
$endgroup$
add a comment |
$begingroup$
Let $L$ and $M$ be finite dimensional linear spaces over the field $F$ and let $g : L times M to F$ be a bilinear mapping. We shall call the set
$L_{0} = {l ∈ L : g(l, m) = 0text{ for all }m ∈ M}$
the left kernel of $g$ and the set
$M_{0} = {m ∈ M : g(l, m) = 0text{ for all }l ∈ L}$
the right kernel of $g$. Prove that $ dim L/L_{0} = dim M/M_{0}$.
linear-algebra
edited Nov 25 '18 at 12:42
Bernard
119k639113
asked Nov 25 '18 at 12:37
LenaLena
163
$endgroup$
$begingroup$
Row rank equals column rank?
$endgroup$
– Charlie Frohman
Nov 25 '18 at 12:43
$begingroup$
Yes, row rank of Gram matrix of bilinear form equals the column rank of Gram matrix.
$endgroup$
– Lena
Nov 25 '18 at 12:46
$begingroup$
Turn that into a proof!
$endgroup$
– Charlie Frohman
Nov 25 '18 at 13:48
add a comment |
$begingroup$
Let $L$ and $M$ be finite dimensional linear spaces over the field $F$ and let $g : L times M to F$ be a bilinear mapping. We shall call the set
$L_{0} = {l ∈ L : g(l, m) = 0text{ for all }m ∈ M}$
the left kernel of $g$ and the set
$M_{0} = {m ∈ M : g(l, m) = 0text{ for all }l ∈ L}$
the right kernel of $g$. Prove that $ dim L/L_{0} = dim M/M_{0}$.
linear-algebra
edited Nov 25 '18 at 12:42
Bernard
119k639113
asked Nov 25 '18 at 12:37
LenaLena
163
$endgroup$
Let $L$ and $M$ be finite dimensional linear spaces over the field $F$ and let $g : L times M to F$ be a bilinear mapping. We shall call the set
$L_{0} = {l ∈ L : g(l, m) = 0text{ for all }m ∈ M}$
the left kernel of $g$ and the set
$M_{0} = {m ∈ M : g(l, m) = 0text{ for all }l ∈ L}$
the right kernel of $g$. Prove that $ dim L/L_{0} = dim M/M_{0}$.
linear-algebra
linear-algebra
edited Nov 25 '18 at 12:42
Bernard
119k639113
asked Nov 25 '18 at 12:37
LenaLena
163
edited Nov 25 '18 at 12:42
Bernard
119k639113
asked Nov 25 '18 at 12:37
LenaLena
163
edited Nov 25 '18 at 12:42
Bernard
119k639113
edited Nov 25 '18 at 12:42
Bernard
119k639113
edited Nov 25 '18 at 12:42
Bernard
119k639113
119k639113
asked Nov 25 '18 at 12:37
LenaLena
163
asked Nov 25 '18 at 12:37
LenaLena
163
asked Nov 25 '18 at 12:37
LenaLena
163
163
$begingroup$
Row rank equals column rank?
$endgroup$
– Charlie Frohman
Nov 25 '18 at 12:43
$begingroup$
Yes, row rank of Gram matrix of bilinear form equals the column rank of Gram matrix.
$endgroup$
– Lena
Nov 25 '18 at 12:46
$begingroup$
Turn that into a proof!
$endgroup$
– Charlie Frohman
Nov 25 '18 at 13:48
add a comment |
$begingroup$
Row rank equals column rank?
$endgroup$
– Charlie Frohman
Nov 25 '18 at 12:43
$begingroup$
Yes, row rank of Gram matrix of bilinear form equals the column rank of Gram matrix.
$endgroup$
– Lena
Nov 25 '18 at 12:46
$begingroup$
Turn that into a proof!
$endgroup$
– Charlie Frohman
Nov 25 '18 at 13:48
$begingroup$
Row rank equals column rank?
$endgroup$
– Charlie Frohman
Nov 25 '18 at 12:43
$begingroup$
Row rank equals column rank?
$endgroup$
– Charlie Frohman
Nov 25 '18 at 12:43
$begingroup$
Yes, row rank of Gram matrix of bilinear form equals the column rank of Gram matrix.
$endgroup$
– Lena
Nov 25 '18 at 12:46
$begingroup$
Yes, row rank of Gram matrix of bilinear form equals the column rank of Gram matrix.
$endgroup$
– Lena
Nov 25 '18 at 12:46
$begingroup$
Turn that into a proof!
$endgroup$
– Charlie Frohman
Nov 25 '18 at 13:48
$begingroup$
Turn that into a proof!
$endgroup$
– Charlie Frohman
Nov 25 '18 at 13:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Maybe you can appreciate a “matrix free” proof.
Define $varphicolon Lto (M/M_0)^*$ by
$$
varphi(x)colon y+M_0mapsto g(x,y)
$$
(where $(M/M_0)^*$ is the dual space to $M/M_0$). For every $xin L$, the linear form $varphi(x)$ is well defined, because if $yin M_0$, then $g(x,y)=0$. The map $varphi$ is clearly linear.
The kernel of $varphi$ is precisely $L_0$, hence $dim(L/L_0)ledim(M/M_0)^*=dim(M/M_0)$.
By symmetry, also the reverse inequality holds.
answered Nov 25 '18 at 14:23
egregegreg
180k1485202
$endgroup$
add a comment |
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$begingroup$
Maybe you can appreciate a “matrix free” proof.
Define $varphicolon Lto (M/M_0)^*$ by
$$
varphi(x)colon y+M_0mapsto g(x,y)
$$
(where $(M/M_0)^*$ is the dual space to $M/M_0$). For every $xin L$, the linear form $varphi(x)$ is well defined, because if $yin M_0$, then $g(x,y)=0$. The map $varphi$ is clearly linear.
The kernel of $varphi$ is precisely $L_0$, hence $dim(L/L_0)ledim(M/M_0)^*=dim(M/M_0)$.
By symmetry, also the reverse inequality holds.
answered Nov 25 '18 at 14:23
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Maybe you can appreciate a “matrix free” proof.
Define $varphicolon Lto (M/M_0)^*$ by
$$
varphi(x)colon y+M_0mapsto g(x,y)
$$
(where $(M/M_0)^*$ is the dual space to $M/M_0$). For every $xin L$, the linear form $varphi(x)$ is well defined, because if $yin M_0$, then $g(x,y)=0$. The map $varphi$ is clearly linear.
The kernel of $varphi$ is precisely $L_0$, hence $dim(L/L_0)ledim(M/M_0)^*=dim(M/M_0)$.
By symmetry, also the reverse inequality holds.
answered Nov 25 '18 at 14:23
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Maybe you can appreciate a “matrix free” proof.
Define $varphicolon Lto (M/M_0)^*$ by
$$
varphi(x)colon y+M_0mapsto g(x,y)
$$
(where $(M/M_0)^*$ is the dual space to $M/M_0$). For every $xin L$, the linear form $varphi(x)$ is well defined, because if $yin M_0$, then $g(x,y)=0$. The map $varphi$ is clearly linear.
The kernel of $varphi$ is precisely $L_0$, hence $dim(L/L_0)ledim(M/M_0)^*=dim(M/M_0)$.
By symmetry, also the reverse inequality holds.
answered Nov 25 '18 at 14:23
egregegreg
180k1485202
$endgroup$
Maybe you can appreciate a “matrix free” proof.
Define $varphicolon Lto (M/M_0)^*$ by
$$
varphi(x)colon y+M_0mapsto g(x,y)
$$
(where $(M/M_0)^*$ is the dual space to $M/M_0$). For every $xin L$, the linear form $varphi(x)$ is well defined, because if $yin M_0$, then $g(x,y)=0$. The map $varphi$ is clearly linear.
The kernel of $varphi$ is precisely $L_0$, hence $dim(L/L_0)ledim(M/M_0)^*=dim(M/M_0)$.
By symmetry, also the reverse inequality holds.
answered Nov 25 '18 at 14:23
egregegreg
180k1485202
answered Nov 25 '18 at 14:23
egregegreg
180k1485202
answered Nov 25 '18 at 14:23
egregegreg
180k1485202
answered Nov 25 '18 at 14:23
egregegreg
180k1485202
180k1485202
add a comment |
add a comment |
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$begingroup$
Row rank equals column rank?
$endgroup$
– Charlie Frohman
Nov 25 '18 at 12:43
$begingroup$
Yes, row rank of Gram matrix of bilinear form equals the column rank of Gram matrix.
$endgroup$
– Lena
Nov 25 '18 at 12:46
$begingroup$
Turn that into a proof!
$endgroup$
– Charlie Frohman
Nov 25 '18 at 13:48