Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $V$, $V^{'}$ be vector spaces over a field $K$ and let $V \times V^{'} \longrightarrow K$ be a bilinear map with left and right kernel $W$, $W^{'}$ respectively. If $V^{'}/W^{'}$ is finite dimensional, then according to theorem 6.4 p. 145 in Lang's Algebra, $V^{'}/W^{'}$ is isomorphic with $\left(V/W\right)^{\vee}$. I understand the arguments based on the two induced injective homomorphisms $V/W \longrightarrow (V^{'}/W^{'})^{\vee}$ and $V^{'}/W^{'} \longrightarrow (V/W)^{\vee}$. What i can't see is why these are inverses of each other, as is stated in the last sentence of the proof. Any insights?

share|improve this question
I don't understand either how these homomorphisms could be inverses of each other since the range of one is not the domain of the other. Anyway I think this is not needed to prove the theorem: if you have an injective homomorphism between two vector spaces of equal finite dimension, then it's an isomorphism. –  LostInMath Sep 11 '11 at 21:16

1 Answer 1

The maps aren't inverses of each other. They don't go between the right spaces to be inverses. Instead, they are dual to each other. To describe the duality, we need the concept of the dual of a linear map. If $g:V\to W$, then $g^{\vee}:W^{\vee}\to V^{\vee}$ is defined by $g^{\vee}(\psi)(v)=\psi(g(v))$. Note that if you are used to fixing a basis and working with matrices, then if $g$ is represented by the matrix $A$, then $g^{\vee}$ is represented by the transpose $A^T$.

By the definition of kernel of a bilinear map, there is an induced pairing $V/W\times V'/W'\to k$, and moreover, this pairing is non degenerate. For the sake of clarity, let $X=V/W, X'=V'/W'$. We want to show the following, which does not require knowing where $X$ came from:

Proposition: If $X$ and $X'$ are vector spaces with a non-degenerate pairing $X\times X'\to k$, then we have natural injections $f:X\to X'^{\vee}$ and $f':X'\to X^{\vee}$ defined by $f(v)(w')=\langle v,w' \rangle$ and $f'(w')(v)=\langle v,w' \rangle$ (note that this implies $f(v)(w')=f'(w')(v)$) .

Moreover, if $\iota_X:X\to X^{\vee\vee}$ is the cannonical inclusion of $X$ into its double dual, defined by $\iota_X(v)(\psi)=\psi(v)$, then $f=f'^{\vee}\circ \iota_X$.

proof: Let $v\in X$ and $v'\in X'$ Then $$[f'^{\vee}\iota_X](v)(v')=[f'^{\vee}\iota_X(v)](v')= \iota_X(v)[f'(v')]=f'(v')(v)=f(v)(v').$$

Since $v'$ was arbitrary, $[f'^{\vee}\iota_X](v)=f(v)$, and because $v$ was arbitrary, $f'^{\vee}\iota_X=f$.

Now, if $X$ is finite dimensional, $\iota_X$ is an isomorphism, and so suppressing the canonical isomorphism, we can write simply $f=f'^{\vee}$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.