Equivalent conditions Isomorphism, dual space and $\det C\neq 0 $ ex

Question

Let $V$ be a finite dimensional vector space over the field $K$, and let $g$ be a bilinear form on $V$, written $\langle,\rangle$. In the previous part: for each $w\in V$, the map $v\to\langle v,w\rangle$ is a functional $L_w$ on $V$, and the map $w\to L_w$ is a linear map of $V$ into the dual space $V^{*}$.(This was already proven in a previous exercise). Show that the following conditions are equivalent:

(i) The kernel of the map $L$ above is {0}

(ii)The map $L$ is an isomorphism between $V$ and $V^{*}$

(iii) If $C$ is the matrix representing the bilinear form with respect to a basis of $V$, then $\det(C)\neq 0$.

A bilinear form satisfying the preceding three conditions is said to be non-degenerate.

I started answering using (ii)

$\dim V=\dim V^{*}$, since w\in V by assumption.

$\ker(L_w)=0$ iff $w=0$

Therefore there exists an isomorphism that was already proven in the book.

(ii) implies (i), since $w\in V$, the $Ker{L}=0$ iff $w=0$ or $v=0$, in which in this case $v=w$. Therefore the $ker{L}=0$ for arbitrary $w$.

I make use of the following theorem for the relation (ii) implies (iii).

Theorem: Let $F:V\to W$ be a linear map, and assume that F is injective and surjective. Then F is invertible.

Proof:Each $w\in W$ matches a v once F is surjective and and the v is uniquely determined for each w once F is injective. If we consider $G:W\to V$, then $G=F^{-1}$, we need to prove G is linear.

If $w_1,w_2\in W$ then $F(v_1)=w_1$ and $F(v_2)=w_2$ and since F is linear $F(v_1+V_2)=F(v_1)+F(v_2)$ Therefore $G(w_1+w_2)=v_1+v_2=G(v_1)+G(v_2)$.

If $c\in K$ then $F(cv_1)=cF(v_1)=cw_1$. Therefore $G(cw_1)=cv_1=cG(w_1)$ $\blacksquare$

Since the map $w\to L_w$ is an isomorphism by (ii) implies the map is invertible. For an arbitrary bilinear matrix C to be invertible it requires $\det(C)\neq 0$, implying (iii).

Questions:

1) Is my usage of the mapping $w\to L_w$ correct? Does it solve the question?

2) Is my poof right?

3) I made used of non-degeneracy in the proof. What is supposed to mean "A bilinear form satisfying the preceding three conditions is said to be non-degenerate."

Thanks in advance!

Answer 1

I'm afraid you did not prove the statement.

Let's go in the usual way.

(i)$\implies$(ii) Since $\dim V=\dim V^*$ and $L\colon V\to V^*$ is injective (having zero kernel), it is also surjective, hence an isomorphism.

(ii)$\implies$(iii) Let $\{v_1,\dots,v_n\}$ be a basis of $V$; the matrix $C=[c_{ij}]$ is defined by $c_{ij}=\langle v_i,v_j\rangle$. Let's compute the matrix of $L$ with respect to $\{v_1,\dots,v_n\}$ and $\{v_1^*,\dots,v_n^*\}$ (the dual basis). By definition, we need to express $L_{v_i}$ as a linear combination of the vectors in the dual basis: $$ L_{v_i}=\alpha_{1}v_1^*+\dots+\alpha_{n}v_n^* $$ Now, for $j=1,2,\dots,n$, $$ L_{v_i}(v_j)=\alpha_1v_1^*(v_j)+\dots+\alpha_nv_n^*(v_j)=\alpha_j $$ so $\alpha_j=\langle v_i,v_j\rangle$.

We conclude the requested matrix is exactly $C$ (or $C^T$ depending on conventions); the matrix associated to an isomorphism is invertible.

(iii)$\implies$(i) Take $v\in\ker L$ and…