Let $g:V\times W\to K$ be a bilinear map written $\langle ,\rangle$. For each $w\in W$, the map of $V$ into $K$ given by:

$L_w:v\mapsto\langle v,w\rangle$ is a functional on $V$, and that map $w\mapsto L_w$ is a linear map of $W$ into the dual space of $V^{*}$.

Show that the following conditions are equivalent:

(i) The map $L:W\to V^*$ is injective (its kernel is $\{0\}$).

(ii) If $C$ is a matrix associated with $g$, then $C$ has a rank $n$, where $n=\dim W$.

(i) $w\mapsto L_w$ is an injective map iff $\ker L=0$

By non degeneracy property

If $v\in V$ and $v\neq 0$.

$L_w(v)=\langle v,w\rangle=0$ iff $w=0$ $\implies \ker(L_w)=0$.

If we consider $\mathscr{B},\mathscr{B}^*$ to be the bases of $V$ and $V^*$

$c_{ij}=g(\mathscr{B},\mathscr{B}^*)=\langle \mathscr{B},\mathscr{B}^*\rangle$

Therefore $c_{ij}$ gives rise to a diagonal matrix of dimension $n$.

By a previous theorem it is proven $\dim V=\dim V^*\implies \dim V^*=n$

As $L:W\to V*$ is injective then $\dim W=n$.


1) How can we know "$L:W\to V^*$ is injective then $\dim W=n$"? We need an isomorphism right? However (I) does not require an isomorphism.

2) Is my proof right? I feel there are steps not well justified as the last one.

3) Is Matrix C a diagonal matrix?

4) Can someone help me correct or provide me an alternative proof?

Thanks in advance!

  • $\begingroup$ You write that $w \mapsto L_w$ is a linear map of $W$ into the dual space of $V^{*}$; did you meant "into the dual space, $V^{*}$? $\endgroup$ – John Hughes Jul 29 '17 at 11:03
  • $\begingroup$ Are you given that $V$ is finite dimensional? $\endgroup$ – John Hughes Jul 29 '17 at 11:03
  • $\begingroup$ @JohnHughes $w\to L_w$ is a map of $W$ into the dual space as you asked. $V$ is finite dimensional. Sorry if there was any misunderstanding. I was away of the discussion but now I am back. I am going to study the case and give feed-back. Thanks for the attention. $\endgroup$ – Pedro Gomes Jul 29 '17 at 12:52

$$ \newcommand{\bv}{{\mathbf v}} \newcommand{\bw}{{\mathbf w}} \newcommand{\bC}{{\mathbf C}} \newcommand{\bz}{{\mathbf 0}} \newcommand{\bx}{{\mathbf x}} $$ Your proof is wrong. It assumes non-degeneracy, which is not given in the hypotheses of the problem. It assumes $V$ is finite dimensional, which is not given in the hypotheses. It refers to "the bases" of $V$ and $V*$, but unless $V$ has dimension zero, or the underlying field is very small (like $F_2$), there are many possible bases for $V$ and $V*$.

As for providing an alternative proof...I might consider providing one, but only once you answer the clarifying questions I've asked.

Suppose that $L$ is injective.

Pick bases $\bw_1, \ldots, \bw_n$ of $W$, and a basis $\bv_1, \ldots, \bv_k$ of $V$ (which we can do because both are finite-dimensional).

Let $c_{ij} = g(\bv_i, \bw_j)$ for $i = 1, \ldots, k; j = 1, \ldots, n$.

Suppose that the rank of $C$ is some number $p$ less than $n$; then there is some nontrivial linear combination of the $n$ columns of $c$ whose value is zero, i.e., there are constants $s_1, \ldots, s_n$, not all zero, with $$ s_1 \bC_1 + s_2 \bC_2 + \ldots + s_n \bC_n = \bz, $$ where $C_i$ denotes the $i$th column of $C$. The $j$th row of the equation above says that $$ s_1 c_{j1} + \ldots + s_n c_{jn} = 0. $$ This must hold for $j = 1, \ldots, k$.

Now substituting the definition of $c_{ij}$, we get $$ s_1 g(\bv_j, \bw_1) + \ldots + s_n g(\bv_j, \bw_n) = 0, $$ which, by bilinearity of $g$, gives $$ g(\bv_j, s_1 \bw_1 + \ldots + s_n \bw_n) = 0 $$ for all $j$.

Letting $\bx = s_1 \bw_1 + \ldots + s_n \bw_n$, which is nonzero because the $\bw$s form a basis and not all $s_j$ are zero, we have $$ g(\bv_j, \bx) = 0 $$ for all $j$, Hence for any coefficients $t_i$, we have $$ g(\sum_i t_j \bv_j , \bx) = 0 $$ and since the $\bv$s are a basis for $V$, this shows that $g(., \bx)$ is everywhere zero on $V$, or to say it differently, shows that $L_\bx$ is the $\bz$ element of $V^{*}$, contradicting the claim that $L$ is injective.

Now forget all the symbols used in the previous part except for the bases $\{\bv_i\}$ and $\{\bw_j\}$.

Now suppose that $L$ is not injective. I'll show that the rank of the matrix is less than $n$.

Because $L$ is not injective, we have $L_{\mathbf a} = L_{\mathbf b}$ for some distinct $\mathbf a, \mathbf b$. Let $\bx = \mathbf b - \mathbf a$. Then $L_{\mathbf x} = L_{\mathbf b} - L_{\mathbf a} = 0$. Write $\bx$ as a linear combination of the elements of our basis for $W$: $$ \bx = s_1 \bw_1 + \ldots + s_k \bw_k, $$ and note that not all the $s_i$ are zero (for if they were, $\bx$ would be zero, which is not, because $\mathbf a \ne \mathbf b$).

For every $\bv \in V$, we have $L_\bx(\bv) = 0$. In particular, for $\bv_1, \bv_2, \ldots$, we have $L_\bx(\bv_i) = 0$, whence $g(\bv_i, \bx) = 0$.

Following the same reasoning as before, but in reverse, we get, for $j = 1, \ldots, k$, that $$ s_1 g(\bv_j, \bw_1) + \ldots + s_n g(\bv_j, \bw_n) = \bz, $$ and hence $$ s_1 \bC_1 + \ldots + s_n \bC_n = \bz, $$ where $\bC_i$ denotes the $i$th column of the matrix $C$. This is a nontrivial relation among the $n$ columns of $C$, hence the set of columns cannot be linearly independent, hence the rank of $C$ (the largest number of linearly independent columns) cannot be $n$.

Note that this proof does not rely on $V$ being finite dimensional, i.e., every step works just fine even if $k = \infty$. It only (as written) requires that $V$ have a countable basis. I suspect it could be adjusted slightly to handle the case where $V$ has an "uncountable basis", but then you get into messy details about just what that term might mean, and I don't want to do that.

  • $\begingroup$ I have been studying your answer and your points have risen some doubts. Concerning only your statement, you actually prove matrix C must have dimension $n$ for the kernel of map $w\to L_w$ to be zero. You pointed out non-degeneracy is not implicit, however I would like to check a previous question in which that property was used in an answer. I want to know if the bilinear form demands non-degeneracy property. math.stackexchange.com/questions/2370353/… $\endgroup$ – Pedro Gomes Jul 29 '17 at 18:37
  • $\begingroup$ I prove nothing about the dimension of C, only its rank. As for your second question: no. There are lots of degenerate bilinear forms. Unless a problem states that the form is nondegenerate, you cannot assume it, just as you cannot assume that an arbitrary function is continuous. $\endgroup$ – John Hughes Jul 29 '17 at 18:41
  • $\begingroup$ Have you checked the question I asked? Non-degeneracy is assumed by the person that answers. This is confusing me. Please check that answer! $\endgroup$ – Pedro Gomes Jul 29 '17 at 18:44
  • $\begingroup$ I don't need to check that...I KNOW the definition of bilinear form. An answer on mse is not gospel... $\endgroup$ – John Hughes Jul 29 '17 at 19:02
  • $\begingroup$ So you are saying that the answer is wrong. The problem is that I do not see an alternative proof without the use of the non-degeneracy property. I was thinking if you could kindly provide me one for the other question as you did here. As I said in a previous comment the book usually refers non-degenerate scalar products everytime they talk about functional. Therefore I am not used to proofs that differ from those. Thanks for your attention! $\endgroup$ – Pedro Gomes Jul 29 '17 at 20:15

Actually the map $L$ is not only injective, but also surjective.


Assume that $\exists w, w' \in W$ such that $L(w) = L(w')$, i.e $L_w = L{w'}$, so $$\forall v \in V \qquad <v, w> = <v, w'> \Rightarrow <v, w - w'> = 0$$, and by the non-degenerate character of $<,>$, $w=w'$, so $L$ is injective.

And as for your second question, since we have a non-degenerate bilinear function between $V$ and $W$, $V $ and $W$ are dual spaces(see the side note at the end of the answer), so $$dim V^* = dim V = dim W$$

In another way, you can also say that, since $L$ is injective $$dim W \leq V^*.$$ So now consider the other way around, i.e let $L' : V \to W^*$, since this map will also be injective, we have $$dim V \leq W^*$$ So, combining two result with the fact that $dim V = dim V^*$, we get $$dim V = dim W$$

A side note, the definition of dual space of a vector space is more general than just the space of linear functionals of that vector space $L(V)$.In fact, the exact definition of a dual vector space of $V$ is a vector space $W$ such that there is a non-degenerate bilinear function defined between $V$ and $W$.

  • $\begingroup$ I don't see the assumption that $V$ is finite dimensional anywhere, but $dim V = dim V^{*}$, which you use, relies on that. $\endgroup$ – John Hughes Jul 28 '17 at 19:41
  • $\begingroup$ @JohnHughes In the question, it is already assumed $W$ to be finite dimensional, and there is a theorem stating that, if one of the vector spaces in a dual pair is finite, then the dual space is also finite and has the same dimension. $\endgroup$ – onurcanbektas Jul 29 '17 at 5:33
  • $\begingroup$ There's an assumption of finite dimension in one of OP's attempted answers, but nothing about it in the question. It only mentions $dim W = n$. (Perhaps you think that "matrix associated with..." implies the dimension is finite but lots of folks work with infinite matrices...) $\endgroup$ – John Hughes Jul 29 '17 at 10:20
  • $\begingroup$ @JohnHughes Generally, whenever I see something like "(something) = n", I always assume that that something is finite. In fact, this is what is taught me in my university as a convention. $\endgroup$ – onurcanbektas Jul 29 '17 at 10:26
  • $\begingroup$ Sure. That makes $W$ finite dimensional. What about $V$? $\endgroup$ – John Hughes Jul 29 '17 at 10:29

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