0
$\begingroup$

Let $E = \{e_1, ..., e_n\}$ be a basis of a $K$-vector space $V$. Let $A: E \times E \to K$ be any map. Then, there is exactly one bilinear form $B: V \times V \to K$ such that $B (e_i, e_j) = A (e_i, e_j)$ for each $e_i, e_j \in E$.

How do you prove it?

$\endgroup$

2 Answers 2

1
$\begingroup$

Can I write it so?:

Let $A(e_{i},e_{j})=a_{ij}$ - any transformation. Since $B(e_{i},e_{j}) = A(e_{i},e_{j})$, so $A(x,y)=\sum_{i, j = 1}^{n}a_{ij}x_{i}y_{j}=B(x,y)$

which proves the uniqueness bilinear form $B:V \times V \rightarrow K$.

$\endgroup$
4
  • $\begingroup$ This is basically right, except that $A$ is not defined on $V \times V$. But writing $x = \sum x_i e_i$ and $y = \sum y_j e_j$, then you are correct that $B(x,y) = \sum x_i y_j B(e_i, e_j) = \sum a_{ij} x_i y_j$. I broke up the argument into two steps - the first uses bilinearity, while the second uses that $B(e_i, e_j) = A(e_i, e_j)$. $\endgroup$ Jan 30, 2013 at 15:25
  • $\begingroup$ Thank you very much ;) Let $E = \{e_{1}, ..., e_{n}\}$ - base of $V$, $A: E \times E \rightarrow K$ - any transformation, such that $A(e_{i}, e_{j}) = B (e_{i}, e_{j})$ for a bilinear form $B: V \times V \rightarrow K$. Then for any $x = \sum {x}_{i} {e} _ {i}$ and $y = \sum y_{j} {e}_{j}$ $A(x,y) = A( \sum {x}_{i} {e}_{i} \sum {y}_{j} {e}_{j}) = \sum x_{i} y_{j} A(e_{i}, e_{j}) = \sum x_{i} y_{j} B(e_{i}, e_{j}) = B(x,y)$ it's ok? $\endgroup$
    – sigma123
    Jan 30, 2013 at 16:21
  • $\begingroup$ In your last equation, $A(x,y)$ is not defined for all $x,y \in V$. So the first two equalities don't make sense. The rest is fine. $\endgroup$ Jan 30, 2013 at 17:28
  • $\begingroup$ I understand now. Once again thank you very much ;) $\endgroup$
    – sigma123
    Jan 30, 2013 at 18:33
0
$\begingroup$

Given two vectors in $V$, expand each as a (unique) linear combination of the basis vectors, then use the bilinearity of $B$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .