I am having trouble understanding when it is correct to use the kronecker delta function. Let me elaborate.

I am given a symmetric bilinear form with its corresponding equation (that tells us what is the image of the form for two vectors x and y). Then I am asked to find the matrix associated to the bilinear form in a basis N and then, in another basis N’.

My problem comes from the fact that in the first case, the plug the basis N right into the equation of the bilinear form to get every entry of the matrix.

However, in the second case, when they want to find the matrix in the basis N’, the plug the basis N’ but instead with the belonging index (1,2 or 3) which indicates the order in the basis, the use indeces i and j. Later they proceed to use the kronecker delta to find the values of the entries of the matrix in the basis N’.

Why is this valid (using the kronecker delta) only for the basis N’ and not for the initial basis N?


1 Answer 1


I'm not sure what you're asking, but I'll give a brief explanation. Let $V$ be a vector space, $B: V \times V \to \Bbb R$ a bilinear form (not necessarily symmetric), and $\mathcal{B} = (v_1,\ldots,v_n)$ and $\mathcal{C} = (w_1,\ldots, w_n)$ be bases for $V$. We have a change of bases matrix $A \doteq [{\rm Id}_{V}]_{\mathcal{C},\mathcal{B}} = (a^i_{\;j})_{i,j=1}^n$, which tells us that $$w_j = \sum_{i=1}^n a^i_{\;j}v_i,\qquad 1 \leq j \leq n.$$If we write $[B]_{\mathcal{B}} = (B(v_i,v_j))_{i,j=1}^n$ and $[B]_{\mathcal{C}} = (B(w_i,w_j))_{i,j=1}^n$, then $$B(w_i,w_j) = B\left(\sum_{k=1}^na^k_{\;i}v_k, \sum_{\ell=1}^na^\ell_{\;j}v_\ell\right) = \sum_{\ell=1}^n \left(\sum_{k=1}^n a^k_{\;i}B(v_k,v_\ell)\right) a^\ell_{\;j}$$tells us that $$\fbox{$[B]_{\mathcal{C}} = A^\top [B]_{\mathcal{B}}A.$}$$The moral of the history is that for finding out $[B]_{\mathcal{C}}$, you only need to know $[B]_{\mathcal{B}}$ and the change of bases matrix $A$.


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