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My lecture notes say that for every bilinear form there exists a linear operator such that $$\tau (v,w) = v.(Tw)$$ and that there must exist some other linear operator $S$ such that $$(Sv).w = v.(Tw).$$ I understand everything up to there but then it says that it's easy to see that in an orthonormal basis, the matrix of S is just the transpose of the matrix of T. I can't get my head around why it has to be an orthonormal basis. Surely, if $A$ is the matrix for $T$ and $B$ is the matrix for $S$ then, $$(Sv).w = v.(Tw)$$ $$(B\underline{v})^T\underline{w} = \underline{v}^T A\underline{w}$$ $$\underline{v}^T B^T \underline{w} = \underline{v}^T A\underline{w}$$ So $B^T = A$ for any basis? Where am I going wrong?

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How come that your "dot product" on the original vectorspace is (in your basis) the euclidian scalarproduct in $\mathbb R^n$? – Blah Apr 6 '12 at 15:28
Because the standard scalar product $\underline{x}.\underline{y} =\underline{x}^T \underline{y}$ so if A is the matrix of the bilinear form, $\tau(v,w) = \underline{v}^T A \underline{w} = v.(Tw)$? – user26069 Apr 6 '12 at 16:15
could you please describe the situation? Is your vectorspace $\mathbb R^n$ and "." the euclidian scalarproduct? – Blah Apr 6 '12 at 16:45
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By using that $v.\omega$ is $\underline{v}^T\underline{w}$, you are assuming that your basis is orthonormal. Because if, say, $V$ is 2-dimensional and $f_1,f_2$ is your basis, then you seem to claim that you see $f_1$ as $[1\ 0]^T$, $f_2=[0\ 1]^T$, and that $$ f_1.f_2=([1\ 0]^T)^T[0\ 1]^T=0. $$

That's of course not necessarily the case. Consider $\mathbb{R}^2$ with the basis $f_1=(1,0)$, $f_2=(1,1)$, with the usual dot product. Then, as vectors in the basis $\{f_1,f_2\}$, we associate $f_1$ with $[1\ 0]^T$ and $f_2$ with $[0\ 1]^T$, but $f_1.f_2=1$, which cannot be obtained by doing the product $f_1^Tf_2$ as matrices in their own basis.

Still in this basis, consider the operator $T$ given by $Tf_1=f_2$, $Tf_2=0$. Then, in the basis $\{f_1,f_2\}$, the operator $T$ is represented by the matrix $$ \begin{bmatrix}0&1\\ 0&0\end{bmatrix}. $$ Let us calculate the entries of $S$ (usually denoted by $T^*$ and called the adjoint of $T$) in the basis $\{f_1,f_2\}$: \begin{eqnarray} S_{11}&=&(Sf_1).f_1=f_1.(Tf_1)=f_1.f_2=1,\ S_{12}&=&(Sf_2).f_1=f_2.Tf_1=f_2.f_2=2,\ S_{21}&=&(Sf_1).f_2=f_1.Tf_2=f_1.0=0,\ S_{22}&=&(Sf_2).f_2=f_2.Tf_2=f_2.0=0. \end{eqnarray} So, in the basis $\{f_1,f_2\}$, the matrix for $S$ is $$ \begin{bmatrix} 1&2\\ 0&0 \end{bmatrix} $$

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There is nothing wrong, and the result is true for any basis.

However, the analysis above skips over the tedious bits by jumping from the vectors ($v, w, S v, T w$) in the vector space $V$ to their representations in $\mathbb{R^n}$ ($\underline{v}, \underline{w}, B \underline{v}, A \underline{w} $). This tedium is considerably reduced by choosing an orthonormal basis.

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My answer is a little glib; to obtain the matrix representation of $T$ or $S$, a linear bijection $L:\mathbb{R^n} \rightarrow V$ is needed. However, you cannot use just any map, you need to preserve the dot (inner) product, so that if $x,y \in \mathbb{R^n}$, then $Lx . Ly = x^T y$. – copper.hat Apr 6 '12 at 23:36

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