# Showing that $(\langle T(b_{i}),b_{j}\rangle )_{1\leq i,j\leq n}=G(b_{1},…,b_{n}) [T^{*}]_{B}^{B}$

Let $V$ be a vector space with a finite Dimension above $\mathbb{C}$ or $\mathbb{R}$.

Let $B=(b_{1},...,b_{n})$ be a basis of V.

I need your help proving this:

$(\langle T(b_{i}),b_{j}\rangle )_{1\leq i,j\leq n}=G(b_{1},...,b_{n}) [T^{*}]_{B}^{B}$

when $G(b_{1},...,b_{n})$ is a Gram matrix, and $T^{*}$ is the adjoint transformation of $T$.

Thank you

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Is this a homework question? If yes, then please tag it! –  Dennis Gulko May 13 '11 at 8:41
No, It's not! I do that for refreshing my memory, it's a homework that I gave myself, If you want to name it somehow. –  user6163 May 13 '11 at 8:43

Calculate $G(b1,..,bn)[T^*]_B^B\cdot e_i$ for any $1\leq i \leq n$, where $e_i=(0,...,0,1,0,...,0)$ ($i$'th place).
You get $[T^*(b_i)]_B=(\alpha_1,...,\alpha_n)$ (i.e. $T^*(b_i)=\alpha_1b_1+...+\alpha_nb_n$ by definition of $[T^*]_B^B$). Multiplying $G(b1,..,bn)(\alpha_1,...,\alpha_n)$ (column), in the $j$'th row we get $$\alpha_1\langle b_1 , b_j \rangle+...+\alpha_n\langle b_n , b_j\rangle= \langle \alpha_1b_1+...+\alpha_nb_n,b_j\rangle=\langle T^*(b_i),b_j\rangle$$ On the other hand, multiply $(\langle T(b_i),b_j \rangle)_{1\leq i,j\leq n} e_i$ and look at the $j$'th row. They should be equal, thus proving the equality.
(Either I messed with the multiplication or you should have $(\langle b_i,T(b_j) \rangle)_{1\leq i,j\leq n}$)