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Let $V$ be a finite-dimensional inner product space, and let $\beta = \{\alpha_{1}, \cdots \alpha_{n}\}$ be an orthonormal basis for $V$. Let $T$ be a linear operator on $V$ and $A$ the matrix of $T$ in the ordered basis $\beta$. Prove $$ A_{ij} = (T\alpha_{j} | \alpha_{i}) $$ where $(\alpha|\beta)$ is the inner product of $\alpha, \beta \in V$


How do I have to approach to solve this problem? My idea is to use the fact that there exists a matrix $G$ such that $$ G_{ij} = (\alpha_{i} | \alpha_{j})\;\;\; $$ Where $\alpha_{i},\;\; i = 1,2, \cdots n$ is orthonormal basis.

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Let $v=\sum_iv_i\alpha_i$ be any vector, then $$(Av)_i=\sum_jA_{ij}v_j=\sum_j(T\alpha_j|\alpha_i)v_j=\Big(\sum_jTv_j\alpha_j|\alpha_i\Big)=(Tv|\alpha_i)=(Tv)_i$$ in which one sees the definition of $A_{ij}$ is consistent.

A more straightforward way, $$A_{ij}=(A\alpha_j|\alpha_i)=(T\alpha_j|\alpha_i)$$

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You approach this by first recalling what it means that a matrix $A$ represents $T$ with respect to $\beta$. You should find that it means that

$$T\alpha_j=\sum_i A_{ij}\alpha_i\quad\text{ for all $j$.}$$

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