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Let $V$ be a Hilbert space with orthonormal basis $e_1,e_2,\ldots$. Let $T:V\to V$ be a linear transformation. Let $A$ be the matrix of $T$ with respect to the basis $e_1,e_2,\ldots$ and let $B$ be the matrix of $T^*$ with respect to the basis $e_1,e_2,\ldots$.

Show that $B=\overline A^t$.

I know that $\langle T(v),v\rangle=\langle v,T^*(v)\rangle$ and I tried to expand it. It turns out that $$a_1\langle A_1,v\rangle+a_2\langle A_2,v\rangle+\cdots+a_n\langle A_n,v\rangle=\overline a_1\langle v,B_1\rangle+\overline a_2\langle v,B_2\rangle+\cdots+\overline a_n\langle v,B_n\rangle$$ where $A_i = i$-th column of $A$ and $B_i = i$-th column of $B$

I'm not sure if I'm on the right track and can someone show me the steps to solve this question?

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First of all, to be able to talk about matrices of linear transformations, your map $T : V \to V$ has to be bounded. This also assures that $T^*$ exists and is bounded, so it also has a matrix.

Let $A = (a_{ij})$ be the matrix of $T$. The coefficients $a_{ij}$ are defined so that for any $j \in \mathbb{N}$ we have:

$$Te_j = \sum_{i=1}^\infty a_{ij}e_i \implies \langle Te_j, e_i\rangle = a_{ij}$$

Similarly, let $B = (b_{ij})$ be the matrix of $T^*$. The coefficients $b_{ij}$ are defined so that for any $j \in \mathbb{N}$ we have:

$$T^*e_j = \sum_{i=1}^\infty b_{ij}e_i \implies \langle T^*e_j, e_i\rangle = b_{ij}$$

Now for any $i, j \in \mathbb{N}$ we have:

$$b_{ij} = \langle T^*e_j, e_i\rangle = \langle e_j, Te_i\rangle = \overline{\langle Te_i, e_j\rangle} = \overline{a_{ji}}$$

Therefore $B = A^* = \overline{A^t}$, i.e. $B$ is the conjugate transpose of $A$.

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