Let $H_1$ and $H_2$ be finite-dimensional (real or complex) Hilbert spaces, let $T \colon H_1 \to H_2$ be a linear operator, [Then $T$ can be shown to be bounded] and let $T^* \colon H_2 \to H_1$ denote the Hilbert adjoint operator for $T$. Let $E \colon= (e_1, \ldots, e_n)$ be an ordered basis for $H_1$, and let $B \colon= (b_1, \ldots, b_m)$ be an ordered basis for $H_2$. Let $[T]$ and $[T^*]$ be the matrices of $T$ and $T^*$, respectively, with respect to the bases $E$ and $B$.

Then how are $[T]$ and $[T^*]$ related?

The bases $E$ and $B$ are not necessarily orthonormal.

It is my feeling that in case $H_1 = H_2 = \mathbb{C}^n$, we have $$[T^*] = \overline{[T]}^t. $$

  • $\begingroup$ $[T^*]=[T]^H= \bar{[T]}^t$ only if the bases are orthonormal. Otherwise you get some extra matrices there. $\endgroup$ – daw Jun 19 '15 at 18:40

Denote $[T]=(a_{ij})$ and $[T^*]=(b_{ij})$. Denote further $X=(x_{ij})$ with $x_{ij}=\langle e_i,e_j\rangle$ and $Y=(y_{ij})$ with $y_{ij}=\langle b_i,b_j\rangle$. Then $ T(e_j)= \sum_k a_{kj} b_k $ and $$ \langle Te_j,b_i\rangle = \sum_k\langle a_{kj} b_k,b_i\rangle = \sum_k a_{kj} y_{ki} . $$ Analogously $T^*(b_i) = \sum_k b_{ki} e_k$, and $$ \langle T^*(b_i), e_j\rangle = \sum_k\langle b_{ki} e_k,e_j\rangle = \sum_k b_{ki} x_{kj} . $$ Hence $$ \sum_k a_{kj} y_{ki} = \langle Te_j,b_i\rangle =\overline{\langle T^*(b_i), e_j\rangle} = \sum_k \overline{ b_{ki}}\overline{x_{kj}} , $$ writing this as matrix equation yields $$ Y^t [T]=\overline{[T^*]}^t \bar X. $$ If both bases are orthonormal then $X$ and $Y$ are identity matrices, and $[T]=\overline{[T^*]}^t$ is obtained.

  • $\begingroup$ can we conclude anything about the invertibility of the matrices $X$ and $Y$? $\endgroup$ – Saaqib Mahmood Jun 19 '16 at 14:07
  • $\begingroup$ they are invertible, as they are generated by scalar products of basis vectors. $\endgroup$ – daw Jun 20 '16 at 17:47

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