Let $T$ be the linear operator on $\mathbb{C}^2$ defined by $$T(a,b)=(2ia+3b,a-b).$$ I am trying to compute the adjoint. The answer is $$T^*(c,d)=(-2ic+d,3c-d)\tag{$1$},$$ which can be seen by (i) computing the matrix of $T$ with respect to the standard ordered basis, then (ii) taking the conjugate transpose of this matrix, and then (iii) using the new matrix to write the formula of $T^*$.

However, I'd like to compute the adjoint by using the definition: $\langle T(v),w\rangle = \langle v, T^*(w) \rangle$. I have $$\langle (a,b),T^*(c,d) \rangle= \langle T(a,b), (c,d) \rangle\\=\langle (2ia+3b,a-b), (c,d)\rangle \\=2aic+3bc+ad-bd\\=a(2ic+d)+b(3c-d)\\=\langle (a,b), (2ic+d, 3c-d)\rangle.$$ However, this implies $T^*(c,d)=(2ic+d,3c-d),$ which disagrees with equation $(1)$.

How can I obtain the correct answer using the definition?

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    $\begingroup$ Well there must be a definition of the inner product given in the question. Or is it the standard inner product? $\endgroup$ – Yadati Kiran Feb 2 at 6:46

You forgot to take the complex conjugate. $$ \langle((2ia + 3b, a-b), (c,d)\rangle=-2aic+3bc+ad-bd $$


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