Let $T \colon \mathbb{C}^n \to \mathbb{C}^n$ be a linear operator, which is bounded of course; so the Hilbert adjoint operator $T^* \colon \mathbb{C}^n \to \mathbb{C}^n$ is a well-defined bounded linear operator with $\Vert T^* \Vert = \Vert T \Vert$.

Suppose that $A$ and $B$ are the matrices of $T$ and $T^*$, respectively, with respect to a given basis for $\mathbb{C}^n$.

Then how do we obtain the following relations? $$\langle Tx, y \rangle = \left( Ax \right)^T \overline{y} \ \mbox{ for all } \ x, y \in \mathbb{C}^n.$$ And, $$\langle x, T^*y \rangle = x^T\overline{By} \ \mbox{ for all } \ x, y \in \mathbb{C}^n.$$

Here we treat any $u \in \mathbb{C}^n$ as a coluumn vector. Moreover, $u^T$ is the row vector that is the transpose of $u$, and $\overline{u}$ is the column vector obtained from $u$ by replacing each component of $u$ by its complex conjugate.

After @Martin Argerami's comment.

Let $\{e_1, \ldots, e_n\}$ be an ordered basis for $\mathbb{C}^n$.

Then each $x \in \mathbb{C}^n$ has a unique representation $$x = \sum_{j=1}^n \xi_j e_j,$$ where $\xi_1, \ldots, \xi_n \in \mathbb{C}$.

For each $j = 1, \ldots, n$, since $T e_j \in \mathbb{C}^n$, there are scalars $\alpha_{ij} \in \mathbb{C}$ for each $i = 1, \ldots, n$ such that $$T e_j = \sum_{i=1}^n \alpha_{ij} e_i.$$

Since $Tx \in \mathbb{C}^n$, we have $$T x = \sum_{i=1}^n \eta_i e_i \ \tag{equation (1) } $$ for a unique $n$-tuple of complex numbers $\eta_1, \ldots, \eta_n$.

But since $x = \sum_{j=1}^n \xi_j e_j$ and since $T$ is linear, we have $$ \begin{align*} Tx & = T( \sum_{j=1}^n \xi_j e_j ) \\ & = \sum_{j=1}^n \xi_j T e_j \\ &= \sum_{j=1}^n ( \xi_j \sum_{i=1}^n \alpha_{ij} e_i ) \\ & = \sum_{j=1}^n \sum_{i=1}^n \xi_j \alpha_{ij} e_i \\ & = \sum_{i=1}^n (\sum_{j=1}^n \alpha_{ij} \xi_j ) e_i. \ \tag{equation (2)} \end{align*}$$

From equation (1) and (2), we obtain $$ \sum_{i=1}^n \eta_i e_i = Tx = \sum_{i=1}^n (\sum_{j=1}^n \alpha_{ij} \xi_j ) e_i. $$ So $$ \sum_{i=1}^n \eta_i e_i - \sum_{i=1}^n (\sum_{j=1}^n \alpha_{ij} \xi_j ) e_i = 0.$$ Or, $$ \sum_{i=1}^n \left( \eta_i - \sum_{j=1}^n \alpha_{ij} \xi_j \right) e_i = 0.$$ Since $e_1, \ldots, e_n$ are linearly independent, we have $$ \eta_i = \sum_{j=1}^n \alpha_{ij} \xi_j \ \mbox{ for each } \ i = 1, \ldots, n. \tag{equation (3)} $$

The matrix $$ A \colon= [\alpha_{ij} ]_{m \times n}$$ is the matrix of the operator $T$ with respcet to the given basis $\{ e_1, \ldots, e_n \}$.

The column vector $$\tilde{x} \colon= [\xi_1, \ldots, \xi_n ]^T$$ is the co-ordinate vector of $x$ with respect to this basis.

And, the column vector $$\tilde{Tx} \colon= [\eta_1, \ldots, \eta_n ]^T$$ is the co-ordinate vector of $Tx $ with respect to the same basis.

Thus we have $$\tilde{Tx } = A \tilde{x} \ \mbox{ for all } \ x \in \mathbb{C}^n. \ \tag{equation (4)} $$

Now let $y \in \mathbb{C}^n$. Then there are unique complex numbers $\lambda_1, \ldots, \lambda_n$ such that $$y = \sum_{i=1}^n \lambda_i e_i.$$ So we have $$\langle Tx, y \rangle = \langle \sum_{i=1}^n \eta_i e_i, \sum_{i=1}^n \lambda_i e_i \rangle = \sum_{i=1}^n \sum_{j=1}^n \eta_i \overline{\lambda_j} \langle e_i, e_j \rangle.$$

Now what next?

If $\{e_1, \ldots, e_n \}$ is an orthonormal set, then we can say that $$\langle Tx, y \rangle = (Ax)^T \overline{y} = x^T A^T \overline{y}.$$ But what if $\{ e_1, \ldots, e_n \}$ is not orthonormal? How do we show the above equality in that case?

  • $\begingroup$ @Daniel Fischer, can you please answer my question in full detail? $\endgroup$ – Saaqib Mahmood Jun 5 '15 at 17:44

It is simply the equality $$ \langle x,y\rangle =x^T\bar y, $$ together with the fact that $ Tx=Ax $, $T^*x=Bx $.

  • $\begingroup$ how is $Tx = Ax$? And, how is $T^*x = Bx$? Can you please explain the whole process in detail? $\endgroup$ – Saaqib Mahmood Jun 4 '15 at 20:15
  • $\begingroup$ Details? It was you, who wrote "Suppose that $A$ and $B$ are the matrices of $T$ and $T^∗$, respectively". What did you mean by that? $\endgroup$ – Martin Argerami Jun 5 '15 at 1:26
  • $\begingroup$ can you please have a look at my question after my latest edit based on your answer above? $\endgroup$ – Saaqib Mahmood Jun 5 '15 at 17:42
  • $\begingroup$ Yes, that's exactly how it goes. As for your last question, we use orthonormal bases precisely because they make our life easier. $\endgroup$ – Martin Argerami Jun 5 '15 at 17:46
  • $\begingroup$ But, @Martin Argerami, Kreyszig states that, a basis for $\mathbb{C}^n$ being given, if $A$ is the matrix of $T$, then $\langle Tx, y \rangle = (Ax)^T \overline{y}$ for all $x, y \in \mathbb{C}^n$. I wonder how this could be true. $\endgroup$ – Saaqib Mahmood Jun 6 '15 at 4:58

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