In Spence's Linear Algebra, 4th Edition book, there's an exercise in chapter 6 who asks to proof that $\langle x, y\rangle = x \cdot A \cdot y^{*}$ is an inner product in $\mathbf{C}^{2}$, with: $$ A = \begin{bmatrix} 1 & i \\ -i & 2 \\ \end{bmatrix}$$

I don't know if there's an easier and faster way to solve this, but basically what I do is prove these four sentences, where $x,y,z \in \mathbf{C}^{2}$ and $\alpha \in \mathbf{C}$:

  1. $ \langle x + z , y \rangle = \langle x , y \rangle + \langle z , y \rangle$
  2. $ \langle \alpha.x, y \rangle = \alpha.\langle x , y \rangle $
  3. $ \langle x, y \rangle = \overline{\langle y , x \rangle}$
  4. $ \langle x, x \rangle \gt 0 \; \text{if} \; x \neq 0$

but I'm stuck on the third item. Here's what I tried so far:

$ \langle x, y\rangle = x \cdot A \cdot y^{*} = \overline{\overline{x \cdot A \cdot y^{*}}} = \overline{\overline{x} \cdot \overline{A} \cdot \overline{y^{*}}} = \overline{\overline{x} \cdot \overline{A} \cdot y^{T}} = \overline{ (y \cdot A^{*} \cdot x^{*})^{T}} = \overline{ (y \cdot A \cdot x^{*})^{T}} = \; ? = \overline{ y \cdot A \cdot x^{*}} = \overline{\langle y, x\rangle}$

I used the fact that $(ABC)^T = C^T B^T A^T$ and $A = A^{*}$ for this problem in particular.

  • $\begingroup$ Since you have $A$, I think you just compute with $x=(x_1,x_2), y=(y_1,y_2)$ you will get the result. $\endgroup$ – TokenToucan Dec 25 '15 at 2:36

Note that $y \cdot A \cdot x^*$ is a scalar. Therefore, $(y \cdot A \cdot x^*)^T = y \cdot A \cdot x^*$.

This justifies the step you have marked with "$?$".

  • $\begingroup$ Ok, I had not noticed that. but if I think in this way, I am applying the transpose operator in a scalar, why I can do that? (I can consider that a scalar is an 1x1 matrix?) $\endgroup$ – Mike Dec 25 '15 at 2:44
  • 2
    $\begingroup$ Yes, a scalar can be thought of as a $1 \times 1$ matrix. $\endgroup$ – JimmyK4542 Dec 25 '15 at 2:46

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