We define $\langle\cdot, \cdot\rangle $: $\mathbb{C}^{n\times n} \times \mathbb{C}^{n\times n}\longrightarrow \mathbb{C} $ by $ \langle A, B\rangle := \operatorname{Tr}(AB^{∗})$. Show that this mapping gives an inner product on the vector space $\mathbb{C}^{n\times n}$.

Does someone have an idea how I can prove that?

  • 3
    $\begingroup$ Can you write the definition of an inner product and tell us which property you are unable to verify? $\endgroup$ Jan 21 '19 at 8:55
  • $\begingroup$ just show that it is positive definite, symmetrical (well in this case conjugate symmetrical) and linear with scalar $\endgroup$
    – user29418
    Jan 21 '19 at 8:57
  • $\begingroup$ see also math.stackexchange.com/questions/476802/… $\endgroup$
    – Widawensen
    Jan 21 '19 at 15:03

An inner product space is a vector space $\mathbb{C}^{n×n} \times \mathbb{C}^{n×n}$ over the field $\mathbb{C}$ together with an inner product, i.e., with a map $$\mathbb{C}^{n×n} \times \mathbb{C}^{n×n}\longrightarrow \mathbb{C}$$ such that satisties

  1. Conjugate symmetry:

    $\langle A,B\rangle ={\overline {\langle B,A\rangle }}$

  2. Linearity in the first argument:

    $\langle aA,B\rangle =a\langle A,B\rangle \\\langle A+B,C\rangle =\langle A,C\rangle +\langle B,C\rangle $

  3. Positive-definiteness:

    $\langle x,x\rangle \geq 0\\\langle x,x\rangle =0\Leftrightarrow x=\mathbf {0} \,.$

where $A,B,C \in \mathbb{C}^{n×n} \times \mathbb{C}^{n×n}$ and $a\in \mathbb{C}$.


Use the following facts.

$$Tr(aA)=aTr(A)$$ $$Tr(A)=Tr(A)^T$$ $$Tr(AB)=Tr(BA)$$ $$Tr(AB^T)=Tr(A^TB)=Tr(B^TA)=Tr(BA^T)$$

  • $\begingroup$ You're right, I just took the inverse equation and apply $Tr$ both sides. My bad. $\endgroup$ Jan 21 '19 at 15:49

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