Fuzhen -Zhang linear algebra Problem Page No.112

Question

For $M_n (\mathbb{C})$, the vector space of all $n \times n $ complex matrices,

if $\langle A, X \rangle \ge 0$ for all $X \ge 0$ in $M_n{\mathbb{C}}$,then $A \ge 0$

which of the following define an inner product on $M_n(\mathbb{C})$?

$1)$$ \langle A, B\rangle = tr(A^*B)$

$2)$$ \langle A, B\rangle = tr(AB^*)$

$3)$$\langle A, B\rangle = tr(BA)$

Taken from Zhang linear algebra books page no .112.

My attempts: I read this Wikipedia article, but could not get any idea on how to clarify these options:

https://en.wikipedia.org/wiki/Inner_product_space

Any hints/solutions will be appreciated, thank you.

Answer 1

Hint: For an inner product, the following statements need to be true:

• $\langle p A, qB \rangle = p \overline{q} \langle A,B \rangle $ for $p,q \in \Bbb C$ ($\bar q$ here denotes the conjugate of $q$)
• $A \neq 0$ implies that $\langle A, A \rangle \neq 0$

These two properties will allow to eliminate the two options that fail to be inner products (in the sense defined by Zhang). The remaining definition indeed yields a valid inner product.

Answer 2

From $(b)$ we can see the positive-definiteness of 1) and 2) 3) isn't positive-definite. Linearity comes from the linearity of trace operator for 1) and 2) Conjugate symmetry: For 1), $N_1(B,A) = tr(B*A) $ but by $5.21$ being a inner-product, $ N_1(B,A) = N_0(A,B) = \overline{N_0(B,A)} = \overline{N_1(A,B)}$ For 2), $Tr(AB) = Tr(BA)$. And 1) is an inner-product. So 2) is an inner-product.