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Exercise 6.1.4

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1.

We prove the formula

⟨ A, B ⟩ = tr (B^{∗} A) = \sum_{j = 1}^{n} {(B^{∗} A)}_{jj} = \sum_{j = 1}^{n} \sum_{i = 1}^{n} B_{ji}^{∗} A_{ij}

= \sum_{j = 1}^{n} \sum_{i = 1}^{n} A_{ij} {\bar{B}}_{ij} = \sum_{i = 1}^{n} \sum_{j = 1}^{n} A_{ij} {\bar{B}}_{ij} = \sum_{i, j} A_{ij} {\bar{B}}_{ij} .

So we may view the space $M_{n \times n} (𝔽)$ to be $F^{n^{2}}$ and the Frobenius inner product is corresponding to the standard inner product in $F^{n^{2}}$ .

2.

Also use the formulae to compute

∥ A ∥ = {(1 + 5 + 9 + 1)}^{\frac{1}{2}} = 4,

∥ B ∥ = {(2 + 0 + 1 + 1)}^{\frac{1}{2}} = 2,

and

⟨ A, B ⟩ = (1 - i) + 0 - 3 i - 1 = - 4 i .

jephianlin

2011-06-27 00:00

Exercise 6.1.4

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