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Exercise 6.1.21

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1.

Observe that

A_{1}^{∗} = {(\frac{1}{2} (A + A^{∗}))}^{∗} = \frac{1}{2} (A^{∗} + A) = A_{1}^{∗},

A_{2}^{∗} = {(\frac{1}{2 i} (A - A^{∗}))}^{∗} = (- \frac{1}{2 i} (A^{∗} - A) = A_{2}^{∗} .

and

A_{1} + i A_{2} = \frac{1}{2} (A + A^{∗}) + \frac{1}{2} (A - A^{∗}) = \frac{1}{2} (A + A) = A .

I don’t think it’s reasonable because $A_{1}$ does not consists of the real part of all entries of $A$ and $A_{2}$ does not consists of the imaginary part of all entries of $A$ . But what’s the answer do you want? Would it be reasonable to ask such a strange question?

2.

If we have

A = B_{1} + i B_{2}

with

B_{1}^{∗} = B_{1}

and

B_{2}^{∗} = B_{2}

, then we have

A^{∗} = B_{1}^{∗} - i B_{2}^{∗}

. Thus we have

B_{1} = \frac{1}{2} (A + A^{∗})

and

B_{2} = \frac{1}{2 i} (A - A^{∗}) .

jephianlin

2011-06-27 00:00

Exercise 6.1.21

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