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Exercise 6.3.18

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For arbitrary matrix $M$ we already have $\det (M) = \det (M^{t})$ . So it’s sufficient to show that $\bar{\det (M)} = \det (\bar{M})$ . We prove this by induction on $n$ , the size of a matrix. For $n = 1$ , we have

\bar{\det (\begin{matrix} a \end{matrix})} = \det (\begin{matrix} \bar{a} \end{matrix}) .

For $n = 2$ , we have

\bar{\det (\begin{matrix} a & b \\ c & d \end{matrix})} = \bar{ad - bc}

= \bar{a} \bar{d} - \bar{b} \bar{c} = \det (\begin{matrix} \bar{a} & \bar{b} \\ \bar{c} & \bar{d} \end{matrix}) .

Suppose the hypothesis is true for $n = k - 1$ . Consider a $k \times k$ matrix $M$ . We have

\begin{array}{l} \bar{\det (M)} & = \bar{\sum_{j = 1}^{k} {(- 1)}^{i + j} M_{ij} \cdot \det ({\tilde{M}}_{ij})} \\ = \sum_{j = 1}^{k} {(- 1)}^{i + j} \bar{M_{ij}} \cdot \bar{\det ({\tilde{M}}_{ij})} \\ = \sum_{j = 1}^{k} {(- 1)}^{i + j} \bar{M_{ij}} \cdot \det ({\tilde{\bar{M}}}_{ij}) = \det (\bar{M}) . \end{array}

This means that

\det (A) = \det (A^{t}) = \det (A^{∗}) .

jephianlin

2011-06-27 00:00

Exercise 6.3.18

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