Exercise 4.5.14

Answers

We could write

δ (\begin{matrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{matrix}) = (A_{11}, A_{12}) \cdot (A_{22} a + A_{21} b, A_{22} c + A_{21} d)

= (A_{21}, A_{22}) \cdot (A_{11} b + A_{12} d, A_{11} a + A_{12} c)

and get the desired result since inner product function is linear.

For the converse, fixed one $n$ -linear function $δ$ and let

a = δ (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}), b = δ (\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}),

c = δ (\begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix}), d = δ (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) .

Now we must have

δ (\begin{matrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{matrix}) = A_{11} δ (\begin{matrix} 1 & 0 \\ A_{21} & A_{22} \end{matrix}) + A_{12} δ (\begin{matrix} 0 & 1 \\ A_{21} & A_{22} \end{matrix})

= A_{11} (A_{21} δ (\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}) + A_{22} δ (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})) + A_{12} (A_{21} δ (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) + A_{22} δ (\begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix}))

= A_{11} A_{22} a + A_{11} A_{21} b + A_{12} A_{22} c + A_{12} A_{21} d .

jephianlin

2011-06-27 00:00