Exercise 3.C.10

Answers

Proof. Use the same notation as in the motivation prior to 3.41.

Let $Z$ be a $1$ -dimensional vector space and $z$ a basis for it. Define $S_{j} : V \to Z$ by

S_{j} v_{r} = A_{j, r} z

for $1 \leq r \leq n$ . Obviously, $M (S_{j}) = A_{j, .}$ . Therefore $A_{j, .} C = M (S_{j}) M (T) = M (S_{j} T)$ . Thus

\begin{aligned} S_{j} T u_{k} & = S_{j} (\sum_{r = 1}^{n} C_{r, k} v_{r}) \\ = \sum_{r = 1}^{n} C_{r, k} S_{j} v_{r} \\ = \sum_{r = 1}^{n} C_{r, k} A_{j, r} z \\ = (AC)_j, kz \end{aligned}

for $1 \leq k \leq p$ . Hence $(AC)_j, . = A_{j, .} C$ . □

celiopassos

2017-10-06 00:00