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Linear Algebra

Exercise 6.8.9

Answers

1.: It comes from the fact that $\dim (M_{n \times n} (𝔽)) = n^{2}$ .
2.: Let $E_{ij}$ be the matrix whose $ij$ -entry is $1$ and other entries are zero. Then we know that ${E_{ij}}_{ij}$ is a basis in $M_{n \times n} (𝔽)$ . Since $ψ_{β}$ is an isomorphism, the set $ψ_{β}^{- 1} ({E_{ij}}_{ij})$ is a basis for $B (V)$ .

jephianlin

2011-06-27 00:00

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