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Exercise 2.6.16

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We have that

rank (L_{A}) = \dim (R (L_{A})) = m - \dim (R {(L_{A})}^{0}) = m - \dim (N ({(L_{A})}^{t}))

= \dim ({(𝔽^{m})}^{∗}) - \dim (N ({(L_{A})}^{t})) = \dim (R ({(L_{A})}^{t})) .

Next, let $α$ , $β$ be the standard basis for $𝔽^{n}$ and $𝔽^{m}$ . Let $α^{∗}$ , $β^{∗}$ be their dual basis. So we have that ${[L_{A})^{t}]}_{β^{∗}}^{α^{∗}} = {({[L_{A}]}_{α}^{β})}^{t} = A^{t}$ by Theorem 2.25. Let $ϕ_{β^{∗}}$ be the isomorphism defined in Theorem 2.21. We get

\dim (R ({(L_{A})}^{t})) = \dim (ϕ_{β^{∗}} (R ({(L_{A})}^{t}))) = \dim (R (L_{A^{t}})) = rank (L_{A^{t}}) .

jephianlin

2011-06-27 00:00

Exercise 2.6.16

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