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Exercise 2.4.22

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It’s linear since

T (f + cg) = ((f + cg) (c_{0}), (f + cg) (c_{1}), \dots (f + cg) (c_{n}))

= (f (c_{0}) + cg (c_{0}), f (c_{1}) + cg (c_{1}), \dots f (c_{n}) + cg (c_{n})) = T (f) + cT (g) .

Since $T (f) = 0$ means $f$ has $n + 1$ zeroes, we know that $f$ must be zero function( This fact can be proven by Lagrange polynomial basis for $P_{n} (𝔽)$ .). So $T$ is injective and it will also be surjective since domain and codomain have same finite dimension.

jephianlin

2011-06-27 00:00

Exercise 2.4.22

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