Consider the function S Pn R n1given by Sfx f0 f1 fn a

Consider the function S : P_n R ⁿ⁺¹given by:

S(f(x)) = (f(0), f(1), . . . , f(n))

(a) Show that S is linear.

(b) Show that S is bijective.

Take two functions f(x) and g(x)

S(f(x)) = (f(0), f(1), . . . , f(n)) and

S(g(x)) = (g(0), g(1), . . . , g(n))

Hence sum

S(f(x))+S(g(x)) = (f(0), f(1), . . . , f(n))+(g(0), g(1), . . . , g(n))

= (f(0)+g(0), f(1)+g(1), . . . , f(n)+g(n))

S(f(x)+g(x)) = (f(0)+g(0), f(1)+g(1), . . . , f(n)+g(n))

Since S(f(x))+S(g(x)) =S(f(x)+g(x)) S is linear

Also S{cf(x)} = (cf(0),c f(1), . . . ,c f(n))

= c (f(0), f(1), . . . , f(n)) = CS{f(x)}

It follows that S is linear

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b) To show that S is bijective.

Let S{f(x)} = S{g(x)}

Then f(0) = g(0), f(1) = g(1)... f(n) = g(n)

Or in other words f(x) = g(x) for all x from 1 to n

So S is one to one.

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For any image {f(0), f(1),....f(n)} in the co domain, we can find preimage as f(x)

So S is onto.

S is bijective.