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Q74E

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Linear Algebra With Applications
Found in: Page 186
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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Short Answer

In Exercise 72 through 74, let Zn be the set of all polynomials of degree n such that f(0) = 0.

74. Define an isomorphism from Zn to Pn-1 (think calculus!).

The transformation Tft=f't is an isomorphism from Zn to Pn-1.

See the step by step solution

Step by Step Solution

Step 1: Definition of Linear transformation

Consider two linear spaces V and W. A function T from V to W is called linear transformation if

  1. T(f+g)=T(f)+T(g)
  2. T(kf)=kT(f)

for all elements f and g of V and for all scalars .

If is finite dimensional, then

dim(V)=rank(T)+nullity(T)=dim(imT)+dim(kerT)

Step 2: Definition of Isomorphism

An invertible linear transformation T is called an isomorphism.

Step 3: Verify whether the linear transformation is an isomorphism

Consider the linear transformation T:ZnPn-1 given by Tft)=f'(t.

Since, ft,gtZn

Tft+gt=Tf+gt =f+g't=f't+g'(t) =Tft+Tgt

Also, consider a scalar c, then

Tcft=Tcft =cf't=cf't =c Tft

Therefore, T is a linear transformation.

Consider any non-zero polynomial,

ft=a0+a1t+...+an-1tn-1 Pn-1

Then,

Tft=f't =a1+2a2+...+nantn-1

Since, it non-zero there exists 0in-1 such that ai0.

Thus, Tft is also a non-zero polynomial in Pn-1.

Hence, ker(T)={0}.

Therefore, dimZn=n=dimpn-1.

Thus, T is an isomorphism.

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