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Expert-verified Found in: Page 186 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # In Exercise 72 through 74, let ${{\mathbf{Z}}}_{{\mathbf{n}}}$ be the set of all polynomials of degree ${\mathbf{\le }}{\mathbf{n}}$ such that f(0) = 0. 74. Define an isomorphism from ${{\mathbf{Z}}}_{{\mathbf{n}}}$ to ${{\mathbf{P}}}_{\mathbf{n}\mathbf{-}\mathbf{1}}$ (think calculus!).

The transformation $T\left(f\left(t\right)\right)=f\text{'}\left(t\right)$ is an isomorphism from ${Z}_{n}$ to ${P}_{n-1}$.

See the 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. ${\mathbf{T}}{\mathbf{\left(}}{\mathbf{f}}{\mathbf{+}}{\mathbf{g}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{T}}\left(f\right){\mathbf{+}}{\mathbf{T}}\left(g\right)$
2. ${\mathbf{T}}\left(\mathrm{kf}\right){\mathbf{=}}{\mathbf{kT}}\left(f\right)$

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

If is finite dimensional, then

${\mathbf{dim}}{\mathbf{\left(}}{\mathbf{V}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{rank}}{\mathbf{\left(}}{\mathbf{T}}{\mathbf{\right)}}{\mathbf{+}}{\mathbf{nullity}}{\mathbf{\left(}}{\mathbf{T}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{dim}}\mathbf{\left(}\mathbf{imT}\mathbf{\right)}{\mathbf{+}}{\mathbf{dim}}\mathbf{\left(}\mathbf{kerT}\mathbf{\right)}$

## 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 $T:{Z}_{n}\to {P}_{n-1}$ given by $T\left(f\left(t\right)\right)=f\text{'}\left(t\right)$.

Since, $f\left(t\right),g\left(t\right)\in {Z}_{n}$

$T\left(f\left(t\right)+g\left(t\right)\right)=T\left(\left(f+g\right)t\right)\phantom{\rule{0ex}{0ex}}=\left(f+g\right)\text{'}t=f\text{'}\left(t\right)+g\text{'}\left(t\right)\phantom{\rule{0ex}{0ex}}=T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)$

Also, consider a scalar c, then

$T\left(cf\left(t\right)\right)=T\left(\left(cf\right)t\right)\phantom{\rule{0ex}{0ex}}=\left(cf\right)\text{'}t=cf\text{'}\left(t\right)\phantom{\rule{0ex}{0ex}}=cT\left(f\left(t\right)\right)$

Therefore, T is a linear transformation.

Consider any non-zero polynomial,

$f\left(t\right)={a}_{0}+{a}_{1}t+...+{a}_{n-1}{t}^{n-1}\in {P}_{n-1}$

Then,

$T\left(f\left(t\right)\right)=f\text{'}\left(t\right)\phantom{\rule{0ex}{0ex}}={a}_{1}+2{a}_{2}+...+n{a}_{n}{t}^{n-1}$

Since, it non-zero there exists $0\le i\le n-1$ such that ${a}_{i}\ne 0$.

Thus, $T\left(f\left(t\right)\right)$ is also a non-zero polynomial in ${P}_{n-1}$.

Hence, $\mathrm{ker}\left(T\right)=\left\{0\right\}$.

Therefore, $\mathrm{dim}\left({\mathrm{Z}}_{\mathrm{n}}\right)=\mathrm{n}=\mathrm{dim}\left({\mathrm{p}}_{\mathrm{n}-1}\right)$.

Thus, T is an isomorphism. ### Want to see more solutions like these? 