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Q73E

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. 73. Is the linear transformation ${\mathbf{T}}\left(f\left(t\right)\right){\mathbf{=}}{{\mathbf{\int }}}_{{\mathbf{0}}}^{{\mathbf{t}}}{\mathbf{f}}\left(x\right){\mathbit{d}}{\mathbit{x}}$ an isomorphism from ${{\mathbf{P}}}_{\mathbf{n}\mathbf{-}\mathbf{1}}$ to ${{\mathbf{Z}}}_{{\mathbf{n}}}$?

The linear transformation $\mathrm{T}\left(\mathrm{f}\left(\mathrm{t}\right)\right)={\int }_{0}^{\mathrm{t}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$ is an isomorphism from ${\mathrm{P}}_{\mathrm{n}-1}$ to ${\mathrm{Z}}_{\mathrm{n}}$.

See the step by step solution

## Step 1: Definition of Linear transformation.

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

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

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

If V is finite dimensional, then

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

## Step 2: Definition of Isomorphism.

An invertible linear transformation T is called an isomorphism.

## Step 3: Verification whether the linear transformation is isomorphism.

Consider the linear transformation $T:{P}_{n-1}\to {Z}_{n}$ given by $T\left(f\left(t\right)\right)={\int }_{0}^{t}f\left(x\right)dx$.

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,

n-1

$T\left(f\left(t\right)\right)={\int }_{0}^{t}\left({a}_{0}t+{a}_{1}t+...+{a}_{n-1}{t}^{n-1}\right)dt\phantom{\rule{0ex}{0ex}}={\left[{a}_{0}t+{a}_{1}\frac{{t}^{2}}{2}+...+{a}_{n-1}\frac{{t}^{n}}{n}\right]}_{0}^{t}$

On applying the limits,

$T\left(f\left(t\right)\right)={a}_{0}t+{a}_{1}\frac{{t}^{2}}{2}+...+{a}_{n-1}\frac{{t}^{n}}{n}$

Since, it non-zero there exists localid="1659421702278" $0\le i\le n-1$ such that ${a}_{i}\ne 0$.

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

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

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.