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Q30E

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Found in: Page 200

### Linear Algebra With Applications

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

# State true or false, the linear transformation T (f) = f (4t - 3 ) from P to P is an isomorphism.

The statement is True

See the step by step solution

## Step 1: Determine the linearity of T .

Consider the function T(f (t)) = f (4t - 3 ) from ${P}_{2}\mathrm{to}{P}_{2}$.

A function D is called a linear transformation on localid="1659425827307" ${\mathbf{ℝ}}$ if the function D satisfies the following property’s.

${\mathbf{\left(}}{\mathbf{a}}{\mathbf{\right)}}{\mathbf{}}{\mathbf{D}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{x}}{\mathbf{+}}{\mathbf{y}}{\mathbf{\right)}}{\mathbf{}}{\mathbf{=}}{\mathbf{D}}{\mathbf{\left(}}{\mathbf{x}}{\mathbf{\right)}}{\mathbf{+}}{\mathbf{D}}{\mathbf{\left(}}{\mathbf{y}}{\mathbf{\right)}}{\mathbf{}}{\mathbf{for}}{\mathbf{}}{\mathbf{all}}{\mathbf{}}{\mathbf{x}}{\mathbf{,}}{\mathbf{y}}{\mathbf{}}{\mathbf{\in }}{\mathbf{ℝ}}\phantom{\rule{0ex}{0ex}}{\mathbf{\left(}}{\mathbf{b}}{\mathbf{\right)}}{\mathbf{}}{\mathbf{D}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{\alpha x}}{\mathbf{\right)}}{\mathbf{}}{\mathbf{=}}{\mathbf{\alpha D}}{\mathbf{\left(}}{\mathbf{x}}{\mathbf{\right)}}{\mathbf{}}{\mathbf{for}}{\mathbf{}}{\mathbf{all}}{\mathbf{}}{\mathbf{constant}}{\mathbf{}}{\mathbf{\alpha }}{\mathbf{\in }}{\mathbf{ℝ}}$

An invertible linear transformation is called isomorphism or dimension of domain and co-domain is not same then the function is not isomorphism.

Assume $f,g\in {P}_{2}\mathrm{then}T\left(f\left(t\right)\right)=f\left(4t-3\right)\mathrm{and}T\left(g\left(t\right)\right)=g\left(4t-3\right)$.

Substitute the value f ( 4t - 3 ) for T ( f (t) ) and g ( 4t - 3 ) for T ( f (t) ) in T ( f (t) ) +T ( f (t) ) as follows.

T ( f (t) ) +T ( f (t) ) =f (4t - 3) + g (4t - 3)

Now, simplify T ( { f + g } (t)) as follows.

T ( { f + g } (t)) = { f + g } ( 4t - 3)

T ( { f + g } (t)) = f ( 4t - 3) +g ( 4t - 3)

T ( { f + g } (t)) = T ( f(t) + T ( g (t))

$\mathrm{Assume}f\in {P}_{2}\mathrm{and}\alpha \in \mathrm{ℝ}\mathrm{then}\mathbit{T}\left(\left(\alpha f\right)\right)=\left(\alpha f\right)\left(4t-3\right).$

Simplify the equation $\mathbit{T}\left(\left(\alpha f\right)\left(t\right)\right)=\left(\alpha f\right)\left(4t-3\right).$as follows.

$\mathbit{T}\left(\left(\alpha f\right)\left(t\right)\right)=\left(\alpha f\right)\left(4t-3\right)\phantom{\rule{0ex}{0ex}}\mathbit{T}\left(\left(\alpha f\right)\left(t\right)\right)=\alpha f\left(4t-3\right)\phantom{\rule{0ex}{0ex}}\mathbit{T}\left(\left(\alpha f\right)\left(t\right)\right)=\alpha T\left(f\left(t\right)\right).$

As $T\left(\left\{f+g\right\}\left(t\right)\right)=T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)$and $T\left(\alpha f\left(t\right)\right)=\alpha T\left(f\left(t\right)\right)$, by the definition of linear transformation T is linear.

## Step 2: Determine the rank of the transformation.

As $\left\{1,t,{t}^{2},...\right\}$is the basis element of P and the matrix corresponding to basis is upper triangular matrix.

Each row and column of upper triangular matrix is linearly independent.

Therefore, the rank (T) is n.

## Step 3: Determine the dimension of kernel.

The Kernel (T) is defined as follows.

dim(T) = dim (Kernal (T)) + rank (T)

n = dim ( Kernel (T)) + 0

dim (Kernel (T)) = 0

Therefore, the dimension of is Kernel (T).

## Step 4: Determine the isomorphism.

Theorem: Consider a linear transformation T defined from $T:V\to W$then the transformation T is an isomorphism if and only if Ker (T) = 0 where .

$ker\left(T\right)=\left\{f\left(x\right)\in P:T\left\{f\left(x\right)\right\}=0\right\}$

As the dimension of Kernel (T) is 0, by the theorem the function T is an isomorphism.

Hence, the statement is True.