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Q58E

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Found in: Page 185

Linear Algebra With Applications

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

Find the image and kernel of the transformation ${\mathbf{T}}$in${\mathbf{T}}\left({x}_{0},{x}_{1},{x}_{2},{x}_{3},...\right){\mathbf{=}}\left(0,{x}_{0},{x}_{2},{x}_{4}\right)$ from ${\mathbf{V}}$to ${\mathbf{V}}$.

The solution is the image consist of all infinite sequence with the initial element as 0 and the kernel contains zero sequence only.

See the step by step solution

Step1: Explanation of the solution

Consider the sequence as follows.

$\mathrm{T}\left({\mathrm{x}}_{0},{\mathrm{x}}_{1},{\mathrm{x}}_{2},{\mathrm{x}}_{3},...\right)=\mathrm{T}\left({\mathrm{x}}_{0},{\mathrm{x}}_{2},{\mathrm{x}}_{4}\right)$

The image of the sequence is as follows.

$\left(0,{\mathrm{x}}_{0},{\mathrm{x}}_{2},{\mathrm{x}}_{4},...\right)$

Therefore, the image consist of all infinite sequence whose initial element is 0.

Step2: Find kernel of the sequence

Consider the sequence as follows.

$\mathrm{T}\left({\mathrm{x}}_{0},{\mathrm{x}}_{1},{\mathrm{x}}_{2},{\mathrm{x}}_{3},...\right)=\mathrm{T}\left({\mathrm{x}}_{0},{\mathrm{x}}_{2},{\mathrm{x}}_{4}\right)$

Now, kernel of the sequence as follows.

$T\left(0,{x}_{0},{x}_{1},{x}_{2},0,{x}_{2},0,{x}_{3},0,...\right)=\left(0,0,0,0,...\right)\phantom{\rule{0ex}{0ex}}{x}_{0}={x}_{1}={x}_{2}=...=0$

Thus, the kernel contains zero sequence only.

Hence, the image consist of all the infinite sequence with initial element as 0 whereas the kernel contains zero sequence only.