Let \( f: \mathbb A_{2 X 2} \rightarrow \mathbb P_3 \) be linear transform defined by:

\[ f( \begin{vmatrix}{p_{1}} && {p_{2}} \\ {p_{3}} && {p_{4}}\end{vmatrix}) = p1 + p2 * x + p3 * x^2 + p4 * x^3 \]

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Let \[ A = \{ \begin{vmatrix}{1} && {2} \\ {3} && {4}\end{vmatrix}, \begin{vmatrix}{-2} && {1} \\ {5} && {3}\end{vmatrix}, \begin{vmatrix}{4} && {-2} \\ {1} && {7}\end{vmatrix}, \begin{vmatrix}{7} && {6} \\ {4} && {3}\end{vmatrix} \} \] be a basis for \( \mathbb A_{2 X 2} \)

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and,

\[ B = \{ 2 - x + 4 * x^2 + 3 * x^3, 5 + 2 * x + x^2 + 4 * x^3, 1 - x + x^2 + 2 * x^3, 2 + x^2 - x^3 \} \] be a basis for \( \mathbb P^3 \)

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If \( M = [a_{ij}]_{4 \: x \: 4} \) represents the linear transform above find \( \displaystyle S = \sum_{i = 1}^{4} \sum_{j = 1}^{4} a_{i j}.\) \(\)

Express the result to four decimal places.

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If your interested you can write a program(in any language) for the General Case below and use it to find the above result.

\(\) General Case:(For program)

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For this case I wrote the matrix in an unconventional manner. \(\)

Let \( f: \mathbb A_{n X n} \rightarrow \mathbb P_{n^2 - 1} \) be linear transform defined by:

\[ f( \begin{vmatrix}{x_{1}} && {x_{2}} && {...} && {x_{n}} \\
{x_{n + 1}} && {x_{n + 2}} && {...} && {x_{2 * n}} \\
{...} && {...} && {...} \\

{x_{(j - 1) * n + 1}} && {x_{(j - 1) * n + 2}} && {...} && {x_{j * n}} \\

{...} && {...} && {...} \\

{x_{(n - 1) * n + 1}} && {x_{(n - 1) * n + 2}} && {...} && {x_{n^2}} \\

\end{vmatrix}) = \sum_{j = 1}^{n^2} p_{j} * x^{j - 1} \]

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Let \[ V_{q} = \begin{vmatrix}{x_{1q}} && {x_{2q}} && {...} && {x_{nq}} \\
{x_{(n + 1)q}} && {x_{(n + 2)q}} && {...} && {x_{(2 * n)q}} \\
{...} && {...} && {...} \\

{x_{((j - 1) * n + 1)q}} && {x_{((j - 1) * n + 2)q}} && {...} && {x_{(j * n)q}} \\

{...} && {...} && {...} \\

{x_{((n - 1) * n + 1)q}} && {x_{((n - 1) * n + 2)q}} && {...} && {x_{(n^2)q}} \\

\end{vmatrix} \]

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and \( A = \{V_{q}|(1 <= q <= n^2)\} \) be a basis for \( \mathbb A_{n X n} \).

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Let \[ W_{q} = \sum_{j = 1}^{n^2} p_{jq} * x^{j - 1} \] \(\)

and \(B = \{W_{q}| (1 <= q <= n^2) \} \) be a basis for \( \mathbb P_{n^2 - 1} \)

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Write a program in any language to find the matrix \( M = [a_{ij}]_{n^2 \: x \:n^2} \) representation of the general linear transform above and the sum \( \displaystyle S = \sum_{i = 1}^{n^2} \sum_{j = 1}^{n^2} a_{i j} \). \(\)

Make certain \( A \) and \( B \) are bases for \( \mathbb A_{n X n} \) and \( \mathbb P_{n^2 - 1} \). \(\)

You can use the program written to find the matrix \( M = [a_{ij}]_{4 \: x \: 4} \) that represents the linear transform above and output \( \displaystyle S = \sum_{i = 1}^{4} \sum_{j = 1}^{4} a_{i j} \).

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Let \[X = \begin{vmatrix}{3} && {-2} \\ {5} && {7}\end{vmatrix} \]

To check the matrix \( M = [a_{ij}]_{4 \: x \: 4} \) found, first find \([f(X)]_B\) without using the matrix \( M = [a_{ij}]_{4 \: x \: 4} \), then find \([f(X)]_B\) using the matrix \( M = [a_{ij}]_{4 \: x \: 4} \).

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