# An algebra problem by Rocco Dalto

Algebra Level 4

Let $$f: \mathbb A_{2 X 2} \rightarrow \mathbb R^4$$ be linear transform defined by:

$f( \begin{vmatrix}{x_{1}} && {x_{2}} \\ {x_{3}} && {x_{4}}\end{vmatrix}) = \left( \begin{array}{cccc} x_{1} - x_{2} + 2 * x_{3} + 4 * x_{4} \\ 2 * x_{1} - x_{2} + 4 * x_{3} + 3 * x_{4} \\ 4 * x_{1} - 3 * x_{2} + 2 * x_{3} + x_{4} \\ 5 * x_{1} - x_{2} + 2 * x_{3} + 2 * x_{4} \\ \end{array} \right)$



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 $$\mathbb A_{2 X 2}$$



and,

$B = \{ \left( \begin{array}{cccc} 2 \\ -1 \\ 4 \\ 3 \\ \end{array} \right), \left( \begin{array}{cccc} 5 \\ 2 \\ 1 \\ 4 \\ \end{array} \right), \left( \begin{array}{cccc} 1 \\ -1 \\ 1 \\ 2 \\ \end{array} \right), \left( \begin{array}{cccc} 2 \\ 0 \\ 1 \\ -1 \\ \end{array} \right) \}$ be a basis for $$\mathbb R^4$$





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.



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)



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

Let $$f: \mathbb A_{n X n} \rightarrow \mathbb R^{n^2}$$ be linear transform defined by:

$f( \begin{vmatrix}{x_{1}} && {x_{2}} && {...} && {x_{n}} \\ {x_{n + 1}} && {x_{n + 2}} && {...} && {x_{2 * n}} \\ {...} && {...} && {...} \\ {x_{(j - 1) * n + 1}} && {x_{(j - 1) * n + 2}} && {...} && {x_{j * n}} \\ {...} && {...} && {...} \\ {x_{(n - 1) * n + 1}} && {x_{(n - 1) * n + 2}} && {...} && {x_{n^2}} \\ \end{vmatrix}) = \left( \begin{array}{cccc} C_{11} * x_{1} + C_{12} * x_{2} + ... + C_{1 * n^2} * x_{n^2} \\ C_{21} * x_{1} + C_{22} * x_{2} + ... + C_{2 * n^2} * x_{n^2} \\ ... \\ C_{j1} * x_{1} + C_{j2} * x_{2} + ... + C_{j * n^2} * x_{n^2} \\ ... \\ C_{n^21} * x_{1} + C_{n^22} * x_{2} + ... + C_{n^2n^2} * x_{n^2} \\ \end{array} \right)$



Let $V_{q} = \begin{vmatrix}{x_{1q}} && {x_{2q}} && {...} && {x_{nq}} \\ {x_{(n + 1)q}} && {x_{(n + 2)q}} && {...} && {x_{(2 * n)q}} \\ {...} && {...} && {...} \\ {x_{((j - 1) * n + 1)q}} && {x_{((j - 1) * n + 2)q}} && {...} && {x_{(j * n)q}} \\ {...} && {...} && {...} \\ {x_{((n - 1) * n + 1)q}} && {x_{((n - 1) * n + 2)q}} && {...} && {x_{(n^2)q}} \\ \end{vmatrix}$



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



Let $W_{q} = \left( \begin{array}{cccccc} a_{1q} \\ a_{2q} \\ ... \\ a_{jq} \\ ... \\ a_{(n^2)q} \\ \end{array} \right)$



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



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 $$\mathbb R^{n^2}$$. 

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



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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