An algebra problem by Rocco Dalto

Algebra Level 5

Let P 2 \mathbb P_{2} denote the set of all polynomials of degree two and R 2 × 2 \mathbb R_{2 \times2} denote the set of all 2 × 2 2\times 2 matrices.

Let f : P 2 → R 2 × 2 f: \mathbb P_{2} \rightarrow \mathbb R_{2\times2} be linear transform defined by

f ( p 0 + p 1 x + p 2 x 2 ) = ∣ p 0 + p 1 p 1 + p 2 p 0 + p 2 p 2 ∣ . f(p_{0} + p_{1} x + p_{2} x^2) =\begin{vmatrix}{p_0 + p_1} && {p_1 + p_2} \\ {p_0 + p_2} && {p_2}\end{vmatrix} .

Denote
A = { 1 + x , x + x 2 , x 2 } A = \{1 + x, x + x^2, x^2 \} be a basis for P 2 \mathbb P_{2}
and
B = { ∣ 1 0 0 1 ∣ , ∣ 0 1 0 1 ∣ , ∣ 1 1 1 0 ∣ , ∣ 0 0 0 1 ∣ } B = \left \{ \begin{vmatrix}{1} && {0} \\ {0} && {1}\end{vmatrix} , \begin{vmatrix}{0} && {1} \\ {0} && {1}\end{vmatrix} , \begin{vmatrix}{1} && {1} \\ {1} && {0}\end{vmatrix} , \begin{vmatrix}{0} && {0} \\ {0} && {1}\end{vmatrix} \right \} a basis for R 2 x 2 \mathbb R_{2 \: x \: 2} .

Let the matrix M = [ a i j ] 4 x 3 M = [a_{ij}]_{4 \: x \: 3} represent the linear transform above and S = ∑ i = 1 4 ∑ j = 1 3 a i j \displaystyle S = \sum_{i = 1}^{4} \sum_{j = 1}^{3} a_{i j} .

Let p = 3 + 4 x − x 2 \textbf{p} = 3 + 4 x - x^2 and [f(p)] B = ( γ 1 γ 2 γ 3 γ 4 ) \textbf{[f(p)]}_{\textbf{B}} = \left( \begin{array}{ccc} \gamma_1 \\ \gamma_2 \\ \gamma_3 \\ \gamma 4 \ \end{array} \right) and T = ∑ j = 1 4 γ j \displaystyle T = \sum_{j = 1}^{4} \gamma_j .

Find S + T S + T .

Refer to previous problem. . .


The answer is 6.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Dec 25, 2016

f ( 1 + x ) = ∣ 2 1 1 0 ∣ = α 1 ∣ 1 0 0 1 ∣ + α 2 ∣ 0 1 0 1 ∣ + α 3 ∣ 1 1 1 0 ∣ + α 4 ∣ 0 1 0 1 ∣ = ∣ α 1 + α 3 α 2 + α 3 α 3 α 1 + α 2 + α 4 ∣ f(1 + x) = \left| \begin{array}{ccc} 2 & 1 \\ 1 & 0 \\ \\ \end{array} \right| = \alpha_{1} \left| \begin{array}{ccc} 1 & 0 \\ 0 & 1 \\ \\ \end{array} \right| + \alpha_{2} \left| \begin{array}{ccc} 0 & 1 \\ 0 & 1 \\ \\ \end{array} \right| + \alpha_{3} \left| \begin{array}{ccc} 1 & 1 \\ 1 & 0 \\ \\ \end{array} \right| + \alpha_{4} \left| \begin{array}{ccc} 0 & 1 \\ 0 & 1 \\ \\ \end{array} \right| =\left| \begin{array}{ccc} \alpha_{1} + \alpha_3 & \alpha_{2} + \alpha_{3} \\ \alpha_3 & \alpha_1 + \alpha_2 + \alpha_4 \\ \\ \end{array} \right|

f ( x + x 2 ) = ∣ 1 2 1 1 ∣ = ∣ β 1 + β 3 β 2 + β 3 β 3 β 1 + β 2 + β 4 ∣ f(x + x^2) = \left| \begin{array}{ccc} 1 & 2 \\ 1 & 1 \\ \\ \end{array} \right| = \left| \begin{array}{ccc} \beta_{1} + \beta_3 & \beta_{2} + \beta_{3} \\ \beta_3 & \beta_1 + \beta_2 + \beta_4 \\ \\ \end{array} \right|

f ( x 2 ) = ∣ 0 1 1 1 ∣ = ∣ λ 1 + λ 3 λ 2 + λ 3 λ 3 λ 1 + λ 2 + λ 4 ∣ f(x^2) = \left| \begin{array}{ccc} 0 & 1 \\ 1 & 1 \\ \\ \end{array} \right| = \left| \begin{array}{ccc} \lambda_{1} + \lambda_3 & \lambda_{2} + \lambda_{3} \\ \lambda_3 & \lambda_1 + \lambda_2 + \lambda_4 \\ \\ \end{array} \right|

⟹ \implies

α 1 + α 3 = 2 \alpha_1 + \alpha_3 = 2

α 2 + α 3 = 1 \alpha_2 + \alpha_3 = 1

α 3 = 1 \alpha_3 = 1

α 1 + α 2 + α 4 = 0 \alpha_1 + \alpha_2 + \alpha_4 = 0

Solving the system we obtain:

( α 1 α 2 α 3 α 4 ) = ( 1 0 1 − 1 ) \left( \begin{array}{ccc} \alpha_1 \\ \alpha_2 \\ \alpha_3 \\ \alpha 4 \ \end{array} \right) = \left( \begin{array}{ccc} 1 \\ 0 \\ 1 \\ -1 \ \end{array} \right)

β 1 + β 3 = 1 \beta_1 + \beta_3 = 1

β 2 + β 3 = 2 \beta_2 + \beta_3 = 2

β 3 = 1 \beta_3 = 1

β 1 + β 2 + β 4 = 1 \beta_1 + \beta_2 + \beta_4 = 1

Solving the system we obtain:

( β 1 β 2 β 3 β 4 ) = ( 0 1 1 0 ) \left( \begin{array}{ccc} \beta_1 \\ \beta_2 \\ \beta_3 \\ \beta 4 \ \end{array} \right) = \left( \begin{array}{ccc} 0 \\ 1 \\ 1 \\ 0 \ \end{array} \right)

λ 1 + λ 3 = 0 \lambda_1 + \lambda_3 = 0

λ 2 + λ 3 = 1 \lambda_2 + \lambda_3 = 1

λ 3 = 1 \lambda_3 = 1

λ 1 + λ 2 + λ 4 = 1 \lambda_1 + \lambda_2 + \lambda_4 = 1

Solving the system we obtain:

( λ 1 λ 2 λ 3 λ 4 ) = ( − 1 0 1 2 ) \left( \begin{array}{ccc} \lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \lambda 4 \ \end{array} \right) = \left( \begin{array}{ccc} -1 \\ 0 \\ 1 \\ 2 \ \end{array} \right)

The matrix M = ∣ 1 0 − 1 0 1 0 1 1 1 − 1 0 2 ∣ M = \left| \begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \\ -1 & 0 & 2\ \end{array} \right|

p = 3 + 4 x − x 2 = a 1 + ( a 1 + a 2 ) x + ( a 2 + a 3 ) x 2 ⟹ p = 3 + 4 x - x^2 = a_1 + (a_1 + a_2) x + (a_2 + a_3) x^2 \implies

[p] A = ( 3 1 − 2 ) ⟹ \textbf{[p]}_\textbf{A} = \left( \begin{array}{ccc} 3 \\ 1 \\ -2 \\ \ \end{array} \right) \implies

[f(p)] B = M [p] A = ( 5 1 2 − 7 ) = ( γ 1 γ 2 γ 3 γ 4 ) \textbf{[f(p)]}_{\textbf{B}} = M \textbf{[p]}_\textbf{A} = \left( \begin{array}{ccc} 5 \\ 1 \\ 2 \\ -7 \\ \ \end{array} \right) = \left( \begin{array}{ccc} \gamma_1 \\ \gamma_2 \\ \gamma_3 \\ \gamma 4 \ \end{array} \right)

⟹ S = ∑ i = 1 4 ∑ j = 1 3 a i j = 5 \implies S = \sum_{i = 1}^{4} \sum_{j = 1}^{3} a_{i j} = 5 and T = ∑ j = 1 4 γ j = 1 ⟹ S + T = 6. T = \sum_{j = 1}^{4} \gamma_j = 1 \implies S + T = 6.

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