An algebra problem by Rocco Dalto

Algebra Level 3

Let f : R 2 → R 3 f: \mathbb R^2 \rightarrow \mathbb R^3 be linear transform defined by:

f ( x 1 x 2 ) = ( 3 x 1 − x 2 x 2 x 1 + x 2 ) f \left( \begin{array}{ccc} x_{1} \\ x_{2} \\ \end{array} \right) = \left( \begin{array}{ccc} 3 x_{1} - x_{2}\\ x_{2} \\ x_{1} + x_{2} \end{array} \right)

and A = { ( − 1 1 ) , ( 2 1 ) } A = \{ \left( \begin{array}{ccc} -1 \\ 1 \\ \end{array} \right), \left( \begin{array}{ccc} 2 \\ 1 \\ \end{array} \right) \} be basis for R 2 \mathbb R^2 and B = { ( 1 − 1 1 ) , ( − 1 1 1 ) , ( 1 0 1 ) } B = \left \{ \left( \begin{array}{ccc} 1 \\ -1 \\ 1 \\ \end{array} \right), \left( \begin{array}{ccc} -1 \\ 1 \\ 1 \ \end{array} \right), \left( \begin{array}{ccc} 1 \\ 0 \\ 1 \ \end{array} \right) \right \} be a basis for R 3 . \mathbb R^3. .

Find a matrix representation for the linear transform above.

If X = ( 1 2 ) ∈ R 2 \: \textbf{X} = \left( \begin{array}{ccc} 1 \\ 2 \\ \end{array} \right) \in \mathbb R^2 and [f(X)] B = ( λ 1 λ 2 λ 3 ) \textbf{[f(X)]}_{\textbf{B}} = \left( \begin{array}{ccc} \lambda_1 \\ \lambda_2 \\ \lambda_3 \ \end{array} \right) ,

Find: λ 1 + λ 2 + λ 3 \lambda_1 + \lambda_2 + \lambda_3 .

Note: Of course you can find [f(X)] B = ( λ 1 λ 2 λ 3 ) \textbf{[f(X)]}_{\textbf{B}} = \left( \begin{array}{ccc} \lambda_1 \\ \lambda_2 \\ \lambda_3 \ \end{array} \right) without finding the matrix, but my intention was to find the matrix.


The answer is 3.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Dec 23, 2016

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

f ( ( 2 1 ) ) = ( 5 1 3 ) f( \left( \begin{array}{ccc} 2 \\ 1 \\ \end{array} \right) ) = \left( \begin{array}{ccc} 5 \\ 1 \\ 3\\ \end{array} \right) = β 1 ( 1 − 1 1 ) + β 2 ( − 1 1 1 ) + β 3 ( 1 0 1 ) \beta_1 \left( \begin{array}{ccc} 1 \\ -1 \\ 1 \\ \end{array} \right) + \beta_2 \left( \begin{array}{ccc} -1 \\ 1 \\ 1 \\ \end{array} \right) + \beta_3 \left( \begin{array}{ccc} 1 \\ 0 \\ 1 \\ \end{array} \right)

⟹ \implies

α 1 − α 2 + α 3 = − 4 \alpha_1 - \alpha_2 + \alpha_3 = -4 − α 1 + α 2 = 1 -\alpha_1 + \alpha_2 = 1 α 1 + α 2 + α 3 = 0 \alpha_1+ \alpha_2 + \alpha_3 = 0

Solving the system we obtain:

( α ! α 2 α 3 ) = ( 1 2 − 3 ) \left( \begin{array}{ccc} \alpha_! \\ \alpha_2 \\ \alpha_3 \\ \end{array} \right) = \left( \begin{array}{ccc} 1 \\ 2 \\ -3 \\ \end{array} \right)

β 1 − β 2 + β 3 = 5 \beta_1 - \beta_2 + \beta_3 = 5 − β 1 + β 2 = 1 -\beta_1 + \beta_2 = 1 β 1 + β 2 + β 3 = 3 \beta_1+ \beta_2 + \beta_3 = 3

Solving the system we obtain:

( β ! β 2 β 3 ) = ( − 2 − 1 6 ) \left( \begin{array}{ccc} \beta_! \\ \beta_2 \\ \beta_3 \\ \end{array} \right) = \left( \begin{array}{ccc} -2 \\ -1 \\ 6 \\ \end{array} \right)

The Matrix M M representation of the linear transform is:

M = ∣ 1 − 2 2 − 1 − 3 6 ∣ M = \left| \begin{array}{ccc} 1 & -2 \\ 2 & -1 \\ -3 & 6 \end{array} \right|

X = ( 1 2 ) = a 1 ( − 1 1 ) + a 2 ( 2 1 ) ⟹ \textbf{X} = \left( \begin{array}{ccc} 1 \\ 2 \\ \end{array} \right) = a_1 \left( \begin{array}{ccc} -1 \\ 1 \\ \end{array} \right) + a_2 \left( \begin{array}{ccc} 2 \\ 1 \\ \end{array} \right) \implies

− a 1 + 2 a 2 = 1 -a_{1} + 2 a_{2} = 1 a 1 + a 2 = 2 a_{1} + a_{2} = 2

⟹ a 1 = a 2 = 1 ⟹ \implies a_1 = a_2 = 1 \implies

[X] A = ( 1 1 ) \textbf{[X]}_{\textbf{A}} = \left( \begin{array}{ccc} 1 \\ 1 \\ \end{array} \right)

⟹ \implies

[f(X)] B = ∣ 1 − 2 2 − 1 − 3 6 ∣ ( 1 1 ) = ( − 1 1 3 ) = ( λ 1 λ 2 λ 3 ) \textbf{[f(X)]}_{\textbf{B}} = \left| \begin{array}{ccc} 1 & -2 \\ 2 & -1 \\ -3 & 6 \end{array} \right| \left( \begin{array}{ccc} 1 \\ 1 \\ \end{array} \right) = \left( \begin{array}{ccc} -1 \\ 1 \\ 3\\ \end{array} \right) = \left( \begin{array}{ccc} \lambda_1 \\ \lambda_2 \\ \lambda_3 \ \end{array} \right)

⟹ λ 1 + λ 2 + λ 3 = 3 \implies \lambda_1 + \lambda_2+ \lambda_3 = \boxed{3}

Without finding the matrix. \textbf{Without finding the matrix.}

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

⟹ \implies

λ 1 − λ 2 + λ 3 = − 1 \lambda_1 - \lambda_2 + \lambda_3 = -1 − λ 1 + λ 2 = 2 -\lambda_1 + \lambda_2 = 2 λ 1 + λ 2 + λ 3 = 3 \lambda_1+ \lambda_2 + \lambda_3 = 3

Solving the system we obtain:

( λ 1 λ 2 λ 3 ) = ( − 1 1 3 ) = [f(X)] B \left( \begin{array}{ccc} \lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \end{array} \right) = \left( \begin{array}{ccc} -1 \\ 1 \\ 3 \\ \end{array} \right) = \textbf{[f(X)]}_{\textbf{B}}

⟹ λ 1 + λ 2 + λ 3 = 3 \implies \lambda_1 + \lambda_2+ \lambda_3 = \boxed{3}

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