An algebra problem by Rocco Dalto

Algebra Level 3

Let P 2 \mathbb P_{2} denote the set of all polynomials of degree two and R 3 \mathbb R^3 denote the set of all vectors in three space.

Let f : P 2 → R 3 f: \mathbb P_{2} \rightarrow \mathbb R^3 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 ) f(p_{0} + p_{1} x + p_{2} x^2) = \left( \begin{array}{ccc} p_{0} + p_{1} \\ p_{1} + p_{2} \\ p_{0} + p_{2} \\ \end{array} \right)

Let
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 − 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. .

Let the matrix M = [ a i j ] 3 x 3 M = [a_{ij}]_{3 \: x \: 3} represent the linear transform above.

If λ 1 , λ 2 , λ 3 \lambda_{1}, \lambda_{2}, \lambda_{3} are the eigenvalues of matrix M M ,

Find: ∑ j = 1 3 λ j \sum_{j = 1}^{3} \lambda_j .

Refer to previous problem. . .


The answer is -0.5.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Dec 27, 2016

f ( 1 + x ) = ( 2 1 1 ) = α 1 ( 1 − 1 1 ) + α 2 ( − 1 1 1 ) + α 3 ( 1 0 1 ) = ( α 1 − α 2 + α 3 − α 1 + α 2 α 1 + α 2 + α 3 ) f(1 + x) = \left( \begin{array}{ccc} 2 \\ 1 \\ 1 \\ \end{array} \right) = \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) = \left( \begin{array}{ccc} \alpha_{1} - \alpha_{2} + \alpha_{3} \\ -\alpha_{1} + \alpha_{2} \\ \alpha_{1} + \alpha_{2} + \alpha_{3} \\ \end{array} \right)

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

f ( x 2 ) = ( 0 1 1 ) = ( γ 1 − γ 2 + γ 3 − γ 1 + γ 2 γ 1 + γ 2 + γ 3 ) f(x^2) = \left( \begin{array}{ccc} 0 \\ 1 \\ 1 \\ \end{array} \right) = \left( \begin{array}{ccc} \gamma_{1} - \gamma_{2} + \gamma_{3} \\ -\gamma_{1} + \gamma_{2} \\ \gamma_{1} + \gamma_{2} + \gamma_{3} \\ \end{array} \right)

Solving the system:

α 1 ∗ − α 2 ∗ + α 3 ∗ = a \alpha_{1}^{*} - \alpha_{2}^{*} + \alpha_{3}^{*} = a − α 1 ∗ + α 2 ∗ = b -\alpha_{1}^{*} + \alpha_{2}^{*} = b α 1 ∗ + α 2 ∗ + α 3 ∗ = c \alpha_{1}^{*} + \alpha_{2}^{*} + \alpha_{3}^{*} = c

⟹ ( α 1 ∗ α 2 ∗ α 3 ∗ ) = ( c − a − 2 b 2 c − a 2 a + b ) ⟹ \implies \left( \begin{array}{ccc} \alpha_{1}^{*} \\ \alpha_{2}^{*} \\ \alpha_{3}^{*} \\ \end{array} \right) = \left( \begin{array}{ccc} \dfrac{c - a - 2b}{2} \\ \dfrac{c - a}{2} \\ a + b \\ \end{array} \right) \implies

( α 1 α 2 α 3 ) = ( − 3 2 − 1 2 3 ) , ( β 1 β 2 β 3 ) = ( − 2 0 3 ) , ( γ 1 γ 2 γ 3 ) = ( − 1 2 1 2 1 ) ⟹ \left( \begin{array}{ccc} \alpha_{1} \\ \alpha_{2} \\ \alpha_{3} \\ \end{array} \right) = \left( \begin{array}{ccc} -\dfrac{3}{2} \\ -\dfrac{1}{2} \\ 3 \\ \end{array} \right), \: \left( \begin{array}{ccc} \beta_{1} \\ \beta_{2} \\ \beta_{3} \\ \end{array} \right) = \left( \begin{array}{ccc} -2 \\ 0 \\ 3 \\ \end{array} \right), \: \left( \begin{array}{ccc} \gamma_{1} \\ \gamma_{2} \\ \gamma_{3} \\ \end{array} \right) = \left( \begin{array}{ccc} -\dfrac{1}{2} \\ \dfrac{1}{2} \\ 1 \\ \end{array} \right) \implies

Matrix M = ∣ − 3 2 − 2 − 1 2 − 1 2 0 1 2 3 3 1 ∣ ⟹ M = \left| \begin{array}{ccc} -\dfrac{3}{2} & -2 & -\dfrac{1}{2} \\ -\dfrac{1}{2} & 0 & \dfrac{1}{2} \\ 3 & 3 & 1 \\ \ \end{array} \right| \implies

M − λ I = ∣ − 3 2 − λ − 2 − 1 2 − 1 2 − λ 1 2 3 3 1 − λ ∣ M - \lambda I = \left| \begin{array}{ccc} -\dfrac{3}{2} - \lambda & -2 & -\dfrac{1}{2} \\ -\dfrac{1}{2} & -\lambda & \dfrac{1}{2} \\ 3 & 3 & 1 - \lambda\\ \ \end{array} \right|

d e t ( M − λ I ) = 0 ⟹ det(M - \lambda I) = 0 \implies

− 4 λ 3 − 2 λ 2 + 10 λ − 4 = 0. -4 \lambda^3 - 2 \lambda^2 + 10 \lambda - 4 = 0.

By inspection λ = 1 \lambda = 1 is a root of − 4 λ 3 − 2 λ 2 + 10 λ − 4 = 0 ⟹ ( λ − 1 ) P ( λ ) = − 4 λ 3 − 2 λ 2 + 10 λ − 4 -4 \lambda^3 - 2 \lambda^2 + 10 \lambda - 4 = 0 \implies (\lambda - 1) P(\lambda) = -4 \lambda^3 - 2 \lambda^2 + 10 \lambda - 4

Dividing we obtain: P ( λ ) = − 4 λ 3 − 2 λ 2 + 10 λ − 4 λ − 1 = − 4 λ 2 − 6 λ + 4 ⟹ 2 λ 2 + 3 λ − 2 = 0 P(\lambda) = \dfrac{-4 \lambda^3 - 2 \lambda^2 + 10 \lambda - 4}{\lambda - 1} = -4 \lambda^2 - 6 \lambda + 4 \implies 2 \lambda^2 + 3 \lambda - 2 = 0 ⟹ ( 2 λ − 1 ) ( λ + 2 ) = 0 ⟹ λ = − 2 , 1 2 \implies (2 \lambda - 1)(\lambda + 2) = 0 \implies \lambda = -2, \dfrac{1}{2}

∴ λ 1 = 1 , λ 2 = − 2 , λ 3 = 1 2 ⟹ ∑ j = 1 3 λ j = − 1 2 \therefore \lambda_{1} = 1, \lambda_{2} = -2, \lambda_{3} = \dfrac{1}{2} \implies \sum_{j = 1}^{3} \lambda_j = \boxed{-\dfrac{1}{2}} .

1 Helpful 0 Interesting 0 Brilliant 1 Confused

The sum of the eigenvalues is the trace of the matrix, which is just − 3 2 + 0 + 1 = − 1 2 -\frac{3}{2} + 0 + 1 = -\frac{1}{2} .

Jon Haussmann - 11 months, 1 week ago

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