Question:

Step 1: Define the given vectors and relationships. Let $v_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$. We are given the following matrix-vector products: $$Pv_1 = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$$ $$P^2v_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$ Let $v_2 = Pv_1$. Then $v_2 = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$. The second given condition can be rewritten as $P(Pv_1) = Pv_2 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$. Step 2: Construct the matrix $P$. We can express the action of $P$ on the basis vectors $v_1$ and $v_2$ in matrix form: $$P \begin{bmatrix} v_1 & v_2 \end{bmatrix} = \begin{bmatrix} Pv_1 & Pv_2 \end{bmatrix}$$ Let $V = \begin{bmatrix} v_1 & v_2 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$. Let $W = \begin{bmatrix} Pv_1 & Pv_2 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 2 & 0 \end{bmatrix}$. Thus, $PV = W$. To find $P$, we compute $P = WV^{-1}$. First, calculate the inverse of $V$: The determinant of $V$ is $\det(V) = (1)(2) - (-1)(-1) = 2 - 1 = 1$. The inverse of $V$ is: $$V^{-1} = \frac{1}{\det(V)} \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$$ Now, calculate $P$: $$P = \begin{bmatrix} -1 & 1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} (-1)(2) + (1)(1) & (-1)(1) + (1)(1) \\ (2)(2) + (0)(1) & (2)(1) + (0)(1) \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 4 & 2 \end{bmatrix}$$ Step 3: Determine the eigenvalues of $P$. The eigenvalues $p$ are the roots of the characteristic equation $\det(P - pI) = 0$. $$P - pI = \begin{bmatrix} -1-p & 0 \\ 4 & 2-p \end{bmatrix}$$ The determinant is: $$\det(P - pI) = (-1-p)(2-p) - (0)(4) = (p+1)(p-2)$$ Setting the determinant to zero to find the eigenvalues: $$(p+1)(p-2) = 0$$ The eigenvalues are $p = -1$ and $p = 2$. Step 4: Identify $p_1$ and $p_2$ and calculate the final expression. Given that $p_1 > p_2$, we assign the values: $$p_1 = 2$$ $$p_2 = -1$$ The required value is $5p_1 + 2p_2$: $$5p_1 + 2p_2 = 5(2) + 2(-1) = 10 - 2 = 8$$