Question

Consider the initial value problem for the vector-valued function x, x′=Ax,A=[1−225],x(0)=[1−1] Find the eigenvalues λ1,λ2 and their corresponding eigenvectors v1,v2 of the coefficient matrix A. (a) Eigenvalues: (if repeated, enter it twice separated by commas)

Answer 1

Answer:

Step-by-step explanation:

Consider the matrix $\begin{matrix} 1 & -2 \\ 2 & 5 \end{matrix}$. We will calculate the correspondent eigenvalues and eigen vector of the matrix. REcall that, to calculate the eigenvalues of a square matrix A, we must solve the following equation $\text{det}(A-\lambda I ) =0$ where I is the identity matrix. In this case we have the following

$\text{det}\left(\begin{matrix} 1-\lambda & -2 \\ 2 & 5-\lambda \end{matrix}\right) =$

which gives us the following polynomial (known as the characteristic polynomial of the matrix A).

$(1-\lambda)(5-\lambda)+4 =0 = \lambda^2-6\lambda + 9 = (\lambda -3)^2$.

Hence, the only eigenvalue of the given matrix is $\lambda = 3$.

For us to calculate the eigenvalue, we want to find a base for the Kernel of matrix $A-\lambda I$ replacing the value of lambda with the value of the eigen value. REcall that finding the base for the Kernel is solving the associated homogeneus system of the matrix

$\left[\begin{matrix} -2 & -2 \\ 2 & 2 \end{matrix}\right]= \left[\begin{matrix} 0 \\ 0 \end{matrix}\right]$

which lead us to the equation $-2x-2y=0$ or equivalently, $y=-x$. Hence, the solution of the system is of the form (x,y) = (x,-x) = x(1,-1). So  the correspondent eigenvector is the vector (1,-1).