a) Finde die Eigenwerte für \(A=\left(\begin{array}{ll}4 & 2 \\ 1 & 3\end{array}\right)\) (b) Zeichne die zwei entsprechenden Eigenvektoren.

The equation to find the eigenvalues of the matrix is given by the determinant of (A - λI) = 0 where A is the given matrix, λ represents the eigenvalues and I is the identity matrix. Therefore, we calculate the determinant of \[ \left(\begin{array}{ll}4-λ & 2 \ 1 & 3-λ\end{array}\right) \]. Set it equals to zero, which results the characteristic equation: (4 - λ)(3 - λ) - (2)(1) = 0. Solve this equation to find the eigenvalues λ.

Rearrange and simplify the characteristic equation to find the eigenvalues. The solved equation is λ^2 - 7λ + 10 = 0. The solutions of this quadratic equation through factorization gives us the eigenvalues λ1=2 and λ2=5.

To find the eigenvectors corresponding to each eigenvalue, we setup and solve the following system of equations for each eigenvalue: (A - λ1I)v1 = 0 and (A - λ2I)v2 = 0, where λ1 and λ2 are the eigenvalues, and v1 and v2 are the corresponding eigenvectors. For λ1=2, this results in the following system of equations: \[ \left(\begin{array}{cc} 2 & 2 \ 1 & 1 \end{array}\right) \left(\begin{array}{c} x1 \ x2 \end{array}\right) = \left(\begin{array}{c} 0 \ 0 \end{array}\right) \] For λ2=5, we get another system of equations: \[ \left(\begin{array}{cc} -1 & 2 \ 1 & -2 \end{array}\right) \left(\begin{array}{c} y1 \ y2 \end{array}\right) = \left(\begin{array}{c} 0 \ 0 \end{array}\right) \] Solve both systems to find the eigenvectors.

After solving the systems of equations, the eigenvector v1 corresponding to λ1=2 is (1, -1), and the eigenvector v2 corresponding to λ2=5 is (2, 1).

Draw both eigenvectors starting from the origin on a 2-D plane. This graphical representation will show the directions along which the transformation A only scales and does not re-direct.