LA1

Find the eigenvalues of the matrix $\begin{bmatrix} 5 & 2 \\ 3 & 6 \\ \end{bmatrix}$ .

First, make sure that the determinant is not zero.

$5\cdot6-2\cdot3=24\,$

Now subtract $\lambda\,$ from each entry of the main diagonal.

$\begin{bmatrix} 5-\lambda & 2 \\ 3 & 6-\lambda \\ \end{bmatrix}$

Build and solve the characteristic equation the same way as for the determinant. Set it equal to zero.

$(5-\lambda)\cdot(6-\lambda)-6=0\,$

$(30-11\lambda+\lambda^2)-6=0\,$

$\lambda^2-11\lambda+24=0\,$

The quadratic formula (or factoring) tells the answer:

$\lambda=\frac{11\pm\sqrt{121-4\cdot1\cdot(24)}}{2}\,$

$\lambda = 3, 8\,$