Question

What eigen value for this matix (1 -2) (-2 0)

Answer 1

I think the answer is A!

Answer 2

You find the eigenvalues of a matrix A by following these steps:

1. Compute the matrix $A' = A-\lambda I$, where I is the identity matrix (1s on the diagonal, 0s elsewhere)
2. Compute the determinant of A'
3. Set the determinant of A' equal to zero and solve for lambda.

So, in this case, we have

$A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]$

The determinant of this matrix is

$\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4$

Finally, we have

$\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}$

So, the two eigenvalues are

$\lambda_1 = \dfrac{1+\sqrt{17}}{2},\quad \lambda_2 = \dfrac{1-\sqrt{17}}{2}$