The matrix $A = \begin{bmatrix}3&k\\8&8\end{bmatrix}$ has two distinct real eigenvalues iff $k > ?$
So I found the determinant by doing:
$(3 – \lambda)(8 – \lambda) – 8k = \lambda^2 – 11\lambda + 24 – 8k \implies \lambda = 8, \lambda = 3$ The thing is, I'm not really sure what they are asking me because I have found what the eigenvalues are: $\lambda_1 = 8, \lambda_2 = 3$.
I'm assuming I need to solve for $k$ somehow but it doesn't seem very straightforward to me, what am I missing here?
Best Answer
Note that the characteristic equation of this matrix is given by $$ \lambda^2 - 11\lambda + (24 - 8k) = 0 $$ The question is to find which values of $k$ produce two distinct real roots $\lambda$ to this equation. It is helpful to consider the discriminant of this quadratic equation, namely $$ b^2 - 4ac = 121 - 4(24 - 8k) $$ recall that a quadratic equation will have distinct real roots iff its discriminant is positive.